JEE Basic Concepts in Chemistry Questions

Mass of compound = 0.25 g, increase in CaCl$$_2$$ tube (water) = 0.15 g, increase in potash tube (CO$$_2$$) = 0.1837 g.

Mass of carbon.

$$\text{Mass of C} = \frac{12}{44} \times 0.1837 = 0.05010$$ g

Mass of hydrogen.

$$\text{Mass of H} = \frac{2}{18} \times 0.15 = 0.01667$$ g

Mass and percentage of oxygen.

Mass of O = $$0.25 - 0.05010 - 0.01667 = 0.18323$$ g

Percentage of oxygen = $$\frac{0.18323}{0.25} \times 100 = 73.3\% \approx 73\%$$

The answer is 73.

The molecular formula of the given antiviral compound is obtained by counting the atoms present in the heterocyclic base and the substituted sugar moiety.

For the nitrogenous base:

• Carbon atoms = 4
• Hydrogen atoms = 2
• Nitrogen atoms = 2
• Oxygen atoms = 2
• Iodine atoms = 1

Hence, the contribution is

$$\mathrm{C_4H_2N_2O_2I}$$

For the sugar moiety:

• Carbon atoms = 5
• Hydrogen atoms = 8
• Oxygen atoms = 3
• Fluorine atoms = 1

Hence, the contribution is

$$\mathrm{C_5H_8O_3F}$$

Therefore, the molecular formula of the compound is

$$\mathrm{C_9H_{10}N_2O_5FI}$$

Using the given atomic masses,

• Carbon: $$9 \times 12 = 108$$
• Hydrogen: $$10 \times 1 = 10$$
• Nitrogen: $$2 \times 14 = 28$$
• Oxygen: $$5 \times 16 = 80$$
• Fluorine: $$1 \times 19 = 19$$
• Iodine: $$1 \times 127 = 127$$

Hence, the molar mass is

$$108 + 10 + 28 + 80 + 19 + 127 = 372\ \mathrm{g\ mol^{-1}}$$

The mass corresponding to $$0.1$$ mole is

$$\text{Mass} = 0.1 \times 372 = 37.2\ \mathrm{g}$$

Expressing this in the required form,

$$37.2\ \mathrm{g} = 372 \times 10^{-1}\ \mathrm{g}$$

Therefore, the required integer value is

$$\boxed{372}$$

We need to find the volume of the sealed container of $$CO_2$$.

We start by writing the chemical reactions involved.

$$CO_2 + Ca(OH)_2 \rightarrow CaCO_3 + H_2O$$

Next, the excess $$Ca(OH)_2$$ is neutralized by hydrochloric acid:

$$Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$$

Then we calculate the moles of excess $$Ca(OH)_2$$ from the titration data.

Moles of HCl = 0.1 M × 0.040 L = 0.004 mol

Moles of excess $$Ca(OH)_2$$ = $$\frac{0.004}{2} = 0.002$$ mol

The problem states that all of the $$CO_2$$ reacted with exactly half of the initial amount of $$Ca(OH)_2$$.

If the initial $$Ca(OH)_2$$ is N mol, then the reacted portion is N/2 mol and the excess is also N/2 mol.

Therefore, $$N/2 = 0.002$$ mol, so $$N = 0.004$$ mol.

It follows that the moles of $$CO_2$$ are equal to the moles of $$Ca(OH)_2$$ that reacted, which is N/2 = 0.002 mol.

Finally, applying the ideal gas law at STP (273 K, 1 atm):

$$V = n \times 22400 \text{ cm}^3/\text{mol} = 0.002 \times 22400 = 44.8 \text{ cm}^3$$

Rounding to the nearest integer gives $$x \approx 45$$ cm³.

Therefore, the volume of the sealed container of $$CO_2$$ is 45 cm³.

Let the hydrocarbon be $$C_xH_y$$. The balanced combustion reaction is:

$$ C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O $$

Given: 11 equivalents of oxygen and 4 equivalents of water.

From water: $$\frac{y}{2} = 4 \implies y = 8$$

From oxygen: $$x + \frac{y}{4} = x + \frac{8}{4} = x + 2 = 11 \implies x = 9$$

The molecular formula is $$C_9H_8$$.

Match List-I with List-II for mole concept calculations.

A. 16 g of CH$$_4$$(g):

Molar mass of CH$$_4$$ = 16 g/mol, so 16 g = 1 mole.

Electrons in CH$$_4$$: C has 6e, 4H have 4e = 10 electrons per molecule.

Total electrons = $$1 \times 6.02 \times 10^{23} \times 10 = 60.2 \times 10^{23}$$ electrons = II

B. 1 g of H$$_2$$(g):

Molar mass of H$$_2$$ = 2 g/mol, so 1 g = 0.5 mol.

Volume at STP = $$0.5 \times 22.4 = 11.2 \approx 11.4$$ L (using 22.7 L/mol for updated STP) = IV

C. 1 mole of N$$_2$$(g):

Mass = $$1 \times 28 = 28$$ g. Weighs 28 g = I

D. 0.5 mol of SO$$_2$$(g):

Mass = $$0.5 \times 64 = 32$$ g. Weighs 32 g = III

Matching: A-II, B-IV, C-I, D-III

The correct answer is Option B: A-II, B-IV, C-I, D-III.

Given: 10.0 mL of 0.01 M $$Ba(OH)_2$$ is to be neutralized by 0.02 M aqueous HBr.

The balanced neutralization reaction is:

$$Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O$$

Moles of $$Ba(OH)_2$$:

$$n_{Ba(OH)_2} = 0.01 \times 10 = 0.1 \text{ mmol}$$

Since $$Ba(OH)_2$$ is dibasic, the moles of $$OH^-$$ ions:

$$n_{OH^-} = 2 \times 0.1 = 0.2 \text{ mmol}$$

For complete neutralization, moles of HBr required = moles of $$OH^-$$ = 0.2 mmol.

Volume of HBr solution needed:

$$V = \frac{n_{HBr}}{M_{HBr}} = \frac{0.2}{0.02} = 10.0 \text{ mL}$$

Therefore, the correct answer is Option C: $$\mathbf{10.0 \text{ mL}}$$.

We need to identify the correct ratio of silica to alumina in good quality cement.

Portland cement is manufactured by heating limestone ($$CaCO_3$$) with clay containing $$SiO_2$$ and $$Al_2O_3$$ in a kiln, and its quality depends on the ratios of these oxides.

For good quality Portland cement, the ratio of silica ($$SiO_2$$) to alumina ($$Al_2O_3$$) should be approximately 2.5 to 4, with an ideal value of about $$3$$, while the ratio of lime ($$CaO$$) to the total acidic oxides ($$SiO_2 + Al_2O_3 + Fe_2O_3$$) is about $$2$$.

When the silica to alumina ratio is too low, the cement sets quickly but lacks strength; conversely, a ratio that is too high causes slow setting.

Hence, the correct answer is Option 4: 3.

Let us analyze both statements:

Statement I: "Permutit process is more efficient compared to the synthetic resin method for the softening of water."

This is incorrect. The synthetic resin (ion exchange resin) method is more efficient than the permutit process. Synthetic resins can remove both cations and anions, while permutit only exchanges cations.

Statement II: "Synthetic resin method results in the formation of soluble sodium salts."

This is incorrect. The synthetic resin method uses cation exchange resins (H⁺ form) and anion exchange resins (OH⁻ form). The cation exchange resin replaces Ca²⁺, Mg²⁺ etc. with H⁺, and the anion exchange resin replaces anions with OH⁻. The final product is deionized water, not sodium salts. (The permutit process forms sodium salts, not the synthetic resin method.)

Since both statements are incorrect, the answer is Option 1: Both the statements I and II are incorrect.

We need to find the molecular formula of hydrocarbon A that requires 9.5 equivalents of O$$_2$$ for combustion and produces 3 equivalents of water.

Write the general combustion equation.

For a hydrocarbon C$$_x$$H$$_y$$:

$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \to x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$$

Set up equations from the given data.

Moles of O$$_2$$ required: $$x + \frac{y}{4} = 9.5$$

Moles of H$$_2$$O produced: $$\frac{y}{2} = 3 \implies y = 6$$

Solve for $$x$$.

$$x + \frac{6}{4} = 9.5$$

$$x + 1.5 = 9.5$$

$$x = 8$$

Identify the molecular formula.

The molecular formula is C$$_8$$H$$_6$$.

The correct answer is Option A: C$$_8$$H$$_6$$.

We need to find the number of water molecules in washing soda and soda ash respectively.

Washing soda: Washing soda is the common name for hydrated sodium carbonate. Its chemical formula is $$Na_2CO_3 \cdot 10H_2O$$ (sodium carbonate decahydrate). The "deca" prefix indicates 10 water molecules. Washing soda is obtained by recrystallizing soda ash from aqueous solution, and the crystals that form contain 10 molecules of water of crystallization per formula unit. Therefore, washing soda has 10 water molecules.

Soda ash: Soda ash is the common name for anhydrous sodium carbonate, $$Na_2CO_3$$. The term "ash" comes from its historical production by burning plant material and extracting the residue. Being anhydrous, soda ash contains 0 (zero) water molecules. It is essentially the same compound as washing soda but without the water of crystallization.

Key distinction: Both washing soda and soda ash are forms of sodium carbonate ($$Na_2CO_3$$). The difference is that washing soda is the hydrated form ($$Na_2CO_3 \cdot 10H_2O$$) while soda ash is the anhydrous form ($$Na_2CO_3$$). When washing soda is heated, it loses its water of crystallization and becomes soda ash.

The correct answer is Option 3: 10 and 0.

We need to identify the minimum dissolved oxygen (DO) for fish growth and the BOD limit for clean water.

For the healthy growth of fish, the concentration of dissolved oxygen in water should be more than 6 ppm. Below this level, fish survival becomes difficult.

Clean water has a Biochemical Oxygen Demand (BOD) of less than 5 ppm. Water with BOD above 5 ppm is considered polluted, and highly polluted water has BOD > 17 ppm.

$$X = 6$$ ppm (dissolved oxygen for fish) and $$Y = 5$$ ppm (BOD for clean water).

The correct answer is Option A: X = 6, Y = 5.

Activated charcoal adsorbs a gas more strongly when the attractive van der Waals forces between its molecules are larger. For the noble gases those forces grow with increasing ease of liquefaction, or equivalently with higher critical temperature $$T_c$$. The critical temperatures are:

He : 5.2 K < Ne : 44.4 K < Ar : 150.9 K < Kr : 209.4 K.

Since krypton has the highest $$T_c$$, 1 g of activated charcoal will adsorb the greatest quantity of Kr. Hence Assertion A is true.

The Reason states that krypton has the highest critical volume $$V_c$$ and critical pressure $$P_c$$, but the lowest compressibility factor at the critical point $$Z_c$$. Experimental critical constants are:

He : $$P_c=2.3\ \text{atm},\;V_c=57\ \text{cm}^3\ \text{mol}^{-1},\;T_c=5.2\ \text{K}$$
Ne : $$P_c=26.3\ \text{atm},\;V_c=41\ \text{cm}^3\ \text{mol}^{-1},\;T_c=44.4\ \text{K}$$
Ar : $$P_c=48.3\ \text{atm},\;V_c=74.6\ \text{cm}^3\ \text{mol}^{-1},\;T_c=150.9\ \text{K}$$
Kr : $$P_c=55.0\ \text{atm},\;V_c=92.0\ \text{cm}^3\ \text{mol}^{-1},\;T_c=209.4\ \text{K}$$

Thus Kr indeed has the highest $$P_c$$ and $$V_c$$ among the four gases. However, the compressibility factor at the critical point is

$$Z_c=\frac{P_cV_c}{RT_c}$$

Using the data above gives $$Z_c\,$$: He 0.302, Ne 0.296, Ar 0.291, Kr 0.294. Krypton does not possess the lowest $$Z_c$$; argon’s value is slightly lower. Therefore the statement “$$Z_c$$ is lowest for krypton” is false, rendering Reason R false.

A true Assertion with a false Reason matches Option A.

Final answer : Option A.

For good quality cement, the ratio:

$$\frac{\text{Lime}}{\text{Silica} + \text{Alumina} + \text{Ferric oxide}}$$

should be close to: $${2}$$

This ratio ensures proper strength and setting properties of cement.
Hence, Option C is correct

We need to evaluate two statements about water quality parameters.

Evaluating Statement I: "If BOD is 4 ppm and dissolved oxygen is 8 ppm, then it is a good quality water."

BOD (Biochemical Oxygen Demand) measures the amount of dissolved oxygen consumed by microorganisms to decompose the organic matter present in water. A lower BOD indicates less organic pollution.

- Clean water typically has a BOD of less than 5 ppm.

- Water is considered polluted when BOD exceeds 17 ppm.

Dissolved oxygen (DO) indicates the amount of oxygen available in water for aquatic organisms. Higher DO values indicate better water quality.

- Good quality water has DO around 6-8 ppm or higher.

Given BOD = 4 ppm (which is less than 5 ppm, so acceptable) and DO = 8 ppm (which is in the good range), this water qualifies as good quality water. Statement I is correct.

Evaluating Statement II: "If the concentration of zinc and nitrate salts are 5 ppm each, then it can be good quality water."

The permissible limits for various ions in drinking water as per standard guidelines are:

- Zinc: The permissible limit is approximately 5 ppm. A concentration of 5 ppm of zinc is within the acceptable range.

- Nitrate (NO$$_3^-$$): The permissible limit for nitrate in drinking water is approximately 45-50 ppm. A concentration of 5 ppm of nitrate is well below this limit.

Since both zinc (5 ppm) and nitrate (5 ppm) are within their respective permissible limits, the water can be considered good quality. Statement II is correct.

Both statements are correct. The correct answer is Option 4: Both the statements I and II are correct.

Matching industries with their waste:

(A) Steel plants → Slag (III): Steelmaking produces slag as a byproduct (mixture of metal oxides and silicon dioxide).

(B) Thermal power plants → Fly ash (II): Burning coal in thermal power plants produces fly ash.

(C) Fertilizer Industries → Gypsum (I): Phosphoric acid production in fertilizer industries produces phosphogypsum as waste.

(D) Paper mills → Bio-degradable wastes (IV): Paper mills produce biodegradable organic waste (lignin, cellulose residues).

Matching: A-III, B-II, C-I, D-IV

The correct answer is Option 1.

Photochemical smog formation requires:

• Sunlight (UV radiation)
• Warm temperature
• NOₓ and hydrocarbons from vehicular emissions
• Stagnant air conditions

Photochemical smog is minimum in cold conditions with less sunlight. Among the given options:

Srinagar, J&K in January: Very cold (below 0°C), minimal sunlight, low temperatures → least favorable for photochemical smog.

This matches option 1.

We need to identify which process does NOT disturb the delicate balance of $$CO_2$$ and $$O_2$$ in the atmosphere.

Understanding the $$CO_2$$-$$O_2$$ balance: In nature, there is a natural cycle that maintains the balance between $$CO_2$$ and $$O_2$$:

- Photosynthesis (by plants): $$6CO_2 + 6H_2O \xrightarrow{sunlight} C_6H_{12}O_6 + 6O_2$$ (consumes $$CO_2$$, produces $$O_2$$)

- Respiration (by all living organisms): $$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O$$ (consumes $$O_2$$, produces $$CO_2$$)

These two processes are complementary and have been occurring for millions of years, maintaining a natural equilibrium.

Analyzing each option:

Option 1: Respiration - Respiration is a natural biological process that is part of the natural carbon cycle. The $$CO_2$$ produced by respiration is consumed by photosynthesis, and the $$O_2$$ consumed by respiration is produced by photosynthesis. This natural cycle does NOT disturb the $$CO_2/O_2$$ balance.

Option 2: Burning of coal - Burning coal is combustion of fossil fuels, which releases large amounts of $$CO_2$$ that were stored underground for millions of years. This adds excess $$CO_2$$ to the atmosphere without a corresponding increase in photosynthesis, thereby disturbing the balance.

Option 3: Deforestation - Cutting down forests reduces the number of trees performing photosynthesis, which means less $$CO_2$$ is absorbed and less $$O_2$$ is produced. This disturbs the balance by shifting it toward higher $$CO_2$$ and lower $$O_2$$.

Option 4: Burning of petroleum - Like coal, burning petroleum releases stored carbon as $$CO_2$$, adding excess $$CO_2$$ to the atmosphere and disturbing the balance.

The correct answer is Option 1: Respiration does NOT disturb the $$CO_2/O_2$$ balance.

BOD measures the amount of dissolved oxygen required by aerobic microorganisms to break down the organic matter present in a given water sample. It serves as a direct indicator of organic water pollution:

• High BOD: Indicates a large amount of organic waste, which consumes significant dissolved oxygen, leaving less for aquatic life (high pollution).
• Low BOD: Indicates very little organic waste or debris, meaning the dissolved oxygen remains abundant and stable.

Standard Water Quality Classification (NCERT Guidelines):

According to environmental chemistry standards, water quality is classified based on its BOD value (measured in ppm or mg/L):

• $$\text{BOD} < 5\text{ ppm}$$: Classified as very clean water.
• $$\text{BOD} \ge 17\text{ ppm}$$: Classified as highly polluted water.

Since the analyzed pond water has a BOD value of 4 (which is less than 5), it falls squarely into the category of unpolluted, healthy water.

Note on the shared web page: The automated green checkmark on the reference page incorrectly labels it as "Highly polluted water." This is factually wrong; a low BOD value of 4 represents clean, unpolluted water.

Conclusion:

A BOD value of 4 indicates minimal organic waste, designating the pond as having clean water.

Answer: Option C — Very clean water

Among the given industrial activities, we need to identify the one least responsible for global warming.

Global warming is primarily caused by greenhouse gas emissions, especially CO₂.

Steel manufacturing: Produces large amounts of CO₂ from coke used in blast furnaces.

Electricity generation in thermal power plants: Burns fossil fuels (coal, gas), releasing enormous quantities of CO₂. This is one of the largest contributors to global warming.

Industrial production of urea: Uses CO₂ as a raw material ($$CO_2 + 2NH_3 \rightarrow (NH_2)_2CO + H_2O$$) but the process itself involves significant energy consumption and emissions.

Manufacturing of cement: While cement production does produce CO₂ (from limestone decomposition and fuel burning), its contribution is relatively lower compared to steel and thermal power generation in terms of the overall industrial share.

Therefore, among the given options, manufacturing of cement is held least responsible for global warming.

We need to find the ratio of densities of fcc to bcc lattices for metal M.

Formula: The density of a crystal lattice is given by:

$$\rho = \frac{Z \times M}{N_A \times a^3}$$

where $$Z$$ is the number of atoms per unit cell, $$M$$ is the molar mass, $$N_A$$ is Avogadro's number, and $$a$$ is the edge length.

Number of atoms per unit cell.

For fcc: $$Z_{fcc} = 4$$

For bcc: $$Z_{bcc} = 2$$

Given edge lengths.

$$a_{fcc} = 2.0$$ Angstrom

$$a_{bcc} = 2.5$$ Angstrom

Calculate the ratio of densities.

Since the metal M is the same in both cases, $$M$$ and $$N_A$$ are the same:

$$\frac{\rho_{fcc}}{\rho_{bcc}} = \frac{Z_{fcc}}{Z_{bcc}} \times \frac{a_{bcc}^3}{a_{fcc}^3}$$

$$= \frac{4}{2} \times \frac{(2.5)^3}{(2.0)^3}$$

$$= 2 \times \frac{15.625}{8.0}$$

$$= 2 \times 1.953125$$

$$= 3.90625$$

Rounding to the nearest integer: $$\approx 4$$

The correct answer is $$\mathbf{4}$$ (Option D).

When 0.5 g of the organic compound X containing 60% carbon undergoes complete combustion, the mass of carbon in the compound can be determined.

The mass of carbon in the compound is calculated as $$\text{Mass of C} = 0.5 \times \frac{60}{100} = 0.3$$ g.

From this mass, the number of moles of carbon is found by dividing by the molar mass of carbon:

$$n_C = \frac{0.3}{12} = 0.025$$ mol.

Under complete combustion, each mole of carbon produces one mole of carbon dioxide according to the reaction:

$$C + O_2 \rightarrow CO_2$$

Thus, the number of moles of CO₂ formed is $$n_{CO_2} = 0.025$$ mol.

The mass of CO₂ produced is then calculated by multiplying by its molar mass:

$$m_{CO_2} = 0.025 \times 44 = 1.1$$ g $$= 11 \times 10^{-1}$$ g.

The correct answer is 11.

We are given that 1.1 g of a carbonate $$\text{M}_2\text{CO}_3$$ on treatment with excess HCl produces 0.01 mol of $$\text{CO}_2$$.

Write the balanced equation: $$ \text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2 $$

Find moles of $$\text{M}_2\text{CO}_3$$. From the balanced equation, 1 mol of $$\text{M}_2\text{CO}_3$$ produces 1 mol of $$\text{CO}_2$$.

Therefore, moles of $$\text{M}_2\text{CO}_3$$ = moles of $$\text{CO}_2$$ = 0.01 mol.

Calculate molar mass: $$ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{1.1}{0.01} = 110 \text{ g mol}^{-1} $$

Verification: If molar mass of $$\text{M}_2\text{CO}_3$$ = 110, then $$2M + 12 + 48 = 110$$, giving $$M = 25$$. This corresponds to a monovalent metal with atomic mass 25.

The molar mass of $$\text{M}_2\text{CO}_3$$ is 110 g mol$$^{-1}$$.

The molar mass of hydrocarbon (A) is (84\ \text{g mol}^{-1}), and it contains (85.8%) carbon.

