JEE Solutions Questions

The relation between moles, molarity and volume is $$n = M \times V$$, where $$n$$ is in moles, $$M$$ in mol L$$^{-1}$$ and $$V$$ in litres.

Step 1 • Moles present before mixing
NaOH: $$M = 2\,\text{M},\; V = 10\,\text{mL}=0.01\,\text{L}$$
$$n_{NaOH} = 2 \times 0.01 = 0.02\,\text{mol}$$

HCl: $$M = 1\,\text{M},\; V = 20\,\text{mL}=0.02\,\text{L}$$
$$n_{HCl} = 1 \times 0.02 = 0.02\,\text{mol}$$

Step 2 • Neutralisation reaction
$$NaOH + HCl \rightarrow NaCl + H_2O$$
Since $$n_{NaOH}=n_{HCl}=0.02\,\text{mol}$$, both are completely consumed and we obtain $$0.02\,\text{mol}$$ NaCl.

Step 3 • Concentration of NaCl in the neutral mixture
Total volume after reaction $$=10\,\text{mL}+20\,\text{mL}=30\,\text{mL}=0.03\,\text{L}$$.
$$M_{NaCl} = \frac{0.02}{0.03} \approx 0.667\,\text{M}$$

Step 4 • Portion taken to the volumetric flask
Volume taken $$=10\,\text{mL}=0.01\,\text{L}$$.
Moles of NaCl transferred $$=0.667 \times 0.01 \approx 6.67\times10^{-3}\,\text{mol}$$.

Step 5 • HCl already present in the flask
The flask initially contains $$2\,\text{mol}$$ of HCl.

Step 6 • Final dilution to 100 mL
Final volume $$=100\,\text{mL}=0.1\,\text{L}$$.

Molarity of HCl:
$$M_{HCl} = \frac{2}{0.1} = 20\,\text{M}$$

Molarity of NaCl:
$$M_{NaCl} = \frac{6.67\times10^{-3}}{0.1} \approx 0.067\,\text{M}$$

Step 7 • Nature of the solution
The solution is a strongly acidic solution of $$20\,\text{M}$$ HCl (with a small amount of NaCl that does not affect acidity).

Hence the correct description is Option B - 20 M HCl solution.

For an ionic solid, the heat (enthalpy) of solution, $$\Delta H_{\text{sol}}$$, is obtained in two steps:
  1. Lattice dissociation: the crystal breaks into gaseous ions. This requires the lattice dissociation energy, $$\Delta H_{\text{lattice(diss)}}$$, which is numerically equal and opposite to the lattice energy of formation.
  2. Hydration: the gaseous ions get surrounded by water molecules, releasing their hydration energies.

Given data (sign conventions):
  • Hydration energy of $$K^+ = -x$$ kJ mol$$^{-1}$$ (exothermic)
  • Hydration energy of $$Cl^- = -y$$ kJ mol$$^{-1}$$ (exothermic)
  • Lattice energy of formation of KCl $$= -z$$ kJ mol$$^{-1}$$ (exothermic for formation from gaseous ions)

Step 1: Lattice dissociation energy
Lattice energy of formation is $$-z$$, so the energy required to dissociate the lattice is the opposite sign:
$$\Delta H_{\text{lattice(diss)}} = +z \text{ kJ mol}^{-1}$$

Step 2: Hydration of the ions
$$\Delta H_{\text{hydration}} = (-x) + (-y) = -(x + y) \text{ kJ mol}^{-1}$$

Overall heat of solution
$$\Delta H_{\text{sol}} = \Delta H_{\text{lattice(diss)}} + \Delta H_{\text{hydration}}$$
$$\Delta H_{\text{sol}} = \bigl(+z\bigr) + \bigl[-(x + y)\bigr]$$
$$\Delta H_{\text{sol}} = z - (x + y)$$

Therefore the heat of solution of KCl is $$z - (x + y)$$ kJ mol$$^{-1}$$.

Hence the correct option is Option C.

Statement I: NaCl is added to ice at 0°C in the ice cream box to prevent melting of ice cream.

When NaCl is added to ice, the freezing point is depressed below 0°C. This means the ice-salt mixture can reach temperatures below 0°C, which helps keep the ice cream frozen. So NaCl does help prevent melting of ice cream. Statement I is TRUE.

Statement II: On addition of NaCl to ice at 0°C, there is a depression in freezing point.

This is a standard colligative property. Adding a solute (NaCl) depresses the freezing point of water. Statement II is TRUE.

Both statements are true. Statement II provides the reason for Statement I.

The correct answer is Option 4: Both Statement I and Statement II are true.

We need to evaluate two statements about the molal depression constant $$K_f$$.

Statement I: Molal depression constant $$K_f$$ is given by $$\frac{M_