The mass of carbon present in one mole of the compound is

$$84 \times 0.858 = 72.072 \approx 72\ \text{g mol}^{-1}.$$

Hence, the number of carbon atoms is

$$\frac{72}{12} = 6.$$

The remaining mass corresponds to hydrogen:

$$84 - 72 = 12\ \text{g mol}^{-1}.$$

Therefore, the number of hydrogen atoms is

$$\frac{12}{1} = 12.$$

Thus, the molecular formula of the hydrocarbon is

$$C_6H_{12}.$$

Hence, the number of hydrogen atoms present in one molecule of the hydrocarbon is

$$\boxed{12}.$$

A real gas occupies 0.15 dm$$^3$$ at 100 atm and 500 K with compressibility factor $$Z_1 = 1.07$$. We need to find its volume at 300 atm and 300 K with $$Z_2 = 1.4$$.

For the two states of the same gas (same $$n$$):

$$\frac{P_1 V_1}{Z_1 T_1} = nR = \frac{P_2 V_2}{Z_2 T_2}$$

$$V_2 = \frac{P_1 V_1 Z_2 T_2}{Z_1 T_1 P_2}$$

$$V_2 = \frac{100 \times 0.15 \times 1.4 \times 300}{1.07 \times 500 \times 300}$$

$$V_2 = \frac{100 \times 0.15 \times 1.4}{1.07 \times 500}$$

$$V_2 = \frac{21}{535} = 0.039252 \text{ dm}^3$$

$$V_2 = 0.039252 \text{ dm}^3 = 392.52 \times 10^{-4} \text{ dm}^3$$

Rounding to the nearest integer: $$\approx 393 \times 10^{-4}$$ dm$$^3$$.

The answer is $$\mathbf{393}$$.

Explanation

According to the kinetic theory of gases,

The root mean square speed is

$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

The most probable speed is

$$v_{mp}=\sqrt{\frac{2RT}{M}}$$

The problem states that

$$(v_{rms})_X=(v_{mp})_Y$$

Substituting the respective expressions,

$$\sqrt{\frac{3RT_X}{M_X}}=\sqrt{\frac{2RT_Y}{M_Y}}$$

Squaring both sides and cancelling $$R$$,

$$\frac{3T_X}{M_X}=\frac{2T_Y}{M_Y}$$

Substituting the given values,

$$\frac{3\times600}{40}=\frac{2\times90}{M_Y}$$

$$\frac{1800}{40}=\frac{180}{M_Y}$$

$$45=\frac{180}{M_Y}$$

$$M_Y=\frac{180}{45}$$

$$M_Y=4\text{ g mol}^{-1}$$

Hence, the molar mass of gas $$Y$$ is

$$4\text{ g mol}^{-1}$$

Therefore, the correct answer is **4**.

Find the volume of monobasic strong acid solution required to neutralize 25 mL of 0.24 M NaOH.

The molarity of the acid solution can be determined from its density and molar mass. With density = 1.21 kg/L = 1210 g/L and molar mass = 24.2 g/mol, the molarity is $$\frac{1210}{24.2} = 50$$ M.

For a monobasic strong acid neutralizing NaOH, the condition $$M_1V_1 = M_2V_2$$ applies. Substituting $$M_1 = 50$$ M, $$M_2 = 0.24$$ M and $$V_2 = 25$$ mL gives $$50 \times V_1 = 0.24 \times 25\,. $$

Solving for $$V_1$$ yields $$V_1 = \frac{0.24 \times 25}{50} = \frac{6}{50} = 0.12\text{ mL}$$ and expressing this in the required form gives $$V_1 = 0.12\text{ mL} = 12 \times 10^{-2}\text{ mL}\,,$$ so the required volume is $$\boxed{12}$$.

We need to find the volume of HCl solution required to completely neutralize the NaOH obtained from the reaction of metallic sodium with water.

Find moles of Na.

Molar mass of Na = 23 g/mol

$$ \text{Moles of Na} = \frac{0.69}{23} = 0.03 \text{ mol} $$

Reaction of Na with water.

$$ 2\text{Na} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 $$

Moles of NaOH formed = Moles of Na = 0.03 mol (1:1 ratio)

Neutralization reaction.

$$ \text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O} $$

Moles of HCl required = Moles of NaOH = 0.03 mol

Find the molarity of HCl solution.

HCl solution contains 73 g/L.

Molar mass of HCl = $$1 + 35.5 = 36.5$$ g/mol

$$ \text{Molarity} = \frac{73}{36.5} = 2 \text{ M} $$

Calculate the required volume.

$$ V = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.03}{2} = 0.015 \text{ L} = 15 \text{ mL} $$

The volume of HCl required is $$\boxed{15}$$ mL.

Reaction 1: $$Na_2O + H_2O \to 2NaOH$$

So X = NaOH (1 oxygen atom per molecule)

Reaction 2: $$Cl_2O_7 + H_2O \to 2HClO_4$$

So Y = HClO₄ (4 oxygen atoms per molecule)

Total oxygen atoms in X and Y = 1 + 4 = 5.

The fuel is $$C_{15}H_{30}$$ with density $$0.756$$ g/mL. We need to find the weight of $$O_2$$ required and $$CO_2$$ released for every litre of fuel.

The balanced combustion reaction is:

$$ C_{15}H_{30} + \frac{45}{2}O_2 \rightarrow 15CO_2 + 15H_2O $$

Verification: C: 15 = 15 ✓; H: 30 = 30 ✓; O: 45 = 30 + 15 = 45 ✓

The mass of 1 litre of fuel is: $$ \text{Mass} = 1000 \text{ mL} \times 0.756 \text{ g/mL} = 756 \text{ g} $$ The molar mass of $$C_{15}H_{30}$$ is: $$ M = 15 \times 12 + 30 \times 1 = 180 + 30 = 210 \text{ g/mol} $$ so the number of moles of fuel in 1 litre is: $$ n = \frac{756}{210} = 3.6 \text{ mol} $$

Each mole of fuel requires $$\frac{45}{2} = 22.5$$ moles of $$O_2$$, so the mass of $$O_2$$ required is: $$ \text{Mass of } O_2 = 3.6 \times 22.5 \times 32 = 3.6 \times 720 = 2592 \text{ g} $$

Each mole of fuel produces 15 moles of $$CO_2$$, so the mass of $$CO_2$$ released is: $$ \text{Mass of } CO_2 = 3.6 \times 15 \times 44 = 3.6 \times 660 = 2376 \text{ g} $$

The weight of oxygen required is 2592 g and $$CO_2$$ released is 2376 g. The correct answer is Option C: 2592 g and 2376 g.

An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with liquid (density 0.95 g/mL), and 40.5 g when filled with an ideal gas at 0.82 atm and 250 K. We need to find the molar mass of the gas.

First, the mass of liquid introduced into the vessel is calculated as $$135.0 - 40.0 = 95.0$$ g. The volume of the vessel is then given by the ratio of this mass to the liquid’s density, namely $$\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{95.0}{0.95} = 100\text{ mL} = 0.1\text{ L}.$$

The mass of the gas is found from the difference between the filled and empty vessel masses, which yields $$40.5 - 40.0 = 0.5$$ g.

Since the gas behaves ideally, we apply the ideal gas law: $$PV = nRT.$$ Solving for the number of moles gives $$n = \frac{PV}{RT} = \frac{0.82 \times 0.1}{0.082 \times 250} = \frac{0.082}{20.5} = 0.004\text{ mol}.$$

The molar mass of the gas is then determined by dividing the mass by the number of moles: $$M = \frac{\text{mass}}{n} = \frac{0.5}{0.004} = 125\text{ g/mol}.$$

The correct answer is Option D: 125 g/mol.

The question asks about the highest industrial consumption of molecular hydrogen ($$H_2$$).

Key Industrial Uses of Hydrogen:

1. Haber-Bosch Process (Ammonia synthesis): $$N_2 + 3H_2 \rightarrow 2NH_3$$ — This is by far the largest industrial consumer of hydrogen gas worldwide. Approximately 50-60% of all industrially produced hydrogen is used to manufacture ammonia.

2. Hydrogenation of oils and fats (carbon compounds) — uses hydrogen but in much smaller quantities.

3. Production of methanol ($$CO + 2H_2 \rightarrow CH_3OH$$) — significant but smaller than ammonia production.

4. Production of HCl (hydrogen + chlorine) — relatively small scale.

Since ammonia ($$NH_3$$) is a compound of nitrogen, the element whose compound consumes the most molecular hydrogen industrially is Nitrogen.

The correct answer is Option D: Nitrogen.

Clark's method of water softening involves adding slaked lime (calcium hydroxide, $$Ca(OH)_2$$) to hard water that contains temporary hardness (bicarbonates of calcium and magnesium).

Reaction with calcium bicarbonate:

Calcium bicarbonate reacts with calcium hydroxide to form insoluble calcium carbonate ($$CaCO_3$$), which precipitates out.

Reaction with magnesium bicarbonate:

Magnesium bicarbonate reacts with calcium hydroxide to form insoluble calcium carbonate and insoluble magnesium hydroxide ($$Mg(OH)_2$$), both of which precipitate out.

Therefore, the metal salts formed during softening of hard water using Clark's method are $$CaCO_3$$ and $$Mg(OH)_2$$.

The correct answer is Option B: $$CaCO_3$$ and $$Mg(OH)_2$$.

Clark's method is used for removing temporary hardness from water. Temporary hardness is caused by the presence of dissolved bicarbonates of calcium and magnesium, i.e., $$Ca(HCO_3)_2$$ and $$Mg(HCO_3)_2$$.

In Clark's method, a calculated amount of slaked lime, $$Ca(OH)_2$$, is added to the hard water. The reactions that occur are:

For calcium bicarbonate: $$Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + 2H_2O$$. Here, calcium carbonate ($$CaCO_3$$) precipitates out since it is insoluble in water.

For magnesium bicarbonate: $$Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2H_2O$$. In this case, magnesium hydroxide ($$Mg(OH)_2$$) precipitates out because $$MgCO_3$$ is slightly soluble in water. The excess $$Ca(OH)_2$$ converts the magnesium compound to the insoluble $$Mg(OH)_2$$.

So the products obtained during the treatment of hard water using Clark's method are $$CaCO_3$$ and $$Mg(OH)_2$$.

Hence, the correct answer is Option C.

We need to evaluate the Assertion and Reason about BOD (Biochemical Oxygen Demand).

Evaluating the Assertion:

BOD of clean water is typically around 1-5 ppm. Polluted water can have a BOD value of 17 ppm or higher, as it contains a large amount of organic matter that requires more oxygen for decomposition.

Therefore, the Assertion is true.

Evaluating the Reason:

BOD (Biochemical Oxygen Demand) is defined as the amount of oxygen required by bacteria to break down only the biodegradable organic matter in water. It does NOT measure the oxygen required for non-biodegradable organic material.

The Reason states that BOD measures oxygen required to oxidise both biodegradable and non-biodegradable organic material, which is incorrect.

Therefore, the Reason is false.

Conclusion:

Assertion is true but Reason is false.

The correct answer is Option C.

Let us verify each statement individually.

Statement I: Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide.

This is correct. Classical smog (also called London smog or reducing smog) occurs in cool, humid climates. It contains smoke, fog, and $$SO_2$$. The $$SO_2$$ acts as a reducing agent, hence it is called reducing smog.

Statement II: Photochemical smog has components: ozone, nitric oxide, acrolein, formaldehyde, PAN, etc.

This is also correct. Photochemical smog (also called Los Angeles smog) is formed by the action of sunlight on nitrogen oxides and hydrocarbons. Its components include ozone ($$O_3$$), nitric oxide (NO), acrolein ($$CH_2=CHCHO$$), formaldehyde ($$HCHO$$), and peroxyacetyl nitrate (PAN).

Since both Statement I and Statement II are correct, the answer is Option A.

We need to evaluate both statements about industrial waste management.

Statement I: The non bio-degradable fly ash and slag from steel industry can be used by cement industry.

This is CORRECT. Fly ash (from thermal power plants) and blast furnace slag (from steel industry) are commonly used as supplementary cementitious materials in the cement industry. Fly ash is used in Portland Pozzolana Cement (PPC), and slag is used in Portland Slag Cement (PSC). This is a well-established practice of industrial waste utilization.

Statement II: The fuel obtained from plastic waste is lead free.

This is CORRECT. When plastic waste is converted into fuel through pyrolysis, the resulting fuel is free from lead. Unlike conventional leaded fuels (which contained tetraethyl lead as an additive), the fuel derived from plastics does not contain lead compounds, making it a cleaner alternative in that regard.

Since both statements are correct:

The correct answer is Option A: Both Statement I and Statement II are correct.

We need to identify the pair in which both substances are herbicides (chemicals used to kill unwanted plants/weeds).

We recall the classification of common pesticides. Aldrin and Dieldrin are well-known insecticides — they belong to the class of chlorinated hydrocarbon insecticides used to kill insects, not weeds.

On the other hand, sodium chlorate ($$NaClO_3$$) and sodium arsenite ($$Na_3AsO_3$$) are both classified as herbicides. Sodium chlorate is a non-selective herbicide that kills plants by oxidative damage, and sodium arsenite has historically been used as a weed killer.

Now let us examine the other options. Option B pairs sodium chlorate (herbicide) with Aldrin (insecticide) — not both herbicides. Option C pairs sodium arsenite (herbicide, noting the question writes "arsinate" but means arsenite) with Dieldrin (insecticide) — again, not both herbicides.

Hence, the only pair where both substances are herbicides is sodium chlorate and sodium arsenite.

Hence, the correct answer is Option D: Sodium chlorate and sodium arsenite.

We need to match the water pollutants/chemicals in List-I with their effects in List-II.

Excess sulphate in drinking water causes a laxative effect (diarrhoea); therefore, A - Sulphate → (III) Laxative effect.

Excess fluoride in water causes fluorosis, resulting in bending of bones and mottling of teeth; hence, B - Fluoride → (II) Bending of bones.

Nicotine has historically been used as a pesticide (insecticide) to kill insects in agriculture, so C - Nicotine → (I) Pesticide.

Sodium arsenite functions as a herbicide (weed killer), giving D - Sodium arsenite → (IV) Herbicide.

This leads to the matching A - III, B - II, C - I, and D - IV, corresponding to Option B.

Eutrophication is the process where a water body becomes excessively enriched with nutrients (especially nitrogen and phosphorus), leading to excessive plant and algal growth.

Understand the eutrophication process: When excess nutrients enter a water body, they promote rapid growth of algae (algal bloom). This dense algal growth covers the surface, blocking sunlight from reaching deeper water.

Consequences of eutrophication: When the algae die, their decomposition by aerobic bacteria consumes large amounts of dissolved oxygen. This leads to:

$$\bullet$$ Depletion of dissolved oxygen (hypoxia or anoxia)

$$\bullet$$ Death of aquatic organisms (fish, invertebrates) that depend on oxygen

$$\bullet$$ Overall loss in biodiversity

Evaluate the options: Option A: Increase in biodiversity — Incorrect; eutrophication kills aquatic organisms due to oxygen depletion.

Option B: Loss in biodiversity — Correct; oxygen depletion causes death of fish and other aquatic life.

Option C: Break down of organic matter — While decomposition does occur, this is part of the process, not the main result of eutrophication.

Option D: Decrease in BOD — Incorrect; BOD actually increases due to the large amount of organic matter from dead algae.

The correct answer is Option B: Loss in biodiversity.

We need to match each pollutant in List-I with its correct source in List-II.

(A) Microorganisms

Microorganisms such as bacteria, viruses, and other pathogens are primarily found in domestic sewage (II). Human waste and household water contain large amounts of harmful microorganisms.

So, A matches with II.

(B) Plant nutrients

Plant nutrients like nitrates, phosphates, and potassium compounds come from chemical fertilizers (III). When excess fertilizers run off into water bodies, they cause eutrophication.

So, B matches with III.

(C) Toxic heavy metals

Toxic heavy metals such as mercury (Hg), lead (Pb), cadmium (Cd), and chromium (Cr) are released from chemical factories (IV) as industrial waste.

So, C matches with IV.

(D) Sediment

Sediment (soil, sand, and rock particles) is produced during strip mining (I). Strip mining removes the surface layer of earth, leading to large amounts of sediment runoff.

So, D matches with I.

The correct matching is: A-II, B-III, C-IV, D-I

Therefore, the correct answer is Option A: A-II, B-III, C-IV, D-I.

We need to match pollutants in water with the diseases they cause at the given concentration levels.

(A) Sulphate (>500 ppm): Excess sulphate in water causes a laxative effect (diarrhoea). This matches with (III).

(B) Nitrate (>50 ppm): Excess nitrate causes methemoglobinemia (also called "blue baby syndrome"). Nitrate gets reduced to nitrite in the body, which converts haemoglobin to methemoglobin, reducing oxygen-carrying capacity. This matches with (I).

(C) Lead (>50 ppb): Lead is a toxic heavy metal that causes kidney damage and also damages the nervous system and liver. This matches with (IV).

(D) Fluoride (>2 ppm): Excess fluoride causes brown mottling of teeth (dental fluorosis). It also causes bone disease (skeletal fluorosis) at higher concentrations. This matches with (II).

A - III, B - I, C - IV, D - II

The correct answer is Option B: A-III, B-I, C-IV, D-II.

We need to identify the water samples that are highly polluted with organic wastes based on their BOD values. The BOD values for the samples are A = 3 ppm, B = 18 ppm, C = 21 ppm, and D = 4 ppm.

Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen needed by aerobic organisms to break down organic material in water. A higher BOD indicates more organic pollution.

Water with BOD < 5 ppm is considered clean, while water with BOD > 17 ppm is considered highly polluted.

A = 3 ppm — Clean water
B = 18 ppm — Highly polluted (BOD > 17 ppm)
C = 21 ppm — Highly polluted (BOD > 17 ppm)
D = 4 ppm — Clean water

The highly polluted samples are B and C.

The correct answer is Option C: B & C.

We need to evaluate both statements about water pollution.

Statement I: "In polluted water, values of both dissolved oxygen and BOD are very low."

Polluted water contains a large amount of organic matter. Microorganisms consume dissolved oxygen to decompose this organic matter. Therefore:

• Dissolved oxygen (DO) is low in polluted water — this part is correct.
• Biochemical Oxygen Demand (BOD) is high in polluted water, because a large quantity of oxygen is needed to break down the organic waste — this part is incorrect.

So Statement I is false.

Statement II: "Eutrophication results in decrease in the amount of dissolved oxygen."

Eutrophication is the excessive growth of algae and aquatic plants due to nutrient enrichment (phosphates, nitrates). When these organisms die, their decomposition by bacteria consumes large amounts of dissolved oxygen, leading to oxygen depletion in the water body.

So Statement II is true.

Therefore, the correct answer is Option D: Statement I is false but Statement II is true.

First extract the data from the table given in the question.

Compound 1 → $$\%P = 50,\; \%Q = 50$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{50}{50}=1$$
Compound 2 → $$\%P = 44.4,\; \%Q = 55.6$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{44.4}{55.6}=0.8=\dfrac45$$
Compound 3 → $$\%P = 40,\; \%Q = 60$$  ⇒ mass ratio $$\dfrac{P}{Q}= \dfrac{40}{60}= \dfrac23$$

Case 1: Checking Option A - “If empirical formula of compound 3 is $$P_3Q_4$$, then empirical formula of compound 2 is $$P_3Q_5$$.”

For compound 3 assume $$P_3Q_4$$.
Weight ratio $$\dfrac{P}{Q}= \dfrac{3M_P}{4M_Q}= \dfrac23$$   $$\Rightarrow 9M_P = 8M_Q$$   $$\Rightarrow \dfrac{M_P}{M_Q}= \dfrac89$$ $$-(1)$$
With the same atomic masses, the ratio for $$P_3Q_5$$ (suggested for compound 2) would be
$$\dfrac{P}{Q}= \dfrac{3M_P}{5M_Q}= \dfrac{3\left(\tfrac89 M_Q\right)}{5M_Q}= \dfrac{24}{45}=0.533$$
The experimental ratio for compound 2 is $$0.8$$, not $$0.533$$. Hence Option A is WRONG.

Case 2: Checking Option B - “If empirical formula of compound 3 is $$P_3Q_2$$ and atomic weight of P is 20, atomic weight of Q is 45.”

Assume $$M_P = 20$$ and empirical formula $$P_3Q_2$$ for compound 3.
Mass of P in one empirical unit = $$3 \times 20 = 60$$
Let $$M_Q = x$$. Mass of Q in one empirical unit = $$2x$$.
Percentage of P: $$\dfrac{60}{60+2x}\times100 = 40\%$$ (from data of compound 3)
$$\dfrac{60}{60+2x}=0.40 \;\Rightarrow\; 60 = 0.40(60+2x)=24+0.8x$$
$$60-24 = 0.8x \;\Rightarrow\; 36 = 0.8x \;\Rightarrow\; x = 45$$
Thus $$M_Q = 45\;\text{u}$$. Option B is CORRECT.

Case 3: Checking Option C - “If empirical formula of compound 2 is $$PQ$$, then empirical formula of compound 1 is $$P_5Q_4$$.”

For compound 2 let empirical formula be $$PQ$$.
Weight ratio $$\dfrac{P}{Q}= \dfrac{M_P}{M_Q}=0.8=\dfrac45$$ ⇒ take $$M_P = 4k,\; M_Q = 5k$$ $$-(2)$$.
With these atomic masses, the ratio for $$P_5Q_4$$ (suggested for compound 1) becomes
$$\dfrac{P}{Q}= \dfrac{5M_P}{4M_Q}= \dfrac{5\,(4k)}{4\,(5k)}= \dfrac{20k}{20k}=1$$
This matches the experimental ratio of compound 1 (50 % : 50 % ⇒ 1 : 1). Hence Option C is CORRECT.

Case 4: Checking Option D - “If atomic weights of P and Q are 70 and 35 respectively, empirical formula of compound 1 is $$P_2Q$$.”

Given $$M_P = 70,\; M_Q = 35$$, compound 1 has equal masses of P and Q.
To get equal masses: number of moles $$n_P : n_Q = \dfrac{1}{70} : \dfrac{1}{35} = 1 : 2$$.
Thus the simplest integer formula is $$PQ_2$$, not $$P_2Q$$. Option D is WRONG.

Therefore the correct choices are:
Option B (atomic weight of Q is 45) and Option C (empirical formula $$P_5Q_4$$).

Final answer: Option B, Option C.

We have 2 L of 0.2 M $$H_2SO_4$$ and 2 L of 0.1 M NaOH solution being mixed together. We need to find the molarity of $$Na_2SO_4$$ formed in the resulting solution.

First, let us calculate the moles of each reactant.

Moles of $$H_2SO_4 = 0.2 \times 2 = 0.4$$ mol

Moles of NaOH $$= 0.1 \times 2 = 0.2$$ mol

The reaction between $$H_2SO_4$$ and NaOH is:

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

From the stoichiometry, 1 mol of $$H_2SO_4$$ requires 2 mol of NaOH. So 0.2 mol of NaOH would react with $$\frac{0.2}{2} = 0.1$$ mol of $$H_2SO_4$$.

Since we have 0.4 mol of $$H_2SO_4$$ but only 0.2 mol of NaOH, the NaOH is the limiting reagent. All 0.2 mol of NaOH reacts with 0.1 mol of $$H_2SO_4$$, and the remaining 0.3 mol of $$H_2SO_4$$ stays unreacted.

From the stoichiometry, 0.1 mol of $$H_2SO_4$$ produces 0.1 mol of $$Na_2SO_4$$.

Now, the total volume of the resulting solution is $$2 + 2 = 4$$ L.

The molarity of $$Na_2SO_4$$ is:

$$\text{Molarity} = \frac{0.1}{4} = 0.025 \text{ M} = 25 \text{ millimolar}$$

Hence, the correct answer is 25.

We have a 10 g mixture of hydrogen ($$H_2$$) and helium ($$He$$) in a vessel of capacity 0.0125 m$$^3$$ at 6 bar and 27 °C and need to find the mass of helium.

Let the mass of helium = $$x$$ g, so the mass of hydrogen = $$(10 - x)$$ g. Hence the moles of helium are $$\frac{x}{4}$$ (molar mass of He = 4 g/mol) and the moles of hydrogen are $$\frac{10 - x}{2}$$ (molar mass of $$H_2$$ = 2 g/mol).

Since we can find the total number of moles using the ideal gas equation, $$PV = nRT$$, we convert units: P = 6 bar = 6 × 10$$^5$$ Pa, V = 0.0125 m$$^3$$, T = 27 °C = 300 K, and R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$ = 8.3 Pa m$$^3$$ K$$^{-1}$$ mol$$^{-1}$$.

Substituting these into $$n = \frac{PV}{RT} = \frac{6 \times 10^5 \times 0.0125}{8.3 \times 300}$$ gives a numerator of 6 × 10$$^5$$ × 0.0125 = 7500 and a denominator of 8.3 × 300 = 2490. This gives $$n = \frac{7500}{2490} = 3.012 \approx 3 \text{ mol}$$.

Therefore, the total moles satisfy $$\frac{x}{4} + \frac{10 - x}{2} = 3$$. Multiplying through by 4 yields $$x + 2(10 - x) = 12$$, which simplifies to $$x + 20 - 2x = 12$$ hence $$-x = -8$$ and $$x = 8$$.

Therefore, the mass of helium in the mixture is 8 g.

We need to find the new pressure of the moist gas when the volume is doubled.

Initially, the total pressure of the moist gas is 4 atm at 27°C, and the vapour pressure of water at this temperature is 0.4 atm.

Subtracting the vapour pressure from the total gives the pressure of the dry gas as Total pressure - Vapour pressure of water = $$4 - 0.4 = 3.6$$ atm.

Applying Boyle’s law to the dry gas, we use $$P_1V_1 = P_2V_2$$, where $$3.6 \times V = P_2 \times 2V$$, yielding $$P_2 = \dfrac{3.6}{2} = 1.8 \text{ atm}$$.

The vapour pressure of water depends only on temperature, not on volume (as long as liquid water is present), so at 27°C it remains 0.4 atm.

In a moist gas problem, we assume water vapour remains saturated (liquid water is present), and therefore the vapour pressure stays at 0.4 atm.

The new total pressure is the sum of the pressure of dry gas and the vapour pressure of water, giving $$1.8 + 0.4 = 2.2 \text{ atm}$$.

Expressing in the required format: $$2.2 \text{ atm} = 22 \times 10^{-1} \text{ atm}$$

Hence, the answer is 22.

The van der Waals equation for a real gas is:

$$\mathrm{\left(P + \frac{an^2}{V^2}\right)(V-nb)=nRT}$$

For one mole of gas:

$$\mathrm{n=1}$$

Therefore,

$$\mathrm{\left(P + \frac{a}{V^2}\right)(V-b)=RT}$$

At very high pressure, volume becomes very small.

Hence, the intermolecular attraction term $$\mathrm{\frac{a}{V^2}}$$ becomes negligible compared to $$\mathrm{P}$$.

Thus,

$$\mathrm{P(V-b)\approx RT}$$

$$\mathrm{PV-Pb=RT}$$

Dividing throughout by $$\mathrm{RT}$$:

$$\mathrm{\frac{PV}{RT}-\frac{Pb}{RT}=1}$$

Since:

$$\mathrm{Z=\frac{PV}{RT}}$$

we get:

$$\mathrm{Z-\frac{Pb}{RT}=1}$$

$$\mathrm{Z=1+\frac{Pb}{RT}}$$

Given:

$$\mathrm{Z=2}$$

$$\mathrm{P=99\ bar}$$

$$\mathrm{R=0.083\ L\ bar\ K^{-1}\ mol^{-1}}$$

$$\mathrm{T=25^\circ C=298\ K}$$

Substituting values:

$$\mathrm{2=1+\frac{99\times b}{0.083\times298}}$$

$$\mathrm{1=\frac{99b}{24.734}}$$

$$\mathrm{b=\frac{24.734}{99}}$$

$$\mathrm{b\approx0.25\ L\ mol^{-1}}$$

Expressing in the form $$\mathrm{\times10^{-2}\ L\ mol^{-1}}$$:

$$\mathrm{0.25 = 25\times10^{-2}}$$

$$\boxed{\mathrm{25}}$$

We need to find the minimum temperature at which a sealed flask will burst.

Volume = $$2 \text{ dm}^3 = 2 \times 10^{-3} \text{ m}^3$$

Mass of propane = 11 g

Burst pressure = 2 MPa = $$2 \times 10^6 \text{ Pa}$$

$$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$$

Calculate moles of propane

Molecular formula of propane: $$C_3H_8$$

Molar mass = $$3 \times 12 + 8 \times 1 = 44 \text{ g/mol}$$

$$n = \frac{11}{44} = 0.25 \text{ mol}$$

Apply the ideal gas equation

$$PV = nRT$$

$$T = \frac{PV}{nR}$$

Substitute values

$$T = \frac{2 \times 10^6 \times 2 \times 10^{-3}}{0.25 \times 8.3}$$

$$T = \frac{4000}{2.075}$$

$$T = 1927.7 \text{ K}$$

Convert to Celsius

$$T = 1927.7 - 273 = 1654.7 \approx 1655 \text{ °C}$$

The answer is 1655.

We consider a body-centred cubic (BCC) unit cell with edge length $$a = 300$$ pm $$= 300 \times 10^{-10}$$ cm $$= 3 \times 10^{-8}$$ cm and density $$\rho = 6.0$$ g cm$$^{-3}$$. For BCC, the number of atoms per unit cell is $$Z = 2$$.

Using the density relation $$\rho = \frac{Z \times M}{N_A \times a^3}$$ and solving for $$M$$ gives $$M = \frac{\rho \times N_A \times a^3}{Z}$$. Substituting the known values yields $$M = \frac{6.0 \times 6.022 \times 10^{23} \times (3 \times 10^{-8})^3}{2}$$.

Since $$a^3 = (3 \times 10^{-8})^3 = 27 \times 10^{-24} = 2.7 \times 10^{-23}$$ cm$$^3$$, substituting this into the expression for $$M$$ gives $$M = \frac{6.0 \times 6.022 \times 10^{23} \times 2.7 \times 10^{-23}}{2} = \frac{6.0 \times 6.022 \times 2.7}{2} = \frac{97.56}{2} = 48.78$$ g/mol.

To find the number of atoms in 180 g of the element, the number of moles is calculated as $$\text{Number of moles} = \frac{180}{48.78} = 3.69$$ mol. Therefore, the number of atoms is $$\text{Number of atoms} = 3.69 \times 6.022 \times 10^{23} = 22.22 \times 10^{23}$$.

Therefore, the number of atoms present in 180 g is $$22 \times 10^{23}$$ (nearest integer).

We are given a two-step reaction where the yield of reaction I is 60% and the yield of reaction II is 50%.

The overall yield of a multi-step reaction is the product of the individual yields of each step.

Overall yield = Yield of Step I $$\times$$ Yield of Step II

$$\text{Overall yield} = \frac{60}{100} \times \frac{50}{100} = 0.6 \times 0.5 = 0.3$$

Converting to percentage:

$$\text{Overall yield} = 0.3 \times 100 = 30\%$$

Therefore, the overall yield of the complete reaction is $$\textbf{30}$$ %.

We need to find the volume of gas adsorbed per gram of the adsorbent.

Find the number of moles of $$H_2$$ gas.

Molar mass of $$H_2$$ = 2 g/mol

$$n = \frac{2.0}{2} = 1 \text{ mol}$$

Calculate the volume of gas using the ideal gas equation.

$$PV = nRT$$

$$V = \frac{nRT}{P} = \frac{1 \times 0.083 \times 300}{1}$$

$$V = 24.9 \text{ L} = 24900 \text{ mL}$$

Find volume per gram of adsorbent.

Mass of adsorbent (platinum powder) = 2.5 g

$$\text{Volume per gram} = \frac{24900}{2.5} = 9960 \text{ mL}$$

Therefore, the volume of gas adsorbed per gram of the adsorbent is $$\textbf{9960}$$ mL.

The compound has the formula $$C_xH_yO_z$$ and weighs 1.80 g. On complete combustion, it produces 2.64 g of $$CO_2$$ and 1.08 g of $$H_2O$$.

Moles of $$CO_2 = \frac{2.64}{44} = 0.06\,\text{mol}$$. Since each mole of $$CO_2$$ contains one mole of carbon, moles of C $$= 0.06$$. Mass of C $$= 0.06 \times 12 = 0.72\,\text{g}$$.

Moles of $$H_2O = \frac{1.08}{18} = 0.06\,\text{mol}$$. Since each mole of $$H_2O$$ contains 2 moles of hydrogen, moles of H $$= 0.12$$. Mass of H $$= 0.12 \times 1 = 0.12\,\text{g}$$.

Mass of oxygen in the compound $$= 1.80 - 0.72 - 0.12 = 0.96\,\text{g}$$.

Percentage of oxygen $$= \frac{0.96}{1.80} \times 100 = 53.33\%$$.

Therefore, the percentage of oxygen in the organic compound is $$53.33\%$$, which corresponds to option (1).

We begin with the van der Waals equation for n moles of a real gas

$$\left(P + \frac{a n^{2}}{V^{2}}\right)\,(V - nb) \;=\; nRT$$

In this expression the pressure-correction term $$\frac{a n^{2}}{V^{2}}$$ is added directly to the pressure $$P$$. Because only like quantities can be added, the whole fraction $$\frac{a n^{2}}{V^{2}}$$ must possess exactly the same unit as $$P$$.

Let us denote the unit of pressure by $$[P]$$, the unit of volume by $$[V]$$ and the unit of amount of substance by $$[n]$$. Correspondingly, let $$[a]$$ be the unknown unit of the parameter $$a$$ which we have to determine.

Since the dimensions of the two addends are equal, we can write

$$\Bigl[\frac{a n^{2}}{V^{2}}\Bigr] \;=\; [P]$$

Substituting the separate units we get

$$[a] \,[n]^{2}\,[V]^{-2} \;=\; [P]$$

Rearranging for $$[a]$$ gives

$$[a] \;=\; [P]\,[V]^{2}\,[n]^{-2}$$

Now we replace each symbol by the commonly used practical units in which the van der Waals constants are usually expressed:

• The convenient laboratory unit for pressure is the atmosphere, so $$[P] = \text{atm}$$.
• The convenient laboratory unit for volume is the cubic decimetre, so $$[V] = \text{dm}^3$$.
• The SI-derived unit for the amount of substance is the mole, so $$[n] = \text{mol}$$.

Substituting these explicit units we arrive at

$$[a] \;=\; (\text{atm})\,(\text{dm}^{3})^{2}\,(\text{mol})^{-2}$$

Simplifying the exponent on the volume unit, we finally have

$$[a] = \text{atm}\,\text{dm}^{6}\,\text{mol}^{-2}$$

This exactly matches Option C in the given list.

Hence, the correct answer is Option C.

At constant T, PV is constant. Thus, the PV graph is a straight line.

First, remember the definition: the biochemical oxygen demand, written as $$\text{BOD}$$, measures the amount of dissolved oxygen required by microorganisms to oxidise the organic matter present in a water sample.

A key point is that a high concentration of biodegradable organic impurities will make microorganisms consume more oxygen, so the value of $$\text{BOD}$$ increases. Conversely, if the water contains very little organic impurity, microorganisms need only a small amount of oxygen, so the value of $$\text{BOD}$$ will be low.

Mathematically we can say

$$\text{Organic pollution} \propto \text{BOD}$$

That single proportionality is enough to compare the four values given. Lower $$\text{BOD}$$ means lower organic pollution, and therefore the water is cleaner.

The four values are $$11\ \text{ppm},\; 15\ \text{ppm},\; 3\ \text{ppm},\; 21\ \text{ppm}$$. Among these, $$3\ \text{ppm}$$ is clearly the smallest.

Because $$3\ \text{ppm}$$ is the minimum value, it represents the least oxygen required by microorganisms, indicating the smallest amount of biodegradable organic matter present.

Therefore, the sample having $$\text{BOD}=3\ \text{ppm}$$ will be the cleanest.

Hence, the correct answer is Option C.

Green chemistry aims to reduce or eliminate the use of hazardous substances in chemical processes. The correct answer is option 4: liquified $$\text{CO}_2$$ for dry cleaning of clothes.

Let us evaluate each option. Chlorine used for bleaching paper is harmful as it produces toxic byproducts such as dioxins and chlorinated compounds, making it environmentally unfriendly. Using a large amount of water alone for washing clothes is wasteful of a precious natural resource, which contradicts green chemistry principles. Tetrachloroethene (perchloroethylene) used in dry cleaning is a volatile organic compound that is toxic and a suspected carcinogen, and it contaminates groundwater — exactly the kind of substance green chemistry seeks to replace.

Liquified $$\text{CO}_2$$ as a solvent for dry cleaning is the green chemistry alternative. $$\text{CO}_2$$ is non-toxic, non-flammable, and the carbon dioxide used is often a byproduct of existing industrial processes, so its use does not add net carbon to the atmosphere. It is an excellent solvent that leaves no harmful residues and is easily recycled in a closed-loop system. This makes it the best example of green chemistry in day-to-day life among the given options.

The correct answer is option 4.

Exerpt from NCERT: "Some of the heat is trapped by gases such as carbondioxide, methane, ozone, chlorofluorocarbon compounds (CFCs) and water vapour in the atmosphere. Thus, they add to the heating of the atmosphere. This causes global warming."

Thus the right option is C (A, C, D).

Statement I says that non-biodegradable wastes are generated by thermal power plants. Thermal power plants burn coal and produce fly ash, bottom ash, and flue gas pollutants. These solid wastes are non-biodegradable. Statement I is true.

Statement II says that biodegradable detergents lead to eutrophication. Even biodegradable detergents contain phosphates, and when these phosphates enter water bodies, they serve as nutrients promoting excessive algal growth, leading to eutrophication. Statement II is true.

Since both statements are true, the correct answer is option (4).

Statement I says that the value of the parameter "Biochemical Oxygen Demand (BOD)" is important for survival of aquatic life. This is true. BOD measures the amount of dissolved oxygen consumed by microorganisms in the biological decomposition of organic matter in water. A high BOD indicates a large amount of organic pollutants, which depletes dissolved oxygen and threatens aquatic life. Therefore, monitoring BOD is indeed important for the survival of aquatic organisms.

Statement II says that the optimum value of BOD is 6.5 ppm. This is false. The value of 6.5 ppm is actually associated with the optimum pH of water or is sometimes confused with dissolved oxygen levels, not BOD. Clean water typically has a BOD of less than 5 ppm. A BOD value of 6.5 ppm would indicate moderately polluted water. The lower the BOD, the cleaner the water and the better conditions for aquatic life.

Therefore, the correct answer is Option (1): Statement I is true but Statement II is false.

We begin with compound (a) carbon monoxide, whose most well-known toxic action is its strong affinity for the blood pigment haemoglobin. The reaction is written as $$\mathrm{CO + Hb \;\longrightarrow\; HbCO}$$ and the product, called carboxy-haemoglobin, blocks the normal transport of oxygen. Thus the only entry in List-II that fits carbon monoxide is “haemoglobin (iii).”

Next, compound (b) sulphur dioxide is an acidic, hygroscopic gas that injures green vegetation; one characteristic injury noted in horticulture is the abnormal hardening or stiffness of developing flower buds. Therefore sulphur dioxide must be matched with “stiffness of flower buds (iv).”

For compound (c) polychlorinated biphenyls (PCBs), textbooks of environmental chemistry state that these chlorinated aromatic compounds are potent human and animal carcinogens. Hence PCBs correspond to “carcinogenic (i).”

Finally, compound (d) the oxides of nitrogen (NO and NO2) can act as a supplementary nitrogen source for certain higher plants; pear trees belonging to the genus Pyrus are specifically reported to absorb and metabolise these oxides. Consequently, oxides of nitrogen are best paired with “metabolized by pyrus plants (ii).”

Collecting all the matches we have found:

$$\begin{aligned} (a) &\;\;\text{Carbon monoxide} \;\longrightarrow\; (iii)\;\text{Haemoglobin} \\ (b) &\;\;\text{Sulphur dioxide} \;\longrightarrow\; (iv)\;\text{Stiffness of flower buds} \\ (c) &\;\;\text{Polychlorinated biphenyls} \;\longrightarrow\; (i)\;\text{Carcinogenic} \\ (d) &\;\;\text{Oxides of nitrogen} \;\longrightarrow\; (ii)\;\text{Metabolized by pyrus plants} \end{aligned}$$

This sequence corresponds exactly to Option A.

Hence, the correct answer is Option A.

Photochemical smog (also called oxidising smog or Los Angeles smog) is formed during the daytime in the presence of sunlight. It involves the photochemical reaction of nitrogen oxides ($$NO_x$$) and volatile organic compounds (VOCs) in the presence of sunlight, producing ozone ($$O_3$$), peroxyacetyl nitrate (PAN), and other secondary pollutants.

The presence of $$O_3$$ is a characteristic feature of photochemical or oxidising smog. Since ozone and other oxidants are the primary components, this type of smog is called oxidising smog. It increases during the daytime because sunlight drives the photochemical reactions that produce these oxidising pollutants.

Reducing smog (London smog), by contrast, occurs in cooler, humid conditions and contains $$SO_2$$ and soot particles. It is not associated with ozone formation.

Therefore, the correct answer is Option (2): Oxidising smog.

We begin by recalling what eutrophication means. Eutrophication is the enrichment of a water body with nutrients, chiefly nitrates ($$\text{NO}_3^-$$) and phosphates ($$\text{PO}_4^{3-}$$). This nutrient loading stimulates excessive algal and plant growth.

Because of the dense algal bloom, sunlight penetration into deeper layers of water is hindered. Aquatic plants in the deeper zone therefore cannot photosynthesise efficiently, so their production of oxygen decreases. Later, when the algal mass dies, its decomposition by bacteria consumes the dissolved oxygen (DO) present in water.

The stoichiometric oxygen demand of the decomposition can be expressed as

$$\text{Organic Matter} + O_2 \longrightarrow \text{CO}_2 + \text{H}_2O + \text{Energy}.$$

The more biomass there is, the more $$O_2$$ is required for complete oxidation. Hence, eutrophication always tends to lower the dissolved oxygen level.

Now we check each statement one by one.

Option A says, “Eutrophication indicates that the water body is polluted.” We have just seen that eutrophication arises from an overload of nutrients, which is indeed a form of water pollution. So this statement is correct.

Option B says, “The dissolved oxygen concentration below 6 ppm inhibits fish growth.” The thumb rule used by environmental engineers is that fish need at least about $$5 \text{-} 6 \text{ mg L}^{-1}$$ (ppm) of DO for normal growth; if $$\text{DO} \lt 6 \text{ ppm},$$ fish become stressed and growth is hampered. Hence Option B is correct.

Option C says, “Eutrophication leads to increase in the oxygen level in water.” We have already explained that decomposition of the excess biomass consumes oxygen according to

$$\text{BOD} = \frac{\text{Mass of } O_2 \text{ consumed}}{\text{Volume of water}},$$

and a high biochemical oxygen demand (BOD) means lower residual $$O_2$$. Therefore, eutrophication does not increase oxygen; it depletes it. Thus Option C is an incorrect statement.

Option D says, “Eutrophication leads to anaerobic conditions.” As the DO falls far below $$1 \text{ ppm},$$ aerobic microorganisms die, and anaerobic bacteria begin to dominate, producing gases like $$\text{H}_2S$$ and $$\text{CH}_4$$. So Option D is a correct statement.

Only Option C contradicts the scientific facts. Hence, the correct answer is Option C.

The reduction of a metal oxide by coke (carbon) depends on the relative positions of the metal and carbon on the Ellingham diagram. Carbon can reduce a metal oxide only if the Gibbs free energy line for the formation of $$CO/CO_2$$ lies below the line for the metal oxide at the operating temperature.

Aluminium oxide ($$Al_2O_3$$) is extremely stable with a very large negative $$\Delta G_f°$$. On the Ellingham diagram, the line for $$Al_2O_3$$ lies well below the line for $$CO$$ at all practical temperatures. This means carbon cannot reduce $$Al_2O_3$$ to aluminium. This is why aluminium is extracted industrially by electrolysis of molten alumina (Hall-Heroult process) rather than by carbon reduction.

In contrast, $$ZnO$$, $$Fe_2O_3$$, and $$Cu_2O$$ can all be reduced by coke at sufficiently high temperatures, as their Ellingham diagram lines cross above the carbon line. Zinc is extracted by carbon reduction at about 1400°C, iron is produced in blast furnaces using coke, and copper oxide is easily reduced by carbon.

Therefore, the reduction that cannot be carried out with coke is $$Al_2O_3 \to Al$$, which is option (1).

The question asks which statement is an incorrect reason for eutrophication. Eutrophication is the excessive enrichment of water bodies with nutrients (especially nitrogen and phosphorus), which triggers rapid algal and plant growth, oxygen depletion, and ecosystem damage.

(A) Excess usage of fertilisers causes nutrient runoff (nitrogen and phosphorus compounds) into rivers and lakes, directly contributing to eutrophication. This is a correct reason.

(B) Excess usage of detergents introduces phosphates into water bodies, which promotes excessive algal growth. This is a correct reason for eutrophication.

(C) Dense plant population in water bodies is directly linked to excess nutrients already present, and such overgrowth further depletes oxygen, worsening eutrophication. This is a valid reason associated with the eutrophication process.

(D) Lack of nutrients in water bodies that prevent plant growth is the exact opposite of what causes eutrophication. Eutrophication occurs due to an excess of nutrients. A nutrient deficiency leads to oligotrophic (nutrient-poor) conditions, not eutrophication. This is clearly an incorrect reason.

Therefore, the incorrect reason for eutrophication is (D) only.

The removal of sulphur from sulphide ores is achieved by the process of roasting. In roasting, the sulphide ore is heated strongly in the presence of excess air (oxygen) below its melting point. During this process, the sulphide ore is converted to the corresponding metal oxide, and sulphur is released as sulphur dioxide ($$SO_2$$) gas.

For example, when zinc sulphide ore is roasted: $$2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$$. Similarly, for copper: $$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$$.

Smelting involves melting the ore with a reducing agent, leaching is a chemical washing process to dissolve the ore, and refining is the purification of the crude metal. None of these specifically target sulphur removal.

Therefore, the correct answer is Option (2): Roasting.

We have been supplied with 100 g of propane, whose molecular formula is $$\mathrm{C_3H_8}$$. The molar mass of propane is obtained by adding the atomic masses: $$3(12\ \text{g mol}^{-1}) + 8(1\ \text{g mol}^{-1}) = 44\ \text{g mol}^{-1}.$$

The general formula for finding moles is first stated:

$$\text{Number of moles} = \dfrac{\text{Given mass}}{\text{Molar mass}}.$$

Using this, the number of moles of propane present is

$$n_{\mathrm{C_3H_8}} = \dfrac{100\ \text{g}}{44\ \text{g mol}^{-1}} = 2.2727\ \text{mol}.$$

Next, the oxygen. The molecular mass of $$\mathrm{O_2}$$ is $$32\ \text{g mol}^{-1}.$$ Hence,

$$n_{\mathrm{O_2,\ given}} = \dfrac{1000\ \text{g}}{32\ \text{g mol}^{-1}} = 31.25\ \text{mol}.$$

The balanced combustion equation for propane is written and will now be used for stoichiometry:

$$\mathrm{C_3H_8 + 5\,O_2 \;\rightarrow\; 3\,CO_2 + 4\,H_2O}.$$

From the coefficients we see that 1 mol of propane needs 5 mol of oxygen. Therefore, the oxygen required for the full consumption of the available propane is

$$n_{\mathrm{O_2,\ reqd}} = 5 \times n_{\mathrm{C_3H_8}} = 5 \times 2.2727 = 11.3636\ \text{mol}.$$

Because $$31.25\ \text{mol} > 11.3636\ \text{mol},$$ oxygen is in excess and propane is the limiting reagent. Hence all propane will be consumed.

By direct proportion from the balanced equation, the amounts of products formed are:

• Carbon dioxide: $$n_{\mathrm{CO_2}} = 3 \times n_{\mathrm{C_3H_8}} = 3 \times 2.2727 = 6.8182\ \text{mol}.$$
• Water vapour: $$n_{\mathrm{H_2O}} = 4 \times n_{\mathrm{C_3H_8}} = 4 \times 2.2727 = 9.0909\ \text{mol}.$$

The oxygen actually consumed equals $$11.3636\ \text{mol},$$ so the oxygen remaining after reaction is

$$n_{\mathrm{O_2,\ left}} = 31.25 - 11.3636 = 19.8864\ \text{mol}.$$

The gaseous mixture after completion therefore contains:

$$\begin{aligned} n_{\mathrm{CO_2}} &= 6.8182\ \text{mol},\\ n_{\mathrm{H_2O}} &= 9.0909\ \text{mol},\\ n_{\mathrm{O_2}} &= 19.8864\ \text{mol}. \end{aligned}$$

The total number of moles in the mixture is obtained by simple addition:

$$n_{\text{total}} = 6.8182 + 9.0909 + 19.8864 = 35.7955\ \text{mol}.$$

We now recall the definition of mole fraction:

$$\chi_i = \dfrac{n_i}{n_{\text{total}}}.$$

Applying this to carbon dioxide,

$$\chi_{\mathrm{CO_2}} = \dfrac{6.8182}{35.7955} = 0.1904.$$

The question states that this value is written as $$x \times 10^{-2},$$ so

$$x = 0.1904 \times 10^{2} = 19.04.$$

Rounding to the nearest integer gives $$x = 19.$$

So, the answer is $$19$$.

We are given the gas-phase reaction $$2\,SO_2(g)+O_2(g)\rightarrow 2\,SO_3(g)$$ which proceeds to completion in a rigid vessel at constant temperature. Because temperature and volume do not change, the partial pressure of each gas is directly proportional to the number of moles present. Hence, we can treat the given partial pressures exactly like the initial “mole” amounts for stoichiometric calculations.

Initially we have

$$P_{SO_2}^{\,\text{initial}} = 250 \text{ mbar},\quad P_{O_2}^{\,\text{initial}} = 750 \text{ mbar},\quad P_{SO_3}^{\,\text{initial}} = 0 \text{ mbar}.$$

According to the balanced equation, the stoichiometric ratio is $$2\;:\;1\;:\;2$$ for $$SO_2 : O_2 : SO_3$$. That is, $$2$$ units of $$SO_2$$ react with $$1$$ unit of $$O_2$$ to form $$2$$ units of $$SO_3$$.

To identify the limiting reactant, we compare the available amounts in the required ratio. The reaction needs half as much $$O_2$$ as $$SO_2$$. Specifically, the amount of $$O_2$$ required for the given $$SO_2$$ is

$$P_{O_2}^{\,\text{needed}} \;=\;\frac{1}{2}\,P_{SO_2}^{\,\text{initial}} \;=\;\frac{1}{2}\,(250\,\text{ mbar}) \;=\;125\,\text{ mbar}.$$

Because the vessel actually contains $$750\,\text{ mbar}$$ of $$O_2$$, which is much larger than the $$125\,\text{ mbar}$$ required, $$SO_2$$ is the limiting reactant. Therefore the reaction will consume all of the initial $$SO_2$$.

Now we calculate how much $$O_2$$ is consumed. From the stoichiometry $$2\,SO_2 \longrightarrow 1\,O_2,$$ every $$2$$ units of $$SO_2$$ use up $$1$$ unit of $$O_2$$, so

$$P_{O_2}^{\,\text{consumed}} \;=\;\frac{1}{2}\,P_{SO_2}^{\,\text{consumed}} \;=\;\frac{1}{2}\,(250\,\text{ mbar}) \;=\;125\,\text{ mbar}.$$

The remaining $$O_2$$ pressure after completion is therefore

$$P_{O_2}^{\,\text{final}} \;=\;P_{O_2}^{\,\text{initial}} \;-\; P_{O_2}^{\,\text{consumed}} \;=\;750\,\text{ mbar}\;-\;125\,\text{ mbar} \;=\;625\,\text{ mbar}.$$

Next, we determine the pressure of $$SO_3$$ formed. The stoichiometry gives $$2\,SO_2 \longrightarrow 2\,SO_3,$$ meaning the same number of units of $$SO_3$$ are produced as the number of $$SO_2$$ units consumed. Hence,

$$P_{SO_3}^{\,\text{formed}} \;=\;P_{SO_2}^{\,\text{consumed}} \;=\;250\,\text{ mbar}.$$

Because no $$SO_3$$ was present initially, the final $$SO_3$$ pressure is simply this amount:

$$P_{SO_3}^{\,\text{final}} = 250\,\text{ mbar}.$$

Finally, we add the partial pressures of all gases remaining in the vessel to obtain the total final pressure:

$$P_{\text{total}}^{\,\text{final}} \;=\;P_{SO_2}^{\,\text{final}} + P_{O_2}^{\,\text{final}} + P_{SO_3}^{\,\text{final}}.$$

The entire initial $$SO_2$$ has reacted, so $$P_{SO_2}^{\,\text{final}} = 0$$. Substituting the values, we get

$$P_{\text{total}}^{\,\text{final}} \;=\;0 + 625\,\text{ mbar} + 250\,\text{ mbar} = 875\,\text{ mbar}.$$

Rounding to the nearest integer is unnecessary because the result is already an integer. Hence, the correct answer is Option A: $$875\,\text{mbar}$$.

The tyre contains nitrogen gas at a fixed volume, so we can apply Gay-Lussac's Law, which states that at constant volume, $$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$.

The initial conditions are $$P_1 = 35$$ psi and $$T_1 = 27°C = 27 + 273 = 300$$ K. The tyre bursts when $$P_2 = 40$$ psi, and we need to find $$T_2$$.

Substituting into the equation: $$\frac{35}{300} = \frac{40}{T_2}$$

Solving for $$T_2$$: $$T_2 = \frac{40 \times 300}{35} = \frac{12000}{35} = 342.86 \text{ K}$$

Converting to Celsius: $$T_2 = 342.86 - 273 = 69.86°C \approx 70°C$$

The temperature at which the car tyre will burst is $$\mathbf{70}$$°C.

We are given that complete combustion of 750 g of an organic compound produces 420 g of $$CO_2$$ and 210 g of $$H_2O$$. We need to find the percentage of hydrogen in the compound.

First, let us find the mass of carbon. The molar mass of $$CO_2$$ is 44 g/mol, and each mole contains 12 g of carbon. So the mass of carbon in 420 g of $$CO_2$$ is $$\frac{12}{44} \times 420 = \frac{5040}{44} = 114.545$$ g.

The percentage of carbon is $$\frac{114.545}{750} \times 100 = 15.27\%$$, which rounds to approximately 15.3%, consistent with the given information.

Now, let us find the mass of hydrogen. The molar mass of $$H_2O$$ is 18 g/mol, and each mole contains 2 g of hydrogen. So the mass of hydrogen in 210 g of $$H_2O$$ is $$\frac{2}{18} \times 210 = \frac{420}{18} = 23.33$$ g.

The percentage of hydrogen is $$\frac{23.33}{750} \times 100 = 3.11\%$$.

Rounding off to the nearest integer, the percentage of hydrogen is $$\mathbf{3}$$.

To determine the percentage yield, first calculate the molar masses of the reactant and product.

For benzoic acid:

$$\mathrm{C_7H_6O_2}$$

$$\mathrm{Molar\ Mass = (7\times12) + (6\times1) + (2\times16)}$$

$$\mathrm{= 84 + 6 + 32 = 122\ g\ mol^{-1}}$$

For m-bromobenzoic acid:

$$\mathrm{C_7H_5BrO_2}$$

$$\mathrm{Molar\ Mass = (7\times12) + (5\times1) + 80 + (2\times16)}$$

$$\mathrm{= 84 + 5 + 80 + 32 = 201\ g\ mol^{-1}}$$

The balanced reaction shows a $$\mathrm{1:1}$$ molar ratio between benzoic acid and m-bromobenzoic acid.

Moles of benzoic acid used:

$$\mathrm{Moles = \frac{6.1}{122}}$$

$$\mathrm{= 0.05\ mol}$$

Therefore,

$$\mathrm{0.05\ mol\ Benzoic\ Acid \rightarrow 0.05\ mol\ Product}$$

Theoretical yield of product:

$$\mathrm{Mass = Moles \times Molar\ Mass}$$

$$\mathrm{= 0.05 \times 201}$$

$$\mathrm{= 10.05\ g}$$

Actual yield obtained:

$$\mathrm{7.8\ g}$$

Percentage yield:

$$\mathrm{\%\ Yield = \frac{Actual\ Yield}{Theoretical\ Yield}\times100}$$

$$\mathrm{= \frac{7.8}{10.05}\times100}$$

$$\mathrm{= 77.61\%}$$

Rounding to the nearest integer:

$$\boxed{\mathrm{78}}$$

For the complete combustion of butane, we first recall and write the balanced chemical equation:

$$2\,\mathrm{C_4H_{10}} \;+\; 13\,\mathrm{O_2}\;\longrightarrow\; 8\,\mathrm{CO_2}\;+\; 10\,\mathrm{H_2O}$$

From this equation we see that:

$$2 \text{ mol } \mathrm{C_4H_{10}} \;\xrightarrow{\text{combustion}}\; 10 \text{ mol } \mathrm{H_2O}$$

Now, we are told that the actual amount of water obtained is $$72.0\ \text{g}$$. Let us convert this mass of water into moles. The molar mass of water $$\mathrm{H_2O}$$ is

$$M(\mathrm{H_2O}) = (2 \times 1) + 16 = 18\ \text{g mol}^{-1}.$$

Therefore, the number of moles of water produced is

$$n(\mathrm{H_2O}) \;=\;\frac{72.0\ \text{g}}{18\ \text{g mol}^{-1}} \;=\; 4.0\ \text{mol}.$$

According to the stoichiometric ratio obtained from the balanced equation, $$2$$ moles of butane give $$10$$ moles of water. Hence, to find the moles of butane required for $$4.0$$ moles of water we write the proportion

$$\frac{2\ \text{mol }\mathrm{C_4H_{10}}}{10\ \text{mol }\mathrm{H_2O}} \;=\;\frac{x\ \text{mol }\mathrm{C_4H_{10}}}{4.0\ \text{mol }\mathrm{H_2O}}.$$

Cross-multiplying gives

$$10\,x = 2 \times 4.0,$$

so

$$x = \frac{2 \times 4.0}{10} = 0.8\ \text{mol}.$$

Now we convert the required moles of butane to mass. The molar mass of butane ($$\mathrm{C_4H_{10}}$$) is

$$M(\mathrm{C_4H_{10}}) = (4 \times 12) + (10 \times 1) = 48 + 10 = 58\ \text{g mol}^{-1}.$$

Hence, the mass of butane consumed is

$$m(\mathrm{C_4H_{10}}) = n \times M = 0.8\ \text{mol} \times 58\ \text{g mol}^{-1} = 46.4\ \text{g}.$$

The question asks us to express this answer in the form $$\_\_\_\_\_\times 10^{-1}\ \text{g}$$ and then give the blank as the nearest integer. We rewrite

$$46.4\ \text{g} = 464 \times 10^{-1}\ \text{g}.$$

So the integer that fills the blank is $$464$$.

Hence, the correct answer is Option 464.

Let the gaseous hydrocarbon be $$C_xH_y$$. The balanced combustion reaction is:

$$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

Since all volumes are measured under the same conditions of temperature and pressure, by Avogadro's law, the ratio of volumes equals the ratio of moles for gaseous species.

From the given information, 1 volume of $$C_xH_y$$ requires 6 volumes of $$O_2$$. So:

$$x + \frac{y}{4} = 6 \quad \cdots (i)$$

Also, 1 volume of $$C_xH_y$$ produces 4 volumes of $$CO_2$$. So:

$$x = 4 \quad \cdots (ii)$$

Substituting $$x = 4$$ in equation (i):

$$4 + \frac{y}{4} = 6$$

$$\frac{y}{4} = 2$$

$$y = 8$$

The hydrocarbon is $$C_4H_8$$ and the value of $$y$$ is $$\textbf{8}$$.

We know that the number of atoms in any given sample can be found with the basic mole concept formula:

$$\text{Number of atoms} \;=\; \text{Number of moles}\;\times\;N_A$$

First, we must evaluate the number of moles present in the 8 g sample of sodium. We use the definition of a mole, which states:

$$\text{Number of moles} \;=\; \frac{\text{Given mass (in g)}}{\text{Molar mass (in g mol}^{-1})}$$

For sodium (Na) the given data are:

Given mass = $$8 \text{ g}$$,   Molar (atomic) mass = $$23.0 \text{ g mol}^{-1}$$.

Substituting these values we get:

$$\text{Number of moles of Na} \;=\; \frac{8}{23.0}$$

Now we multiply this mole value by Avogadro’s constant to obtain the number of atoms:

$$\text{Number of atoms} \;=\; \left(\frac{8}{23.0}\right)\times (6.02\times10^{23})$$

Carrying out the multiplication in the numerator first:

$$8 \times 6.02 = 48.16$$

So we have:

$$\text{Number of atoms} \;=\; \frac{48.16}{23.0}\times10^{23}$$

Next we perform the division $$48.16 \div 23.0$$:

$$\frac{48.16}{23.0} \approx 2.093$$

Thus,

$$\text{Number of atoms} \;\approx\; 2.093\times10^{23}$$

The question states that this number of atoms can be written as $$x\times10^{23}$$, so by comparison we have $$x \approx 2.093$$. Rounding to the nearest integer gives:

$$x = 2$$

So, the answer is $$2$$.

We have to find the number of significant figures present in the decimal number $$0.00340$$. To do this, let us first recall the rules that govern significant figures in any measured quantity.

Rule 1: All non-zero digits $$\,(1,2,3,4,5,6,7,8,9)\,$$ are always significant.

Rule 2: Any zeros lying between two non-zero digits are significant. This is sometimes called the “captive zero” rule.

Rule 3: Leading zeros, that is, zeros written only to set the position of the decimal point, are not significant.

Rule 4: Trailing zeros in a number containing a decimal point are significant because they show the measured precision.

Now we apply these rules one by one to the given number $$0.00340$$.

First, let us write out the individual digits, underlining them one by one to decide whether each is significant or not:

$$0.\;0\;0\;3\;4\;0$$

In words, the figure reads: “zero point zero zero three four zero.” We label the positions so that we can reason clearly:

$$0.(\text{decimal})\underbrace{00}_{\text{leading zeros}}3\,4\,0$$

Step by step, we apply each rule:

1. The first two zeros after the decimal point are leading zeros. According to Rule 3, such zeros merely fix the position of the decimal point; hence they are not counted as significant.

2. Next comes the digit $$3$$. Since $$3$$ is a non-zero digit, Rule 1 tells us it is always significant. Thus we have our first significant figure.

3. Immediately after $$3$$ we have the digit $$4$$. Again, $$4$$ is non-zero, so by Rule 1 it is significant. Now we have two significant figures counted so far.

4. Finally, we find a zero written after the digit $$4$$, and importantly, the number has a decimal point. This zero is a trailing zero in a decimal. Rule 4 tells us that such zeros are significant because they indicate the precision up to that decimal place. Hence this zero is also counted.

No more digits remain. Summarising, the digits that are significant are:

$$\color{blue}{3},\quad\color{blue}{4},\quad\color{blue}{0}$$

That gives a total count of

$$\text{Number of significant figures}=3$$

So, the answer is $$3$$.

We are given 35 mL of 0.15 M lead nitrate ($$Pb(NO_3)_2$$) and 20 mL of 0.12 M chromic sulphate ($$Cr_2(SO_4)_3$$). The precipitation reaction is: $$Pb^{2+} + SO_4^{2-} \rightarrow PbSO_4 \downarrow$$.

Moles of $$Pb^{2+}$$ ions = $$0.035 \times 0.15 = 0.00525$$ mol.

Since each formula unit of $$Cr_2(SO_4)_3$$ provides 3 sulphate ions, the moles of $$SO_4^{2-}$$ ions = $$0.020 \times 0.12 \times 3 = 0.0072$$ mol.

The reaction between $$Pb^{2+}$$ and $$SO_4^{2-}$$ is in a 1:1 molar ratio. Since 0.00525 mol of $$Pb^{2+}$$ is less than 0.0072 mol of $$SO_4^{2-}$$, lead is the limiting reagent.

Therefore, moles of $$PbSO_4$$ precipitated = $$0.00525$$ mol = $$525 \times 10^{-5}$$ mol.

The answer is $$525$$.

In an equimolar mixture, both NaOH and Na$$_2$$CO$$_3$$ are present in equal numbers of moles. Let the number of moles of each compound be $$n$$. The total mass of the mixture is 4 g:

$$n \times M_{\text{NaOH}} + n \times M_{\text{Na}_2\text{CO}_3} = 4 \text{ g}$$

The molar masses are: NaOH = 23 + 16 + 1 = 40 g mol$$^{-1}$$, and Na$$_2$$CO$$_3$$ = 2(23) + 12 + 3(16) = 106 g mol$$^{-1}$$.

$$n(40 + 106) = 4$$

$$n = \frac{4}{146} = 0.02740 \text{ mol}$$

The mass of NaOH in the mixture is:

$$x = n \times 40 = \frac{4}{146} \times 40 = \frac{160}{146} \approx 1.096 \text{ g}$$

Rounding to the nearest integer, the value of $$x$$ is $$\mathbf{1}$$ g.

The volume of methane consumed in one year is given as $$4.00 \times 10^3\ \text{m}^3$$. We recall that $$1\ \text{m}^3 = 1000\ \text{L}$$, so

$$V = 4.00 \times 10^3\ \text{m}^3 \times 1000\ \frac{\text{L}}{\text{m}^3} = 4.00 \times 10^6\ \text{L}.$$

For an ideal gas the relation $$PV = nRT$$ holds, where $$P$$ is the pressure, $$V$$ is the volume, $$T$$ is the absolute temperature, $$R$$ is the universal gas constant and $$n$$ is the number of moles.

We are told that $$P = 1.0\ \text{atm}$$ and $$T = 300\ \text{K}$$, with $$R = 0.083\ \text{L atm K}^{-1}\text{mol}^{-1}$$. Substituting these values we get

$$n = \frac{PV}{RT} = \frac{(1.0\ \text{atm})(4.00 \times 10^6\ \text{L})}{(0.083\ \text{L atm K}^{-1}\text{mol}^{-1})(300\ \text{K})}.$$

First we multiply the denominator: $$0.083 \times 300 = 24.9.$$

Now we perform the division:

$$n = \frac{4.00 \times 10^6}{24.9}\ \text{mol} \approx 1.606 \times 10^5\ \text{mol}.$$

The molar mass of methane CH$$_4$$ is obtained by adding atomic masses, $$12\ (\text{for C}) + 4 \times 1\ (\text{for H}) = 16\ \text{g mol}^{-1}.$$

Hence the mass of gas used is

$$m = n \times M = (1.606 \times 10^5\ \text{mol})(16\ \text{g mol}^{-1}) = 2.5696 \times 10^6\ \text{g}.$$

We rewrite this mass in the required form $$x \times 10^5\ \text{g}$$:

$$2.5696 \times 10^6\ \text{g} = 25.696 \times 10^5\ \text{g}.$$

Taking the nearest integer, $$x \approx 26.$$

Hence, the correct answer is Option 26.

We have an LPG cylinder that contains gas initially at a pressure of $$P_1 = 300 \,\text{kPa} = 3 \times 10^5 \,\text{Pa}$$ and an initial temperature of $$t_1 = 27^\circ\text{C}$$. To work with gas-law equations we must always convert Celsius to Kelvin, because the Kelvin scale starts from absolute zero. The conversion formula is

$$T (\text{K}) = t (^\circ\text{C}) + 273.$$

Applying this conversion, the initial absolute temperature is

$$T_1 = 27 + 273 = 300 \,\text{K}.$$

The cylinder can safely withstand a maximum pressure of $$P_2 = 1.2 \times 10^6 \,\text{Pa}.$$ We are asked to find the temperature $$t_2$$ (in °C) at which the cylinder will reach this pressure and burst.

An LPG cylinder is a rigid container; therefore its volume remains constant during heating. For a fixed amount of an ideal gas kept at constant volume, pressure is directly proportional to absolute temperature. This relation is expressed by Gay-Lussac’s (or the Isochoric) gas law:

$$\frac{P}{T} = \text{constant} \qquad\Longrightarrow\qquad \frac{P_1}{T_1} = \frac{P_2}{T_2}.$$

Rearranging this formula to solve for the unknown absolute temperature $$T_2$$, we write

$$T_2 = \frac{P_2 \, T_1}{P_1}.$$

Now we substitute each known value carefully:

$$T_2 = \frac{1.2 \times 10^6 \,\text{Pa} \times 300 \,\text{K}}{3 \times 10^5 \,\text{Pa}}.$$

First, notice that the pascal (Pa) units cancel out. Next, perform the arithmetic with the numbers and powers of ten:

$$T_2 = \frac{1.2 \times 10^6}{3 \times 10^5} \times 300 \,\text{K}.$$

The fraction of the coefficients is $$\frac{1.2}{3} = 0.4$$, and the fraction of the powers of ten is $$\frac{10^6}{10^5} = 10^{6-5} = 10^1 = 10$$. Therefore

$$T_2 = 0.4 \times 10 \times 300 \,\text{K}.$$

Multiplying step by step, $$0.4 \times 10 = 4$$, and then $$4 \times 300 = 1200.$$ Hence

$$T_2 = 1200 \,\text{K}.$$

We now convert this absolute temperature back to the Celsius scale by reversing the earlier formula:

$$t_2 (^\circ\text{C}) = T_2 (\text{K}) - 273.$$

Substituting $$T_2 = 1200 \,\text{K}$$, we get

$$t_2 = 1200 - 273 = 927^\circ\text{C}.$$

This temperature is the minimum room temperature at which the cylinder pressure will reach $$1.2 \times 10^6 \,\text{Pa}$$ and consequently burst.

So, the answer is $$927^\circ\text{C}.$$

First, we find the mean burette reading from the five readings: $$(4.5 + 4.5 + 4.4 + 4.4 + 4.4)/5 = 22.2/5 = 4.44$$ mL. This is the mean volume of NaOH used.

Oxalic acid ($$H_2C_2O_4$$) is a diprotic acid, so the reaction with NaOH is: $$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$$.

The millimoles of oxalic acid taken are: $$10.0 \times 1.25 = 12.5$$ mmol.

From the stoichiometry, the millimoles of NaOH required are: $$2 \times 12.5 = 25.0$$ mmol.

The molarity of NaOH is: $$M = \frac{25.0}{4.44} = 5.63$$ M.

Rounding to the nearest integer, the molarity of the NaOH solution is $$\boxed{6}$$ M.

We use the ideal gas equation $$PV = nRT$$ to find the volume of acetylene gas.

The molar mass of acetylene ($$C_2H_2$$) is $$2(12) + 2(1) = 26$$ g/mol.

Number of moles: $$n = \frac{4.75}{26} = 0.1827$$ mol.

Temperature: $$T = 50 + 273 = 323$$ K.

Pressure: $$P = \frac{740}{760}$$ atm $$= 0.9737$$ atm.

Applying the ideal gas law:

$$V = \frac{nRT}{P} = \frac{0.1827 \times 0.0826 \times 323}{0.9737}$$

$$V = \frac{0.1827 \times 26.68}{0.9737} = \frac{4.874}{0.9737} = 5.006 \text{ L}$$

Rounding off to the nearest integer, the volume is $$\textbf{5}$$ L.

We are given that the empty (tare) mass of the cylinder is $$14.8\ \text{kg}.$$

When the cylinder is completely filled with LPG its total mass becomes $$29.0\ \text{kg}.$$

Hence the initial mass of LPG present is obtained by simple subtraction:

$$m_1 \;=\; 29.0\ \text{kg} - 14.8\ \text{kg} = 14.2\ \text{kg}.$$

At this stage the gauge shows an initial pressure

$$P_1 = 3.47\ \text{atm}.$$

During usage the mass of the cylinder (shell + remaining LPG) falls to $$23.0\ \text{kg}.$$

Therefore the mass of LPG that still remains is

$$m_2 \;=\; 23.0\ \text{kg} - 14.8\ \text{kg} = 8.2\ \text{kg}.$$

We treat LPG as an ideal gas. The ideal-gas equation is first stated:

$$PV = nRT,$$

where  $$n = \dfrac{m}{M}$$  is the number of moles, $$m$$ is the mass of gas, and $$M$$ its molar mass. For a rigid steel cylinder, the volume $$V$$ is constant. The problem says the ambient temperature is unchanged, so $$T$$ is also constant. Hence, for two different states of the same gas in the same cylinder, the ratio

$$\dfrac{P}{n} = \dfrac{RT}{V}$$

remains fixed. Substituting $$n = m/M$$, we get

$$P = \dfrac{RT}{VM}\,m.$$

Everything in the prefactor $$\dfrac{RT}{VM}$$ is constant, so pressure is directly proportional to mass when temperature and volume stay fixed:

$$\dfrac{P_1}{P_2} = \dfrac{m_1}{m_2}.$$

We now substitute the numerical values:

$$P_2 = P_1 \times \dfrac{m_2}{m_1} = 3.47\ \text{atm} \times \dfrac{8.2\ \text{kg}}{14.2\ \text{kg}}.$$

First compute the mass ratio:

$$\dfrac{8.2}{14.2} = 0.57746\;(\text{approximately}).$$

Then multiply by the initial pressure:

$$P_2 = 3.47 \times 0.57746 = 2.0038\ \text{atm}\;(\text{approximately}).$$

Rounding to the nearest whole number, we have

$$P_2 \approx 2\ \text{atm}.$$

So, the answer is $$2$$.

First, calculate the moles of each gas. The molar mass of methane ($$CH_4$$) is $$12 + 4 \times 1 = 16$$ g/mol, so moles of methane = $$\frac{6.4}{16} = 0.4$$ mol. The molar mass of carbon dioxide ($$CO_2$$) is $$12 + 2 \times 16 = 44$$ g/mol, so moles of carbon dioxide = $$\frac{8.8}{44} = 0.2$$ mol.

The total moles of gas in the mixture is $$n_{total} = 0.4 + 0.2 = 0.6$$ mol. The temperature in Kelvin is $$27 + 273 = 300$$ K.

Using the ideal gas law $$PV = nRT$$, the total pressure is $$P = \frac{nRT}{V} = \frac{0.6 \times 8.314 \times 300}{10} = \frac{1496.52}{10} = 149.652$$ kPa. Rounding to the nearest integer gives $$P \approx 150$$ kPa.

In titrimetric analysis, we first identify and discard any outlier readings. The five readings are 4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL, and 5.0 mL. By convention, the concordant readings (those that agree closely) are considered reliable. The three identical values of 5.0 mL are concordant, and the accepted titre value is taken as 5.0 mL.

Using titre = 5.0 mL of 0.2 M HCl for 10.0 mL of $$\text{Na}_2\text{CO}_3$$ solution, the reaction is $$\text{Na}_2\text{CO}_3 + 2\text{HCl} \to 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$.

Millimoles of HCl used $$= 0.2 \times 5.0 = 1.0$$ mmol. Since 2 moles of HCl react with 1 mole of $$\text{Na}_2\text{CO}_3$$, millimoles of $$\text{Na}_2\text{CO}_3 = \frac{1.0}{2} = 0.5$$ mmol. This 0.5 mmol is present in 10.0 mL of solution, so the concentration is $$\frac{0.5}{10.0} = 0.05$$ M $$= 50$$ mM.

The answer is $$50$$ mM.

First, find the moles of NaOH and HCl present:

Moles of NaOH = $$0.5 \text{ M} \times 0.250 \text{ L} = 0.125 \text{ mol}$$

Moles of HCl = $$1.0 \text{ M} \times 0.500 \text{ L} = 0.500 \text{ mol}$$

The neutralization reaction is: NaOH + HCl → NaCl + H₂O. NaOH is the limiting reagent, so 0.125 mol of HCl is consumed.

Moles of unreacted HCl = $$0.500 - 0.125 = 0.375 \text{ mol}$$

Number of HCl molecules = $$0.375 \times 6.022 \times 10^{23} = 2.258 \times 10^{23}$$

Expressing as $$p \times 10^{21}$$: $$p = \frac{2.258 \times 10^{23}}{10^{21}} = 225.8 \approx 226$$

Therefore, $$p = \boxed{226}$$.

First, let us recall Dalton’s atomic theory (1808). Dalton proposed that:

$$1.$$ Matter is made up of tiny, indivisible particles called atoms.
$$2.$$ All atoms of a given element are identical in mass and all other properties, while atoms of different elements have different masses and properties.
$$3.$$ Atoms are indestructible and conserve their identity in every ordinary chemical change; during a chemical reaction atoms are merely rearranged.
$$4.$$ Compounds are formed when atoms of different elements combine in simple whole-number ratios.

Now we compare each option with the above list.

Option A states that “chemical reactions involve reorganization of atoms which are neither created nor destroyed.”
This is exactly Dalton’s third postulate. So Option A was proposed by Dalton.

Option B says “all the atoms of a given element have identical properties including identical mass, whereas atoms of different elements differ in mass.”
This matches Dalton’s second postulate. Hence Option B was also proposed by Dalton.

Option C states “when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure.”
This statement is known as Gay-Lussac’s law of combining volumes, formulated in 1808-1809 after Dalton had already announced his atomic theory. Dalton did not include any reference to volumes of gases in his postulates. Therefore Option C was not proposed by Dalton.

Option D says “matter consists of indivisible atoms.”
This is identical to Dalton’s first postulate, so Option D was indeed proposed by Dalton.

Comparing all the options, only Option C is outside the scope of Dalton’s original atomic theory.

Hence, the correct answer is Option C.

First we recall the balanced chemical reaction between urea and aqueous sodium hydroxide, which furnishes ammonia as one of the products:

$$$\mathrm{NH_2CONH_2 + 2\,NaOH \;\longrightarrow\; Na_2CO_3 + 2\,NH_3}$$$

This equation tells us that

$$$1\;\text{mol urea} \;\xrightarrow{\phantom{2\,NaOH}}\; 2\;\text{mol NH}_3$$$

Now we calculate how many moles of urea are present in the given 0.6 g sample. The molar mass of urea is obtained by adding the individual atomic masses:

$$$M(\mathrm{NH_2CONH_2}) = 2\times14 \;(\text N) + 4\times1 \;(\text H) + 12 \;(\text C) + 16 \;(\text O) = 28 + 4 + 12 + 16 = 60\;\text{g mol}^{-1}$$$

Hence,

$$$n(\text{urea}) = \frac{m}{M} = \frac{0.6\;\text g}{60\;\text{g mol}^{-1}} = 0.01\;\text{mol}$$$

Using the stoichiometric ratio obtained earlier, the moles of ammonia liberated are

$$$n(\mathrm{NH_3}) = 2 \times n(\text{urea}) = 2 \times 0.01 = 0.02\;\text{mol}$$$

Next, ammonia is a base that reacts with hydrochloric acid according to the simple 1 : 1 neutralisation equation,

$$\mathrm{NH_3 + HCl \;\longrightarrow\; NH_4Cl}$$

Therefore, the number of moles (or equivalents, because HCl is monoprotic) of HCl needed equals the moles of NH$$_3$$ present:

$$n(\mathrm{HCl\;required}) = 0.02\;\text{mol}$$

Let us now examine each option by converting its volume and normality into moles of HCl supplied, using the relation $$n = N \times V$$ with $$V$$ in litres.

For Option A: $$$n = 0.4\;\text N \times 0.200\;\text L = 0.080\;\text{mol}$$$

For Option B: $$$n = 0.2\;\text N \times 0.200\;\text L = 0.040\;\text{mol}$$$

For Option C: $$$n = 0.2\;\text N \times 0.100\;\text L = 0.020\;\text{mol}$$$

For Option D: $$$n = 0.1\;\text N \times 0.100\;\text L = 0.010\;\text{mol}$$$

The requirement is 0.02 mol, and only Option C furnishes exactly this amount of acid.

Hence, the correct answer is Option C.

The ideal gas equation can be rewritten to express density by combining the basic gas law with molecular weight properties:

$$PV = nRT = \frac{m}{M}RT$$

Rearranging the equation to solve for density ($$d = \frac{m}{V}$$) gives:

$$P = \frac{m}{V}\frac{RT}{M} \implies P = \frac{dRT}{M}$$

Isolating the density ($$d$$), we get our working formula:

$$d = \left(\frac{M}{R}\right)\frac{P}{T}$$

Step-by-Step Analysis of Each Graph:

• Graph I: $$d$$ vs $$T$$ (at constant $$P$$)

  • From the formula, keeping $$P$$ constant means density is inversely proportional to temperature: $$d \propto \frac{1}{T}$$.
  • The graph of an inverse relationship is a rectangular hyperbola.
  • Status: Graph I is correct.



• Graph II: $$d$$ vs $$T$$ (at constant $$P$$)

  • As established above, $$d$$ and $$T$$ share an inverse relationship ($$d \propto \frac{1}{T}$$). As temperature increases, gas volume expands, causing the density to decrease.
  • Graph II depicts a linear increase with a positive slope ($$d \propto T$$), which is physically impossible for an ideal gas.
  • Status: Graph II is incorrect.



• Graph III: $$d$$ vs $$\frac{1}{T}$$ (at constant $$P$$)

  • Since $$d \propto \frac{1}{T}$$, plotting $$d$$ directly against the inverse of temperature ($$\frac{1}{T}$$) yields a linear relationship ($$y = mx$$).
  • The graph should be a straight line passing through the origin.
  • Status: Graph III is correct.



• Graph IV: $$d$$ vs $$P$$ (at constant $$T$$)

  • Keeping $$T$$ constant means density is directly proportional to pressure: $$d \propto P$$.
  • The graph should be a straight line passing through the origin with a positive slope.
  • Status: Graph IV is correct.

Conclusion:

Because density is inversely related to temperature at a constant pressure, Graph II is geometrically and scientifically flawed.

Answer: Option B — II

First we note the atomic (molar) mass of zinc:

$$M_{\text{Zn}} = 65 \ \text{g mol}^{-1}$$

The mass of zinc given is

$$m_{\text{Zn}} = 5 \ \text{g}$$

Hence the number of moles of zinc present is obtained from the relation

$$n = \dfrac{m}{M}$$

Substituting the values, we get

$$n_{\text{Zn}} = \dfrac{5 \ \text{g}}{65 \ \text{g mol}^{-1}} = \dfrac{5}{65} \ \text{mol} = \dfrac{1}{13} \ \text{mol}$$

Now we consider the two separate reactions.

Reaction (a): Zinc with dilute hydrochloric acid

The balanced chemical equation is

$$\text{Zn} + 2\,\text{HCl} \longrightarrow \text{ZnCl}_2 + \text{H}_2$$

From the equation we see that

1 mol Zn → 1 mol H2

Therefore, the moles of hydrogen evolved in this case will be

$$n_{\text{H}_2,\; \text{from HCl}} = n_{\text{Zn}} = \dfrac{1}{13} \ \text{mol}$$

Reaction (b): Zinc with aqueous sodium hydroxide

The balanced chemical equation is

$$\text{Zn} + 2\,\text{NaOH} \longrightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2$$

(The product Na2ZnO2 is called sodium zincate.)

Again, we observe from the equation that

1 mol Zn → 1 mol H2

Thus the moles of hydrogen evolved here are also

$$n_{\text{H}_2,\; \text{from NaOH}} = n_{\text{Zn}} = \dfrac{1}{13} \ \text{mol}$$

Since both reactions produce the same number of moles of hydrogen gas, and equal moles of any gas occupy equal volumes under identical conditions (Avogadro’s law), the volumes of hydrogen obtained from the two experiments are equal.

Therefore,

$$V_{\text{H}_2\;(\text{from HCl})} : V_{\text{H}_2\;(\text{from NaOH})} = 1 : 1$$

Hence, the correct answer is Option B.

• Primary vs. Secondary Standards: Sodium hydroxide ($$\text{NaOH}$$) is hygroscopic and absorbs carbon dioxide ($$\text{CO}_2$$) from the air, making it a secondary standard. Its concentration cannot be calculated directly by weighing. It must be standardized against a stable primary standard like oxalic acid ($$\text{H}_2\text{C}_2\text{O}_4$$).
• Concentrated Acids: Concentrated $$\text{H}_2\text{SO}_4$$ is not a primary standard and requires dilution and prior standardization, making it unsuitable for an accurate direct titration reference.
• Apparatus Configuration: In standard laboratory practice, the highly alkaline, caustic sodium hydroxide solution is placed in the burette. This prevents the solution from sticking to or etching the glass of a volumetric pipette and allows the accurately measured primary standard (oxalic acid) to remain cleanly inside the conical flask.

Conclusion:

The concentration of the aqueous $$\text{NaOH}$$ solution is most accurately verified when it is placed in the burette and titrated against a known concentration of aqueous oxalic acid measured out into a conical flask.

Answer: Option B — Aq. NaOH in a burette and aqueous oxalic acid in a conical flask

We first recall what happens in the zeolite, also called the permutit, process. Natural or artificial zeolite is represented ideally by the formula $$\text{Na}_{2}\text{Z}$$ where $$\text{Z}^{2-}$$ is the aluminosilicate framework. When hard water containing $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ ions passes through the bed, the following cation‐exchange takes place:

$$\text{Na}_{2}\text{Z}\;+\;\text{Ca}^{2+}\;\longrightarrow\;\text{CaZ}\;+\;2\text{Na}^{+}$$

$$\text{Na}_{2}\text{Z}\;+\;\text{Mg}^{2+}\;\longrightarrow\;\text{MgZ}\;+\;2\text{Na}^{+}$$

So, in the zeolite method only cations are exchanged; the accompanying anions such as $$\text{HCO}_{3}^{-}$$, $$\text{SO}_{4}^{2-}$$, $$\text{Cl}^{-}$$ remain in solution. After the exchange, although $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ are removed, the water still contains the anions, now accompanied by harmless $$\text{Na}^{+}$$ ions. Thus the zeolite process softens water, but it does not produce completely de-ionised water.

Now we look at the synthetic resin, or ion-exchange, method. In this technique two different resins are employed consecutively:

(i) A cation-exchange resin, generally denoted $$\text{R}\!-\!\text{SO}_{3}\text{H}$$, reacts as

$$\text{R}\!-\!\text{SO}_{3}\text{H}\;+\;\text{M}^{n+}\;\longrightarrow\;\bigl(\text{R}\!-\!\text{SO}_{3}\bigr)_{n}\text{M}\;+\;n\text{H}^{+}$$

where $$\text{M}^{n+}$$ stands for all cations present (not only $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$, but also $$\text{Na}^{+},\; \text{K}^{+},\; \text{Fe}^{3+}$$, etc.).

(ii) Immediately after that, the water passes through an anion-exchange resin, commonly written $$\text{R}\!-\!\text{N}\bigl(\text{CH}_{3}\bigr)_{3}\text{OH}$$, which functions according to

$$\text{R}\!-\!\text{N}\bigl(\text{CH}_{3}\bigr)_{3}\text{OH}\;+\;\text{A}^{-}\;\longrightarrow\;\text{R}\!-\!\text{N}\bigl(\text{CH}_{3}\bigr)_{3}\text{A}\;+\;\text{OH}^{-}$$

with $$\text{A}^{-}$$ representing all anions present ($$\text{Cl}^{-},\;\text{SO}_{4}^{2-},\;\text{HCO}_{3}^{-},\;\dots$$). The $$\text{H}^{+}$$ ions from the first resin and the $$\text{OH}^{-}$$ ions from the second resin combine to give $$\text{H}_{2}\text{O}$$, so the final output is nearly completely de-ionised water.

Because the synthetic resin method eliminates both cations and anions, the degree of purification it achieves is much higher than that of the zeolite process, which eliminates only cations. Therefore we say that, in comparison with the zeolite process, the synthetic resin method is more efficient since it can exchange both kinds of ions.

The statement that exactly conveys this idea among the given choices is:

“more efficient as it can exchange both cations as well as anions.”

Hence, the correct answer is Option B.

Water is said to be hard when it contains appreciable amounts of the ions $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$. These ions can be present either as their bicarbonates or as their chlorides / sulphates.

We classify hardness into two types:

• Temporary hardness: caused mainly by $$\mathrm{Ca(HCO_3)_2}$$ and $$\mathrm{Mg(HCO_3)_2}$$.
• Permanent hardness: caused by $$$\mathrm{CaCl_2,\; CaSO_4,\; MgCl_2,\; MgSO_4}$$$ and similar salts.

Let us recall what each of the given treatments does.

Clark’s method (also called the lime-soda cold-lime process) removes temporary hardness by adding calculated amounts of lime, $$\mathrm{Ca(OH)_2}$$. We first state the reactions used:

$$$\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \;\longrightarrow\; 2\,CaCO_3 \downarrow + 2\,H_2O}$$$

$$$\mathrm{Mg(HCO_3)_2 + 2\,Ca(OH)_2 \;\longrightarrow\; 2\,CaCO_3 \downarrow + Mg(OH)_2 \downarrow + 2\,H_2O}$$$

The insoluble $$\mathrm{CaCO_3}$$ and $$\mathrm{Mg(OH)_2}$$ precipitate and can be filtered off, so the bicarbonate (temporary) hardness is removed. However, chlorides and sulphates of calcium and magnesium do not react with lime in this way, so permanent hardness remains untreated by Clark’s method.

Ion-exchange (zeolite) method works for both kinds of hardness. In the sodium-zeolite bed, the exchange reaction is

$$$\mathrm{Na_2Z + Ca^{2+} \;\longrightarrow\; CaZ + 2\,Na^{+}}$$$

which removes $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$ regardless of their accompanying anions. Hence it is perfectly suitable for removing permanent hardness.

Calgon’s method uses sodium hexametaphosphate, $$\mathrm{(NaPO_3)_6}$$, popularly called Calgon. The anion $$\mathrm{[Na_2P_6O_{18}]^{2-}}$$ complexes $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$:

$$$\mathrm{[P_6O_{18}]^{6-} + Ca^{2+} \;\longrightarrow\; [CaP_6O_{18}]^{4-}}$$$

This sequestration keeps the metal ions in solution but non-interfering, thus removing permanent hardness effectively.

Treatment with sodium carbonate (washing soda) also removes permanent hardness. Stating the precipitation reactions:

$$\mathrm{Ca^{2+} + CO_3^{2-} \;\longrightarrow\; CaCO_3 \downarrow}$$

$$\mathrm{Mg^{2+} + CO_3^{2-} \;\longrightarrow\; MgCO_3 \downarrow}$$

The insoluble carbonates settle out, and therefore chlorides and sulphates of $$\mathrm{Ca^{2+}}$$/$$\mathrm{Mg^{2+}}$$ are eliminated.

We can now compare:

• Clark’s method ⇒ removes only temporary hardness.
• The other three methods ⇒ remove permanent hardness very effectively.

Because the question asks for the method NOT suitable for removing permanent hardness, Clark’s method is the odd one out.

Hence, the correct answer is Option A.

To determine the correct condition that indicates a polluted environment, we evaluate each option based on standard environmental chemistry principles.

Eutrophication is the excessive enrichment of a water body with nutrients such as nitrates and phosphates, usually from fertilizers or sewage. This causes rapid algal growth, and the subsequent decomposition of these algae consumes dissolved oxygen, making the water unsuitable for aquatic life. Hence, eutrophication is a clear indicator of water pollution.

A concentration of $$0.03%$$ of $$CO_2$$ in the atmosphere represents the normal natural level and is not considered an indication of pollution.

A Biochemical Oxygen Demand (BOD) value of $$5\ \text{ppm}$$ is generally associated with clean or only slightly polluted water. Highly polluted water typically has a much higher BOD value.

The normal pH of rainwater is approximately $$5.6$$ due to the dissolution of atmospheric $$CO_2$$ forming weak carbonic acid. Rainwater is considered acid rain only when its pH falls significantly below $$5.6$$.

Therefore, among the given options, eutrophication is the only condition that inherently signifies a polluted environment.

Hence, the correct answer is Option A.

First, we recollect what is meant by the two thermal treatments used in metallurgy:

$$\text{Calcination}$$ is the process in which an ore, generally a carbonate or a hydrated oxide, is heated in limited or no supply of air. The typical chemical change is the decomposition of a metal carbonate into its oxide with liberation of carbon dioxide gas. We state the generic equation:

$$\text{MCO}_3 \; \xrightarrow{\;\text{heat}\;} \; \text{MO} \;+\; \text{CO}_2$$

Now, carbon dioxide $$(\text{CO}_2)$$ is a well-known greenhouse gas. When it accumulates in the atmosphere it traps infrared radiation, producing the greenhouse effect. This effect raises the average temperature of the planet, a phenomenon commonly referred to as global warming. Hence, the chief environmental consequence of calcination is global warming.

Next, $$\text{Roasting}$$ is the process in which a sulphide ore is heated in excess of air so that the sulphide is converted into oxide with the simultaneous evolution of sulphur dioxide gas. The simplified form of the reaction is written as:

$$2\,\text{MS} \;+\; 3\,\text{O}_2 \;\xrightarrow{\;\text{heat}\;}\; 2\,\text{MO} \;+\; 2\,\text{SO}_2$$

Here, $$(\text{SO}_2)$$ escapes into the atmosphere. Sulphur dioxide is an acidic oxide. In moist air it is oxidised further and combines with water to form sulphurous and sulphuric acids as shown below:

$$\text{SO}_2 + \dfrac{1}{2}\,\text{O}_2 \longrightarrow \text{SO}_3$$

$$\text{SO}_3 + \text{H}_2\text{O} \longrightarrow \text{H}_2\text{SO}_4$$

These acids dissolve in atmospheric water droplets and fall back to the earth in the form of rain, popularly termed acid rain. Therefore, roasting contributes mainly to acid rain rather than to global warming or photochemical smog.

Putting both results side by side, we have:

• Calcination ⟶ liberation of $$\text{CO}_2$$ ⟶ global warming
• Roasting ⟶ liberation of $$\text{SO}_2$$ ⟶ acid rain

Among the given options, the pair “Global warming and acid rain” matches this sequence exactly.

Hence, the correct answer is Option B.

First, we recall the precise definition of Biochemical Oxygen Demand, abbreviated as BOD. In environmental chemistry, BOD is that amount of dissolved oxygen which is consumed by aerobic micro-organisms while they oxidise the biodegradable organic matter present in a definite volume of a water sample, generally over a fixed time (5 days) at a fixed temperature (20 °C).

Stating the definition in mathematical form, we write

$$\text{BOD} = \frac{\text{Mass of }O_2\text{ consumed (mg)}}{\text{Volume of sample (L)}},$$

and because $$1\;\text{mg}\,\text{L}^{-1}=1\;\text{ppm},$$ the numerical value of BOD is usually expressed in $$\text{ppm}$$ (parts per million).

Now we compare each option with this definition:

Option A speaks of oxygen “for sustaining life in a water body.” That is a description of the dissolved oxygen content, not of the oxygen consumed. Hence it does not match the definition of BOD.

Option B states that oxygen is required “by bacteria to break-down organic waste in a certain volume of a water sample.” This wording exactly mirrors the definition above: aerobic bacteria (micro-organisms) oxidise organic waste, and the oxygen they consume per volume is BOD. So Option B is consistent.

Option C mentions “photochemical breakdown.” Photochemical reactions involve light, not bacteria, so this does not coincide with the biochemical process that defines BOD.

Option D refers to “anaerobic bacteria” acting on “inorganic waste.” BOD involves aerobic (oxygen-using) processes and organic matter, so Option D is also incorrect.

Only Option B satisfies all parts of the formal definition stated above.

Hence, the correct answer is Option B.

We have an initial spherical balloon of radius $$r_1 = 3\;\text{cm}$$ which contains helium gas at a pressure $$P_1 = 48 \times 10^{-3}\;\text{bar}$$. A second balloon of radius $$r_2 = 12\;\text{cm}$$ is filled with the same amount of helium gas and is kept at the same temperature. For an ideal gas the relation $$PV = nRT$$ holds. Because the temperature $$T$$ and the amount of gas $$n$$ remain unchanged, the product $$PV$$ is constant, so

$$P_1 V_1 = P_2 V_2.$$

The volume of a sphere is given by the formula $$V = \dfrac{4}{3}\pi r^{3}$$. Substituting the two radii, we write

$$V_1 = \dfrac{4}{3}\pi r_1^{3}, \qquad V_2 = \dfrac{4}{3}\pi r_2^{3}.$$

Dividing the second volume by the first, the common factor $$\dfrac{4}{3}\pi$$ cancels:

$$\frac{V_2}{V_1} \;=\; \frac{r_2^{3}}{r_1^{3}}.$$

Now we substitute the numerical radii:

$$\frac{V_2}{V_1} = \left(\frac{12}{3}\right)^{3} = 4^{3} = 64.$$

Hence $$V_2 = 64\,V_1.$$ Putting this result in $$P_1 V_1 = P_2 V_2$$, we have

$$P_1 V_1 = P_2 (64\,V_1).$$

The volume factor $$V_1$$ now cancels from both sides, yielding

$$P_2 = \frac{P_1}{64}.$$

Substituting $$P_1 = 48 \times 10^{-3}\;\text{bar}$$, we get

$$P_2 = \frac{48 \times 10^{-3}}{64}\;\text{bar}.$$

We first carry out the division of the numerical coefficients. Writing the fraction explicitly,

$$\frac{48}{64} = \frac{48 \div 16}{64 \div 16} = \frac{3}{4} = 0.75.$$

Therefore,

$$P_2 = 0.75 \times 10^{-3}\;\text{bar}.$$

The answer is required in the form “$$\dots \times 10^{-6}\;\text{bar}$$”. To convert $$0.75 \times 10^{-3}$$ into a $$\times 10^{-6}$$ expression, we shift the factor $$10^{-3}$$ three powers lower:

$$0.75 \times 10^{-3} \;=\; 0.75 \times 10^{3} \times 10^{-6}.$$\; (since $$10^{-3} = 10^{3} \times 10^{-6}$$)

Now, $$0.75 \times 10^{3} = 0.75 \times 1000 = 750.$$ Hence

$$P_2 = 750 \times 10^{-6}\;\text{bar}.$$

So, the answer is $$750$$.

First we note that the volume of oxygen required by the person is given as 492 L at a pressure of 1 atm and a temperature of 300 K. To convert this volume into the corresponding amount (in moles) of O2, we use the Ideal-Gas equation, which is stated as $$PV = nRT$$ where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is absolute temperature.

Re-arranging the equation to obtain $$n$$ gives $$n = \dfrac{PV}{RT}\;.$$

Substituting the numerical values, we have

$$n = \dfrac{(1\;\text{atm})(492\;\text{L})}{(0.082\;\text{L atm mol}^{-1}\text{K}^{-1})(300\;\text{K})}\;.$$

Simplifying the denominator first, $$0.082\times 300 = 24.6\;.$$

Hence,

$$n = \dfrac{492}{24.6} = 20\;\text{mol}\;.$$

Thus the daily requirement of pure oxygen for the person is $$20$$ moles.

Now we turn to the chemical reaction that produces the oxygen:

$$\text{NaClO}_3(s) + \text{Fe}(s) \;\rightarrow\; \text{O}_2(g) + \text{NaCl}(s) + \text{FeO}(s)\;.$$

We check the atom balance in this equation. On the left there are 1 Na, 1 Cl, 1 Fe and 3 O atoms. On the right there are 1 Na (in NaCl), 1 Cl (in NaCl), 1 Fe (in FeO) and $$2+1=3$$ oxygen atoms (two in O2 and one in FeO). Because every element balances exactly, the coefficients in the written equation are already the smallest whole numbers. Hence one mole of NaClO3 produces one mole of O2.

Therefore, to obtain $$20$$ moles of O2, we need exactly $$20$$ moles of NaClO3.

Next we convert these moles into grams. The molar mass of NaClO3 is calculated as

$$M(\text{NaClO}_3) = M(\text{Na}) + M(\text{Cl}) + 3M(\text{O}) = 23 + 35.5 + 3(16) = 23 + 35.5 + 48 = 106.5\;\text{g mol}^{-1}\;.$$

Multiplying the molar mass by the number of moles gives the required mass:

$$m = n\,M = 20 \times 106.5\;\text{g} = 2130\;\text{g}\;.$$

So, the answer is $$2130\;\text{g}\;.$$

We begin with the balanced chemical equation for the synthesis of ammonia:

$$\mathrm{N_2 + 3\,H_2 \;\longrightarrow\; 2\,NH_3}$$

From this equation we note the stoichiometric mole ratio:

$$\mathrm{N_2 : H_2 : NH_3 \;=\; 1 : 3 : 2}$$

Now we convert each given mass into moles because stoichiometric calculations are always performed in terms of moles.

The molar mass of dinitrogen is $$\mathrm{28\;g\,mol^{-1}}$$, so for $$2.8\;\text{kg}$$ (which equals $$2800\;\text{g}$$) the number of moles of dinitrogen is

$$n_{\mathrm{N_2}}=\dfrac{2800\;\text{g}}{28\;\text{g\,mol}^{-1}}=100\;\text{mol}$$

The molar mass of dihydrogen is $$\mathrm{2\;g\,mol^{-1}}$$, so for $$1\;\text{kg}$$ (which equals $$1000\;\text{g}$$) the number of moles of dihydrogen is

$$n_{\mathrm{H_2}}=\dfrac{1000\;\text{g}}{2\;\text{g\,mol}^{-1}}=500\;\text{mol}$$

Next we identify the limiting reagent by comparing the available mole ratio with the required stoichiometric ratio.

According to the balanced equation, $$1$$ mole of $$\mathrm{N_2}$$ needs $$3$$ moles of $$\mathrm{H_2}$$. Therefore, for the $$100$$ moles of $$\mathrm{N_2}$$ present, the hydrogen required would be

$$100 \times 3 = 300\;\text{mol of } \mathrm{H_2}$$

We actually have $$500\;\text{mol of } \mathrm{H_2}$$, which is more than the $$300\;\text{mol}$$ required. Hence hydrogen is present in excess, and nitrogen is the limiting reagent.

Since $$\mathrm{N_2}$$ is the limiting reagent, the amount of ammonia formed depends entirely on the moles of $$\mathrm{N_2}$$ available. The balanced equation tells us that $$1$$ mole of $$\mathrm{N_2}$$ produces $$2$$ moles of $$\mathrm{NH_3}$$. Therefore, for the $$100$$ moles of $$\mathrm{N_2}$$ the moles of ammonia produced will be

$$n_{\mathrm{NH_3}} = 100 \times 2 = 200\;\text{mol}$$

Finally, we convert these moles of ammonia into mass. The molar mass of ammonia is $$\mathrm{17\;g\,mol^{-1}}$$, so

$$m_{\mathrm{NH_3}} = 200\;\text{mol} \times 17\;\text{g\,mol}^{-1} = 3400\;\text{g}$$

This value can also be expressed as $$3.4\;\text{kg}$$, but since the question asks for grams, we keep it as $$3400\;\text{g}$$.

So, the answer is $$3400 \text{ g}$$.

Let the molecular formula of the saturated, open-chain organic compound be $$\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z$$.

Its molar mass (in atomic-mass units) is $$12x + y + 16z$$.

Mass percentage of any element is obtained from
$$\text{mass\% of element}=\dfrac{\text{mass of that element in 1 mole}}{\text{molar mass}}\times 100.$$

We have the given ratio of the mass percentages of carbon and hydrogen

$$\dfrac{\%\,\mathrm{C}}{\%\,\mathrm{H}}=\dfrac{4}{1}.$$

Using the formula, this ratio becomes

$$\dfrac{\dfrac{12x}{12x+y+16z}}{\dfrac{y}{12x+y+16z}}=\dfrac{4}{1}.$$

The common denominator cancels, giving

$$\dfrac{12x}{y}=4.$$

So

$$12x = 4y \;\;\Longrightarrow\;\; y = 3x.$$

Next, the ratio of the mass percentages of carbon and oxygen is supplied as

$$\dfrac{\%\,\mathrm{C}}{\%\,\mathrm{O}}=\dfrac{3}{4}.$$

This becomes

$$\dfrac{\dfrac{12x}{12x+y+16z}}{\dfrac{16z}{12x+y+16z}}=\dfrac{3}{4}.$$

Again the denominator cancels, leaving

$$\dfrac{12x}{16z}=\dfrac{3}{4}.$$

Cross-multiplying gives

$$12x \times 4 = 16z \times 3 \;\;\Longrightarrow\;\; 48x = 48z \;\;\Longrightarrow\;\; x = z.$$

Substituting $$y=3x$$ and $$z=x$$, the empirical formula is obtained as

$$\mathrm{C}_x\mathrm{H}_{3x}\mathrm{O}_x.$$

Dividing every subscript by the common factor $$x$$ gives the simplest whole-number ratio

$$\mathrm{CH}_3\mathrm{O}.$$

Because a real molecule must satisfy the valencies exactly, we multiply this unit by the smallest integer that produces a fully saturated, acyclic structure with single bonds only. Taking $$n=2$$ delivers the molecule

$$\mathrm{C}_2\mathrm{H}_6\mathrm{O}_2,$$

which is indeed a saturated open-chain compound (for instance, ethane-1,2-diol).

Now we balance its complete combustion. For one mole, write

$$\mathrm{C}_2\mathrm{H}_6\mathrm{O}_2 + a\,\mathrm{O}_2 \;\longrightarrow\; 2\,\mathrm{CO}_2 + 3\,\mathrm{H}_2\mathrm{O}.$$\;

On the right there are $$2\times2=4$$ oxygen atoms in $$\mathrm{CO}_2$$ and $$3\times1=3$$ oxygen atoms in $$\mathrm{H}_2\mathrm{O},$$ a total of $$4+3=7$$ oxygen atoms.

The organic molecule already supplies $$2$$ oxygen atoms, so the remaining $$7-2 = 5$$ oxygen atoms must come from molecular oxygen:

$$\dfrac{5}{2}\,\mathrm{O}_2.$$

Therefore, for one mole of $$\mathrm{C}_2\mathrm{H}_6\mathrm{O}_2$$ we need $$\dfrac{5}{2}$$ moles of $$\mathrm{O}_2.$$

For two moles of the organic compound the amount of oxygen required is

$$2 \times \dfrac{5}{2} = 5 \text{ moles of } \mathrm{O}_2.$$

So, the answer is $$5$$.

We have a water sample in which the concentration of magnesium sulphate is given as $$10^{-3}\ \text{mol L}^{-1}$$.

First, let us convert this molar concentration into the mass of $$MgSO_4$$ present in one litre of the water sample.

The molar mass of $$MgSO_4$$ is given to be $$120.37\ \text{g mol}^{-1}$$. Using the relation

$$\text{mass (g)} = \text{molarity (mol L}^{-1}\text{)}\times\text{molar mass (g mol}^{-1}\text{)},$$

we obtain

$$\text{mass of }MgSO_4 = 10^{-3}\ \text{mol L}^{-1}\times120.37\ \text{g mol}^{-1} = 0.12037\ \text{g L}^{-1}.$$

Changing grams to milligrams (since $$1\ \text{g}=1000\ \text{mg}$$):

$$0.12037\ \text{g L}^{-1} = 0.12037\times1000\ \text{mg L}^{-1} = 120.37\ \text{mg L}^{-1}.$$

The hardness of water is always expressed as the equivalent concentration of $$CaCO_3$$. To carry out this conversion we must compare the equivalent amounts of the two compounds.

For any species,

$$\text{number of equivalents} = \text{moles}\times\text{valency},$$

and the corresponding mass in terms of any reference compound equals

$$\text{Number of equivalents}\times\text{equivalent weight of reference compound}.$$

Here, the hardness-producing ion from $$MgSO_4$$ is $$Mg^{2+}$$, which possesses a valency of 2. Thus, one mole of $$MgSO_4$$ provides two equivalents. For the present sample:

$$\text{equivalents of }MgSO_4 = 10^{-3}\ \text{mol L}^{-1}\times2 = 2\times10^{-3}\ \text{equiv L}^{-1}.$$

The reference compound is $$CaCO_3$$. Its molar mass is $$100\ \text{g mol}^{-1}$$ and its valency with respect to hardness (coming from $$Ca^{2+}$$) is also 2, so its equivalent weight is

$$\frac{\text{molar mass}}{\text{valency}} = \frac{100\ \text{g mol}^{-1}}{2} = 50\ \text{g equiv}^{-1}.$$

The mass of $$CaCO_3$$ that possesses the same number of equivalents as the given $$MgSO_4$$ is therefore

$$\text{mass of }CaCO_3 = 2\times10^{-3}\ \text{equiv L}^{-1}\times50\ \text{g equiv}^{-1} = 0.1\ \text{g L}^{-1}.$$

Again converting grams to milligrams,

$$0.1\ \text{g L}^{-1} = 0.1\times1000\ \text{mg L}^{-1} = 100\ \text{mg L}^{-1}.$$

Since 1 milligram per litre is numerically equal to 1 part per million (ppm) for aqueous solutions, the hardness of the given water sample is

$$100\ \text{ppm}.$$

So, the answer is $$100$$.

We first recall the meaning of “10 ppm”. In foods, parts per million for solids is interpreted as

$$1\ \text{ppm}=1\ \text{mg of solute per kg of food}.$$

So, if we want $$10\ \text{ppm}$$ of iron (Fe) in wheat, we need

$$10\ \text{mg Fe per kg wheat}.$$

Now we have $$100\ \text{kg}$$ of wheat. Therefore the total mass of iron required is

$$10\ \text{mg Fe}\times 100\ \text{kg}=1000\ \text{mg Fe}.$$

Since $$1000\ \text{mg}=1\ \text{g},$$ we need

$$1\ \text{g of elemental Fe}.$$

The fortifying chemical available is ferrous sulphate heptahydrate, whose formula is $$\mathrm{FeSO_4\cdot7H_2O}.$$ To find out how much of this salt provides 1 g of iron, we must calculate its molar mass.

Total molar mass of $$\mathrm{FeSO_4\cdot7H_2O}:$$

Fe: $$55.85\ \text{g mol}^{-1}$$

S: $$32.00\ \text{g mol}^{-1}$$

O in $$\mathrm{SO_4}: 4\times16.00 = 64.00\ \text{g mol}^{-1}$$

Water of crystallisation $$7\mathrm{H_2O}: 7\times(2\times1.00+16.00)=7\times18.00=126.00\ \text{g mol}^{-1}$$

Adding all parts,

$$\begin{aligned} M(\mathrm{FeSO_4\cdot7H_2O}) &= 55.85+32.00+64.00+126.00\\ &=277.85\ \text{g mol}^{-1}. \end{aligned}$$

In one mole (277.85 g) of this salt there is exactly one mole of iron, i.e. 55.85 g of Fe. Therefore, the mass fraction of iron in the salt is

$$\text{Mass fraction of Fe}= \frac{55.85}{277.85}.$$

To obtain $$1\ \text{g}$$ of iron, the mass of salt needed is found by the relation

Salt required $$= \frac{\text{desired mass of Fe}}{\text{mass fraction of Fe}} = \frac{1\ \text{g} }{55.85/277.85} = \frac{277.85}{55.85}\ \text{g} .$$

Evaluating the ratio,

$$\frac{277.85}{55.85}\approx 4.97\ \text{g}.$$

Rounding off $$\approx$$ 5g

We begin by recalling the fundamental principle for neutralisation reactions carried out in terms of equivalents. The number of equivalents of acid consumed must exactly equal the number of equivalents of base supplied. Mathematically, this is stated as

$$N_1 V_1 \;=\; N_2 V_2,$$

where $$N$$ represents the normality and $$V$$ the volume of the respective solutions. Here the subscripts 1 and 2 denote the acid and the base, respectively.

Now we identify each quantity from the data given in the problem:

• The phosphinic acid (the acid side) has a normality $$N_1 = 0.1\; \text{N}$$ and a volume $$V_1 = 10\; \text{mL}.$$

• The sodium hydroxide solution (the base side) has a normality $$N_2 = 0.1\; \text{N}$$, while its volume $$V_2$$ is what we have to determine.

Substituting these numerical values directly into the neutralisation equation, we get

$$0.1 \times 10 \;=\; 0.1 \times V_2.$$

To isolate $$V_2$$, we divide both sides of the equation by $$0.1$$:

$$V_2 \;=\; \dfrac{0.1 \times 10}{0.1}.$$

Simplifying the numerator, we first find $$0.1 \times 10 = 1.0$$, so the fraction becomes

$$V_2 \;=\; \dfrac{1.0}{0.1}.$$

Carrying out this final division, we obtain

$$V_2 = 10\; \text{mL}.$$

Hence the volume of $$0.1 \text{ N}$$ NaOH required to completely neutralise $$10 \text{ mL}$$ of $$0.1 \text{ N}$$ phosphinic acid is $$10 \text{ mL}$$.

So, the answer is $$10$$.

We have a solution in which $$0.27\ \text{g}$$ of the long-chain fatty acid is dissolved in $$100\ \text{cm}^3$$ (that is, $$100\ \text{mL}$$) of hexane.

First we calculate the concentration of fatty acid in the solution. By definition

$$\text{Concentration}=\frac{\text{mass of solute}}{\text{volume of solution}} =\frac{0.27\ \text{g}}{100\ \text{cm}^3}=0.0027\ \text{g cm}^{-3}.$$

Now, only $$10\ \text{mL}=10\ \text{cm}^3$$ of this solution is taken. The mass of fatty acid present in this aliquot is therefore

$$m=\left(0.0027\ \text{g cm}^{-3}\right)\left(10\ \text{cm}^3\right)=0.027\ \text{g}.$$

When the hexane evaporates, this entire mass spreads out on the water surface to form a monomolecular layer. To find the thickness (height) of this layer, we first need the volume it occupies. Using the relation

$$\text{Volume}=\frac{\text{mass}}{\text{density}},$$

and the given density $$\rho =0.9\ \text{g cm}^{-3},$$ we get

$$V=\frac{0.027\ \text{g}}{0.9\ \text{g cm}^{-3}}=0.03\ \text{cm}^3.$$

Next we determine the area over which the fatty acid spreads. The watch glass has a circular water surface whose radius is the distance from edge to centre, namely $$r=10\ \text{cm}.$$ The area of a circle is given by

$$A=\pi r^{2}.$$

With the approximation $$\pi =3,$$ we have

$$A=3\,(10\ \text{cm})^{2}=3\times100\ \text{cm}^2=300\ \text{cm}^2.$$

The thickness (height) $$h$$ of the monolayer is the volume divided by the area:

$$h=\frac{V}{A}=\frac{0.03\ \text{cm}^3}{300\ \text{cm}^2}=0.0001\ \text{cm}=1\times10^{-4}\ \text{cm}.$$

To express this height in metres we use $$1\ \text{cm}=10^{-2}\ \text{m},$$ so

$$h=1\times10^{-4}\ \text{cm}\times10^{-2}\ \text{m cm}^{-1}=1\times10^{-6}\ \text{m}.$$

Hence, the correct answer is Option B.

We recall that the compression factor is defined as $$Z=\dfrac{pV}{RT}.$$

For the gases in the problem the experimentally observed equation of state neglects the attraction term and is written as $$p=\dfrac{RT}{V-b},$$ where $$b$$ is the van der Waals excluded-volume constant.

We first manipulate this equation to obtain $$V$$ in terms of $$p$$.

Starting with $$p(V-b)=RT,$$ we multiply out to get $$pV-pb=RT.$$

Now we isolate $$V$$. Adding $$pb$$ to both sides and dividing by $$p$$, we obtain $$V=b+\dfrac{RT}{p}.$$

We substitute this expression for $$V$$ in the definition of the compressibility factor: $$Z=\dfrac{pV}{RT}= \dfrac{p\left(b+\dfrac{RT}{p}\right)}{RT}.$$

We separate the two terms in the numerator: $$Z=\dfrac{pb}{RT}+\dfrac{p}{RT}\cdot \dfrac{RT}{p}.$$

The second term simplifies immediately to 1, giving the very convenient linear relation $$Z=1+\dfrac{b}{RT}\,p.$$

Thus, at the fixed temperature $$T$$ of the experiment, the plot of $$Z$$ versus $$p$$ is a straight line with slope $$\text{slope}= \dfrac{b}{RT}.$$

Because $$R$$ and $$T$$ are common to all the gases, the steepness of the line is governed solely by the magnitude of $$b$$. The larger the value of $$b$$, the steeper the increase of $$Z$$ with pressure.

The van der Waals constant $$b$$ reflects the effective molecular size; for the noble gases it increases with atomic size in the order $$\text{Ne} < \text{Ar} < \text{Kr} < \text{Xe}.$$

Therefore, xenon (Xe) has the largest $$b$$ and consequently the largest value of $$\dfrac{b}{RT}$$, giving the greatest slope in the $$Z$$ vs. $$p$$ plot.

Hence, the correct answer is Option D.

The formula for the most probable velocity (Vmp) of a gas molecule based on the Maxwell-Boltzmann distribution is:

Vmp = square root of (2 * R * T / M)

Where:

• R is the universal gas constant.
• T is the absolute temperature in Kelvin (K).
• M is the molar mass of the gas.

From this formula, we can see that:

• Vmp is directly proportional to the square root of Temperature (T).
• Vmp is inversely proportional to the square root of Molar Mass (M).

Therefore, Vmp increases when Temperature (T) increases or Molar Mass (M) decreases.

Given Data

We need to compare the values of Vmp for three specific cases:

1. N2 at 300 K (Molar mass of N2 = 28 g/mol)
2. O2 at 400 K (Molar mass of O2 = 32 g/mol)
3. H2 at 300 K (Molar mass of H2 = 2 g/mol)

Let's look at the ratio of (T / M) for each option to determine the relative order of their speeds:

• For N2 at 300 K: Ratio = 300 / 28 = 10.71
• For O2 at 400 K: Ratio = 400 / 32 = 12.50
• For H2 at 300 K: Ratio = 300 / 2 = 150.00

Comparing the Values

Comparing the ratios calculated above: 10.71 < 12.50 < 150.00

This gives us the order of increasing most probable velocity: Vmp(N2 at 300 K) < Vmp(O2 at 400 K) < Vmp(H2 at 300 K)

Matching with the Graph

Looking at the provided Maxwell-Boltzmann distribution curves along the x-axis (Speed, v):

• Point I has the lowest speed value.
• Point II has an intermediate speed value.
• Point III has the highest speed value.

Therefore:

• Point I corresponds to Vmp of N2 at 300 K
• Point II corresponds to Vmp of O2 at 400 K
• Point III corresponds to Vmp of H2 at 300 K

For real gases we use the relation $$PV = Z\,nRT$$, where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$T$$ is the temperature, $$R$$ is the universal gas constant and $$Z$$ is the compressibility factor.

Both gases A and B are taken at the same temperature and contain an equal number of moles, so $$n$$, $$R$$ and $$T$$ are common to both. Hence we can write

$$P_A V_A = Z_A nRT \quad\text{and}\quad P_B V_B = Z_B nRT.$$

Dividing the first equation by the second we get

$$\dfrac{P_A V_A}{P_B V_B} = \dfrac{Z_A}{Z_B}.$$

The problem tells us that the volume of gas A is twice that of gas B, so $$V_A = 2V_B.$$ Substituting this into the ratio gives

$$\dfrac{P_A (2V_B)}{P_B V_B} = \dfrac{Z_A}{Z_B}.$$

Simplifying the left side, $$V_B$$ cancels:

$$\dfrac{2P_A}{P_B} = \dfrac{Z_A}{Z_B}.$$

We are also told that the compressibility factor of gas A is three times that of gas B, so $$Z_A = 3Z_B.$$ Substituting this value yields

$$\dfrac{2P_A}{P_B} = \dfrac{3Z_B}{Z_B} = 3.$$

Now solving for the ratio of the pressures, we have

$$2P_A = 3P_B.$$

This is exactly the relation asked for between the two pressures. Hence, the correct answer is Option A.

We start with the ideal-gas equation, which states that for any ideal gas the relation $$PV = nRT$$ holds, where $$P$$ is the pressure, $$V$$ is the volume, $$n$$ is the total number of moles present, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

In the vessel we have two gases together, so the total number of moles is the sum of the individual moles: $$n = 0.5 + x$$.

The numerical data given are: $$P = 200\ \text{Pa}, \quad V = 10\ \text{m}^3, \quad T = 1000\ \text{K}$$, while $$R$$ is to be kept as a symbol. Substituting all these directly into $$PV = nRT$$ gives

$$200 \times 10 = (0.5 + x)\, R \times 1000.$$

Simplifying the left-hand side first, we obtain

$$200 \times 10 = 2000,$$

so the equation becomes

$$2000 = (0.5 + x)\, 1000 R.$$

To isolate the bracket, we divide both sides by $$1000R$$:

$$\frac{2000}{1000R} = 0.5 + x.$$

The fraction on the left simplifies because $$2000/1000 = 2$$, hence

$$\frac{2}{R} = 0.5 + x.$$

Now we solve for $$x$$ by subtracting $$0.5$$ from both sides:

$$x = \frac{2}{R} - 0.5.$$

Writing the decimal $$0.5$$ as the fraction $$\dfrac{1}{2}$$ and bringing the two fractions to a common denominator $$2R$$, we get

$$x = \frac{2}{R} - \frac{1}{2} = \frac{4}{2R} - \frac{R}{2R} = \frac{4 - R}{2R}.$$

This expression matches Option C in the list. Hence, the correct answer is Option C.

nRT1 = ( n - 2n/5 ) R T2

T1=27+273 = 300k

Hence upon solving we get that T2 = 500k.

We recall from elementary water chemistry that hardness is produced by dissolved salts of calcium and magnesium.

First, we separate hardness into two kinds.

$$\text{Hardness} = \text{Temporary hardness} + \text{Permanent hardness}$$

Temporary hardness arises from those salts that can be removed simply by boiling. The salts that decompose on heating are the bicarbonates of calcium and magnesium.

For example, when we boil water containing calcium bicarbonate, the following decomposition occurs:

$$Ca(HCO_3)_2 \xrightarrow{\text{boil}} CaCO_3 \downarrow + CO_2\uparrow + H_2O$$

The formation of the insoluble precipitate $$CaCO_3$$ takes the calcium ion out of solution, so hardness is removed. Because this hardness disappears on heating, it is called temporary.

Now we look at each option.

Option A gives $$Na_2SO_4$$. Sodium ions do not cause hardness, and sulphates of sodium remain soluble on boiling, so this cannot account for temporary hardness.

Option B offers $$NaCl$$. Again, sodium chloride contains no calcium or magnesium, so it does not cause hardness at all.

Option C presents $$CaCl_2$$. Although the salt has calcium, chlorides do not decompose on heating:

$$CaCl_2 \xrightarrow{\text{boil}} \text{no change}$$

Therefore the hardness produced by $$CaCl_2$$ is not removed by boiling; this belongs to permanent hardness.

Option D lists $$Ca(HCO_3)_2$$, the very bicarbonate that breaks down on boiling and hence is the textbook cause of temporary hardness.

Among all the choices, only $$Ca(HCO_3)_2$$ satisfies the requirement for temporary hardness.

Hence, the correct answer is Option D.

First, let us recall the meaning of the unit “ppm”. The abbreviation “ppm” stands for “parts per million”. When we say that a substance has a concentration of so many ppm in water, we mean that out of one million parts by mass of the solution, that many parts are the solute. Mathematically we write this definition as $$1\;\text{ppm}= \dfrac{1\;\text{part}}{10^{6}\;\text{parts}}.$$

In the context of drinking-water standards, national and international agencies such as the Bureau of Indian Standards (BIS) and the World Health Organization (WHO) specify upper limits for different metal ions to protect human health. Copper is an essential trace element, but excess copper can cause gastrointestinal irritation and other toxic effects. Hence the allowable limit is fixed after careful toxicological studies.

The value prescribed in the BIS standard IS 10500 : 2012 for copper in potable water is $$3\;\text{ppm},$$ which is also written as $$3\;\text{mg L}^{-1}$$ because $$1\;\text{ppm}=1\;\text{mg per litre}$$ for dilute aqueous solutions (density ≈ 1 g mL−1). Therefore, the maximum permitted concentration of copper that should not be exceeded in drinking water is $$3\;\text{ppm}.$$

Comparing this value with the given options - 3 ppm, 5 ppm, 0.5 ppm and 0.05 ppm - we see that the correct numerical match is 3 ppm.

Hence, the correct answer is Option A.

Step 1: Calculate the total number of moles of surfactant Volume of the surfactant solution = 10 mL = 0.01 L Concentration of the solution = 1 mM = 0.001 mol/L Number of moles = Concentration * Volume Number of moles = 0.001 mol/L * 0.01 L = 10^-5 moles

Step 2: Calculate the total number of surfactant molecules Using Avogadro's number (6 * 10^23 molecules/mol): Number of molecules = 10^-5 mol * 6 * 10^23 molecules/mol = 6 * 10^18 molecules

Step 3: Calculate the area occupied by a single molecule Total Area = 0.24 cm^2 Area per molecule = Total Area / Number of molecules Area per molecule = 0.24 cm^2 / (6 * 10^18) = 4 * 10^-20 cm^2

Step 4: Find the edge length of the cubic polar head Since the polar head is approximated as a cube, the area it occupies on the substrate is the area of one face (a^2), where a is the edge length: a^2 = 4 * 10^-20 cm^2

Taking the square root of both sides: a = 2 * 10^-10 cm

Step 5: Convert the unit to picometers (pm) Convert centimeters to meters: a = 2 * 10^-12 m

Since 1 pm = 10^-12 m: a = 2 pm

Final Answer: The edge length of the polar head is 2 pm (or 2 * 10^-12 m).

We first recall the definition of BOD (Biological Oxygen Demand). It is the amount of dissolved oxygen, measured in parts per million (ppm), that microorganisms require to decompose the organic matter present in a given volume of water.

Environmental standards state that:

$$\text{If BOD} \lt 5\ \text{ppm, the water is regarded as \textit{clean}.}$$

$$\text{If BOD} \gt 5\ \text{ppm, the water is regarded as \textit{polluted}.}$$

Furthermore, when the BOD value rises well beyond $$5\ \text{ppm}$$, the pollution is often termed high. Typical textbooks label water with BOD values above roughly $$17\text{-}18\ \text{ppm}$$ as highly polluted.

Now we compare the given samples with these benchmarks.

For the first sample: $$\text{BOD} = 4\ \text{ppm}.$$

Since $$4\ \text{ppm} \lt 5\ \text{ppm},$$ we classify this water as clean.

For the second sample: $$\text{BOD} = 18\ \text{ppm}.$$

Here $$18\ \text{ppm} \gt 17\ \text{ppm},$$ so the water is definitely in the highly polluted category.

Putting the two results together, we have:

$$\bigl(4\ \text{ppm},\,18\ \text{ppm}\bigr) \;=\; (\text{Clean},\,\text{Highly polluted}).$$

Among the listed choices, this description corresponds to Option D: “Clean and Highly polluted.”

Hence, the correct answer is Option D.

First, recall the definition: the Biological Oxygen Demand (BOD) of a water sample is the mass of dissolved oxygen (in milligrams per litre) that aerobic microorganisms need to oxidise the organic matter present in the water over a standard period (usually 5 days at 20 °C). In simple words, higher BOD means more biodegradable organic pollutants are present, so more oxygen is consumed and the water quality is poorer.

Mathematically we can write the direct proportionality

$$\text{Pollution level} \;\propto\; \text{BOD}$$

and the usual drinking-water guideline is

$$\text{BOD}_{\text{permissible}} < 5 \;\text{mg\,L}^{-1}.$$

Now, for the two glasses we have the numerical data

$$\text{BOD}_A = 10\;\text{mg\,L}^{-1}, \qquad \text{BOD}_B = 20\;\text{mg\,L}^{-1}.$$

Both of these individual values clearly violate the $$<5\;\text{mg\,L}^{-1}$$ criterion, so neither sample qualifies as potable (drinking) water. To compare their relative pollution levels, simply look at the difference:

$$\text{BOD}_B - \text{BOD}_A = 20 - 10 = 10\;\text{mg\,L}^{-1} > 0,$$

which shows $$\text{BOD}_B > \text{BOD}_A.$$ Because BOD is directly proportional to the concentration of organic pollutants, the larger value corresponds to the more polluted sample. Thus glass B, with the higher BOD of $$20\;\text{mg\,L}^{-1},$$ is more polluted than glass A.

Among the given options, the only statement that matches this conclusion is “B is more polluted than A.”

Hence, the correct answer is Option B.

We begin with the thermal decomposition of sodium chlorate. The balanced chemical equation is first written because it supplies the mole ratio that will later connect the loss of oxygen to the amount of NaCl formed:

$$2\,\text{NaClO}_{3} \;\xrightarrow{\Delta}\; 2\,\text{NaCl} \;+\; 3\,\text{O}_{2}$$

This equation tells us that every $$2$$ moles of $$\text{NaClO}_{3}$$ give $$2$$ moles of $$\text{NaCl}$$ and $$3$$ moles of $$\text{O}_{2}$$. The loss in mass recorded in the experiment comes only from the oxygen that escapes as gas.

The question states that $$0.16\;\text{g}$$ of oxygen is lost. First, we must convert this mass to moles. The formula we use is:

$$\text{Moles} \;=\; \dfrac{\text{Given mass}}{\text{Molar mass}}$$

The molar mass of molecular oxygen $$\text{O}_{2}$$ is $$32\;\text{g mol}^{-1}$$, so

$$n(\text{O}_{2}) \;=\; \dfrac{0.16\;\text{g}}{32\;\text{g mol}^{-1}} \;=\; 0.005\;\text{mol}$$

Now we link these moles of $$\text{O}_{2}$$ to the moles of $$\text{NaCl}$$ produced by employing the stoichiometric coefficients from the balanced decomposition equation. From the equation we read that $$3$$ moles of $$\text{O}_{2}$$ correspond to $$2$$ moles of $$\text{NaCl}$$. Therefore, using proportion:

$$n(\text{NaCl}) \;=\; n(\text{O}_{2}) \times \dfrac{2}{3} \;=\; 0.005 \times \dfrac{2}{3} \;=\; 0.003333\;\text{mol}$$

The residue containing this $$\text{NaCl}$$ is dissolved in water and treated with a solution of silver nitrate. The precipitation reaction that occurs is

$$\text{NaCl} \;+\; \text{AgNO}_{3} \;\longrightarrow\; \text{AgCl} \;+\; \text{NaNO}_{3}$$

This reaction is a simple double displacement and, importantly, the mole ratio of $$\text{NaCl}$$ to $$\text{AgCl}$$ is $$1\!:\!1$$. Hence, the moles of $$\text{AgCl}$$ formed equal the moles of $$\text{NaCl}$$ present:

$$n(\text{AgCl}) = n(\text{NaCl}) = 0.003333\;\text{mol}$$

Finally, we convert these moles of silver chloride to mass, again using the same fundamental formula. The molar mass of $$\text{AgCl}$$ is given as $$143.5\;\text{g mol}^{-1}$$, so

$$\text{Mass of AgCl} \;=\; n(\text{AgCl}) \times M(\text{AgCl}) \;=\; 0.003333\;\text{mol} \times 143.5\;\text{g mol}^{-1}$$

$$\text{Mass of AgCl} \;=\; 0.4783\;\text{g}$$

Rounding this value to two significant figures, as is typical for laboratory data, we obtain $$0.48\;\text{g}$$.

Hence, the correct answer is Option D.

We are told that chlorine constitutes 3.55 % of the mass of the unknown chlorohydrocarbon. This means that, out of every 100 g of the substance, 3.55 g is chlorine. We wish to deal with just 1 g of the substance, so we scale the mass of chlorine proportionally.

For 1 g of the chlorohydrocarbon the mass of chlorine present is

$$\text{mass of Cl in 1 g} \;=\;\frac{3.55}{100}\times 1\ \text{g}\;=\;0.0355\ \text{g}.$$

Now we convert this mass of chlorine to moles. First, we state the molar-mass relation:

Formula: $$n = \frac{m}{M},$$ where $$n$$ is the number of moles, $$m$$ is the given mass, and $$M$$ is the molar mass.

Using the atomic mass of chlorine, $$M_{\text{Cl}} = 35.5\ \text{g mol}^{-1},$$ we have

$$n_{\text{Cl}} \;=\;\frac{0.0355\ \text{g}}{35.5\ \text{g mol}^{-1}}.$$

Carrying out the division,

$$n_{\text{Cl}} = 0.001\ \text{mol}.$$

Each mole contains Avogadro’s number of entities. We therefore multiply the moles of chlorine by Avogadro’s constant to find the number of chlorine atoms.

Formula: $$N = n \times N_{\!A},$$ where $$N$$ is the number of entities and $$N_{\!A} = 6.023 \times 10^{23}\ \text{mol}^{-1}.$$

Substituting the values,

$$N_{\text{Cl atoms}} \;=\;0.001\ \text{mol}\times 6.023 \times 10^{23}\ \text{mol}^{-1}.$$

Multiplying,

$$N_{\text{Cl atoms}} = 6.023 \times 10^{20}.$$

Therefore, 1 g of the chlorohydrocarbon contains $$6.023 \times 10^{20}$$ chlorine atoms.

Hence, the correct answer is Option D.

We have to find, for every one gram of the original solid or gas taken as reactant, what mass of nitrogen gas $$N_2$$ is liberated. In other words, for each reaction we shall evaluate

$$\frac{\text{mass of }N_2\text{ obtained}}{\text{mass of reactant consumed}}$$

The reaction giving the largest value of this ratio will be the one that produces the “maximum quantity” of $$N_2$$ per gram of reactant. Let us examine the four reactions one by one.

Reaction (A)  $$\mathrm{Ba(N_3)_2(s)\;\longrightarrow\;Ba(s)+3\,N_2(g)}$$

Molar mass of the reactant $$\mathrm{Ba(N_3)_2$$:

We use the given atomic masses: $$\mathrm{Ba}=137\;u,\;N=14\;u.$$

$$\begin{aligned} M_r[\mathrm{Ba(N_3)_2}]&=&137 \;(\text{from Ba})+2\times(3\times14)\;(\text{from 2 azide ions})\\ &=&137+2\times42\\ &=&137+84\\ &=&221\;\text{g mol}^{-1}. \end{aligned}$$

From the balanced equation, $$1$$ mole of the azide produces $$3$$ moles of $$N_2$$. Hence the mass of $$N_2$$ formed by 1 mole of reactant is

$$m_{N_2}=3\times28=84\;\text{g}.$$ So,

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{84}{221}=0.380\;(\text{approximately}).$$

Reaction (B)  $$\mathrm{(NH_4)_2Cr_2O_7(s)\;\longrightarrow\;N_2(g)+4H_2O(g)+Cr_2O_3(s)}$$

Molar mass of $$\mathrm{(NH_4)_2Cr_2O_7}$$:

Using $$\mathrm{N}=14,\;H=1,\;Cr=52,\;O=16$$,

$$\begin{aligned} M_r&=&2\times14 + 8\times1 + 2\times52 + 7\times16\\ &=&28 + 8 + 104 + 112\\ &=&252\;\text{g mol}^{-1}. \end{aligned}$$

One mole of the dichromate gives exactly one mole of $$N_2$$, i.e.

$$m_{N_2}=1\times28=28\;\text{g}.$$

Therefore

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{28}{252}=0.111.$$

Reaction (C)  $$2NH_3(g)\;\longrightarrow\;N_2(g)+3H_2(g)$$

Molar mass of $$NH_3$$ is

$$M_r[NH_3]=14+3\times1=17\;\text{g mol}^{-1}.$$

The balanced equation shows that $$2$$ moles of $$NH_3$$ (mass $$2\times17=34\;\text{g}$$) produce $$1$$ mole of $$N_2$$ (mass $$28\;\text{g}$$). Hence, for the stoichiometric amount,

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{28}{34}=0.8235\;(\text{approximately}).$$

Because this ratio is for 34 g of reactant, for 1 g the value is the same decimal 0.8235. This is clearly larger than the corresponding ratios found so far.

Reaction (D)  $$2NH_4NO_3(s)\;\longrightarrow\;2N_2(g)+4H_2O(g)+O_2(g)$$

Molar mass of $$NH_4NO_3$$:

$$\begin{aligned} M_r[NH_4NO_3]&=&(2\times14)+(4\times1)+(3\times16)\\ &=&28+4+48\\ &=&80\;\text{g mol}^{-1}. \end{aligned}$$

The reaction consumes $$2$$ moles of nitrate (mass $$2\times80=160\;\text{g}$$) and yields $$2$$ moles of $$N_2$$ (mass $$2\times28=56\;\text{g}$$). Therefore,

$$\frac{\text{mass of }N_2}{\text{mass of reactant}} =\frac{56}{160}=0.35.$$

Now let us collect the four ratios:

$$ \begin{aligned} \text{Reaction A}:&\;0.380\\ \text{Reaction B}:&\;0.111\\ \text{Reaction C}:&\;\mathbf{0.824}\;(\text{highest})\\ \text{Reaction D}:&\;0.35 \end{aligned} $$

Clearly, Reaction (C) produces the largest mass of $$N_2$$ per gram of reactant.

Hence, the correct answer is Option C.

An organic compound has the general formula $$C_{x}H_{y}O_{z}$$.

According to the statement, the mass-percent ratio of carbon to hydrogen is 6 : 1. Using atomic masses $$12\;{\rm u}$$ for carbon and $$1\;{\rm u}$$ for hydrogen, the ratio of their masses present in one empirical-formula unit is

$$\dfrac{12x}{1\cdot y}= \dfrac{6}{1}.$$

Hence,

$$12x = 6y \quad\Longrightarrow\quad y = 2x.$$

So, inside the empirical formula the number of hydrogen atoms is twice the number of carbon atoms.

Now we use the second piece of information. First state the combustion equation of the hydrocarbon $$C_{x}H_{y}:$$

$$C_{x}H_{y} + \left(x + \dfrac{y}{4}\right)O_{2}\; \longrightarrow\; x\,CO_{2} + \dfrac{y}{2}\,H_{2}O.$$

Therefore, one molecule of $$C_{x}H_{y}$$ needs $$\left(x + \dfrac{y}{4}\right)$$ molecules of $$O_{2}$$, that is

$$2\left(x + \dfrac{y}{4}\right)=2x+\dfrac{y}{2}$$ oxygen atoms in total.

The problem says that one molecule of $$C_{x}H_{y}O_{z}$$ contains half of this oxygen. So the oxygen atoms present in the organic molecule satisfy

$$z = \dfrac{1}{2}\left(2x+\dfrac{y}{2}\right) = x+\dfrac{y}{4}.$$

We already have $$y = 2x,$$ so substituting this into the above relation gives

$$z = x + \dfrac{2x}{4} = x + \dfrac{x}{2} = \dfrac{3x}{2}.$$

For $$z$$ to be an integer, $$x$$ must be even. Choosing the smallest even value, $$x = 2,$$ we obtain

$$y = 2x = 4, \qquad z = \dfrac{3(2)}{2} = 3.$$

Thus the empirical formula becomes

$$C_{2}H_{4}O_{3}.$$

Comparing with the given options, this corresponds to Option A.

Hence, the correct answer is Option A.

For any ideal gas, the density $$\rho$$ at a given temperature $$T$$ and pressure $$P$$ is related to its molar mass $$M$$ by the ideal-gas formula rearranged for density

$$PV = nRT \;\;\Longrightarrow\;\; P = \dfrac{nRT}{V} = \dfrac{\left(\dfrac{m}{M}\right)RT}{V} = \dfrac{mRT}{MV}$$

Dividing both sides by $$RT$$, we get

$$\dfrac{P}{RT} = \dfrac{m}{MV} = \dfrac{\rho}{M}$$  because $$\rho = \dfrac{m}{V}$$.

So, explicitly,

$$\rho = \dfrac{P}{RT}\,M.$$

We have the same $$P$$ and the same $$T$$ for both gases, therefore the factor $$\dfrac{P}{RT}$$ is common and cancels out when we take a ratio. Hence, the ratio of densities equals the ratio of molar masses:

$$\dfrac{\rho_{\text{NH}_3}}{\rho_{\text{HCl}}} = \dfrac{M_{\text{NH}_3}}{M_{\text{HCl}}}.$$

Now we calculate each molar mass step by step.

For ammonia, NH3:

$$M_{\text{NH}_3} = M_{\text{N}} + 3M_{\text{H}} = 14.0 \,\text{u} + 3(1.0 \,\text{u}) = 14.0 \,\text{u} + 3.0 \,\text{u} = 17.0 \,\text{u}.$$

For hydrogen chloride, HCl:

Atomic mass of H = 1.0 u (given implicitly), atomic mass of Cl = 35.5 u (given).

$$M_{\text{HCl}} = M_{\text{H}} + M_{\text{Cl}} = 1.0 \,\text{u} + 35.5 \,\text{u} = 36.5 \,\text{u}.$$

Substituting these two molar masses into the ratio expression, we obtain

$$\dfrac{\rho_{\text{NH}_3}}{\rho_{\text{HCl}}} = \dfrac{17.0}{36.5}.$$

Now we carry out the division numerically:

$$\dfrac{17.0}{36.5} = 0.46575\ldots \approx 0.46.$$

So, the density of ammonia is about 0.46 times that of hydrogen chloride at the same temperature and pressure.

Hence, the correct answer is Option C.

We are told that a sample of mass 1 g of the carbonate $$M_2CO_3$$ reacts completely with excess hydrochloric acid and that this reaction liberates $$0.01186$$ moles of carbon dioxide.

First, let us write the balanced chemical equation for the reaction between the carbonate and hydrochloric acid:

$$M_2CO_3 + 2\,HCl \rightarrow 2\,MCl + H_2O + CO_2$$

From the stoichiometry of this equation we notice that one mole of the carbonate $$M_2CO_3$$ produces one mole of carbon dioxide $$CO_2$$.

Therefore, the number of moles of the carbonate present in the 1 g sample is exactly the same as the number of moles of carbon dioxide formed.

So we have

$$n(M_2CO_3)=n(CO_2)=0.01186\;\text{mol}$$

Now, the definition of molar mass is

$$\text{Molar mass}=\frac{\text{Mass of sample}}{\text{Number of moles}}$$

Substituting the given mass (1 g) and the calculated moles (0.01186 mol) into this formula we get

$$\text{Molar mass of }M_2CO_3=\frac{1\;\text{g}}{0.01186\;\text{mol}}$$

Carrying out the division,

$$\frac{1}{0.01186}=84.3\;\text{g mol}^{-1}$$

Hence, the molar mass of $$M_2CO_3$$ is $$84.3\;\text{g mol}^{-1}$$.

Hence, the correct answer is Option A.

We begin by writing the balanced chemical reaction between ferric chloride and an excess of sodium hydroxide: $$FeCl_3 \;+\; 3\,NaOH \;\rightarrow\; Fe(OH)_3 \;+\; 3\,NaCl$$. From this equation we see that one mole of $$FeCl_3$$ produces one mole of the precipitate $$Fe(OH)_3$$.

The mass of the precipitate obtained in the experiment is given as 2.14 g. To convert this mass into moles, we first need the molar mass of $$Fe(OH)_3$$. We list every atomic contribution:

$$\text{Molar mass of }Fe(OH)_3 = M_{Fe} + 3\,(M_O + M_H).$$

Substituting the atomic molar masses $$M_{Fe}=56\;g\,mol^{-1},\; M_O=16\;g\,mol^{-1},\; M_H=1\;g\,mol^{-1}$$, we obtain

$$ \begin{aligned} M_{Fe(OH)_3} &= 56 + 3\,(16+1)\\ &= 56 + 3\times17\\ &= 56 + 51\\ &= 107\;g\,mol^{-1}. \end{aligned} $$

Now we calculate the moles of $$Fe(OH)_3$$ precipitated using the definition $$\text{moles} = \dfrac{\text{mass}}{\text{molar mass}}$$:

$$ n_{Fe(OH)_3} = \dfrac{2.14\;g}{107\;g\,mol^{-1}} = 0.0200\;mol. $$

Because the stoichiometric ratio between $$FeCl_3$$ and $$Fe(OH)_3$$ is 1 : 1, the moles of $$FeCl_3$$ originally present are also $$0.0200\;mol$$.

The volume of the $$FeCl_3$$ solution used is 100 mL, which we convert to litres for molarity calculations: $$100\;mL = 0.100\;L$$.

Molarity is defined as $$M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$$. Substituting the values we have just obtained gives

$$ M_{FeCl_3} = \dfrac{0.0200\;mol}{0.100\;L} = 0.200\;M. $$

Hence, the correct answer is Option B.

The body of the person is said to have a total mass of $$75\;\text{kg}$$.

We are also told that hydrogen contributes $$10.0\%$$ of the body mass. The definition of percentage by mass is:

$$\text{mass of element} = \left(\dfrac{\text{percentage}}{100}\right)\times \text{total mass}$$

Substituting the given numbers for hydrogen, we get

$$m_{^{1}\!H} = \left(\dfrac{10.0}{100}\right)\times 75\;\text{kg} = 0.10 \times 75\;\text{kg} = 7.5\;\text{kg}.$$

Each ordinary hydrogen atom, $$^{1}\!H$$, has an atomic mass of approximately $$1\;\text{u}$$, whereas deuterium, $$^{2}\!H$$, has an atomic mass of approximately $$2\;\text{u}$$. Replacing every $$^{1}\!H$$ with $$^{2}\!H$$ therefore doubles the mass of every hydrogen atom. Mathematically, the new mass of hydrogen becomes

$$m_{^{2}\!H} = 2 \times m_{^{1}\!H} = 2 \times 7.5\;\text{kg} = 15\;\text{kg}.$$

The increase in the person’s total weight is the difference between the new hydrogen mass and the original hydrogen mass:

$$\Delta m = m_{^{2}\!H} - m_{^{1}\!H} = 15\;\text{kg} - 7.5\;\text{kg} = 7.5\;\text{kg}.$$

So, the weight gained after the complete replacement of $$^{1}\!H$$ by $$^{2}\!H$$ is $$7.5\;\text{kg}$$.

Hence, the correct answer is Option B.

For any ideal gas, we use the relation between density ($$\rho$$), pressure ($$P$$), molar mass ($$M$$) and temperature ($$T$$):

$$\rho = \frac{PM}{RT}$$

Here $$R$$ is the universal gas constant. Notice that at the same temperature $$T$$ and with the same units for pressure, the factor $$\dfrac{1}{RT}$$ is common for every gas we compare.

We are told that at 300 K, the density of the unknown gas (let us call it X) at 2 bar is double the density of nitrogen ($$N_2$$) at 4 bar. Writing this mathematically,

$$\rho_X = 2\,\rho_{N_2}$$

Using the density formula for each gas and keeping $$RT$$ common we have

$$\frac{P_X M_X}{RT} = 2 \left(\frac{P_{N_2} M_{N_2}}{RT}\right)$$

The $$RT$$ terms cancel out immediately:

$$P_X M_X = 2\,P_{N_2}\,M_{N_2}$$

Now substitute the given pressures and the known molar mass of dinitrogen ($$M_{N_2}=28\ \text{g mol}^{-1}$$):

$$\bigl(2\ \text{bar}\bigr)\,M_X = 2 \times \bigl(4\ \text{bar}\bigr)\times (28\ \text{g mol}^{-1})$$

Divide both sides by the 2 bar on the left to isolate $$M_X$$:

$$M_X = \frac{2 \times 4 \times 28}{2}\ \text{g mol}^{-1}$$

Simplify step by step:

$$M_X = \frac{8 \times 28}{2}\ \text{g mol}^{-1}$$

$$M_X = 4 \times 28\ \text{g mol}^{-1}$$

$$M_X = 112\ \text{g mol}^{-1}$$

Hence, the correct answer is Option B.

First, recall the basic definition: a greenhouse gas is one that absorbs infrared radiation and contributes to the greenhouse effect, thereby warming the Earth’s atmosphere. Classic examples, stated in every standard textbook, are $$CO_2$$, $$CH_4$$, $$N_2O$$, $$O_3$$, water vapour, CFCs, and a few others. In contrast, gases like $$N_2$$ and $$O_2$$ make up most of the air but do not absorb infrared radiation appreciably, while gases such as $$SO_2$$ and $$Cl_2$$ may cause pollution or participate in other atmospheric chemistry without being major greenhouse contributors.

Now we examine each option one by one and test whether every gas listed truly belongs to the greenhouse category.

Option A presents $$O_3$$, $$NO_2$$, $$SO_2$$, $$Cl_2$$. We recognise $$O_3$$ as a greenhouse gas, but $$NO_2$$ is chiefly an air pollutant tied to photochemical smog, $$SO_2$$ leads to acid rain, and $$Cl_2$$ is not a greenhouse gas at all. Because three of the four gases fail the definition, the entire set is disqualified.

Option B gives $$CH_4$$, $$O_3$$, $$N_2$$, $$SO_2$$. Here $$CH_4$$ and $$O_3$$ do fit, yet $$N_2$$ is the principal inert component of air with negligible infrared absorption, and $$SO_2$$ again is mainly a pollutant. Since two entries are non-greenhouse, this set is also incorrect.

Option C lists $$CO_2$$, $$CH_4$$, $$N_2O$$, $$O_3$$. Each one of these gases is explicitly named in the Intergovernmental Panel on Climate Change (IPCC) reports as a greenhouse gas: $$CO_2$$ is the benchmark, $$CH_4$$ is the second-most discussed, $$N_2O$$ ranks third in potency, and $$O_3$$ acts both in the troposphere and stratosphere. Because every member of the set satisfies the required property, Option C stands correct.

Option D proposes $$O_3$$, $$N_2$$, $$CO_2$$, $$NO_2$$. While $$O_3$$ and $$CO_2$$ are acceptable, $$N_2$$ once again is not a greenhouse gas, and $$NO_2$$ plays only a marginal infrared role compared with its importance in smog chemistry. Therefore this set is also invalid.

Comparing all four possibilities, only Option C contains a collection where each gas is undeniably a greenhouse gas. Hence, the correct answer is Option 3.

We are told that the compound contains carbon, hydrogen and sulphur, and that the percentage by mass of sulphur in the compound is 8 %. Our aim is to find the smallest (minimum) molecular mass that is consistent with this percentage.

First recall the general definition of mass percent. For any element present in a compound, the percentage by mass is given by the formula

$$\text{Mass percent of an element} \;=\; \frac{\text{Mass of that element in one mole of the compound}}{\text{Molar mass of the compound}}\times 100.$$

Applying this to sulphur, we have

$$\text{Percentage of S}=\frac{\text{Mass of S present in one mole}}{\text{Molar mass of compound}}\times 100.$$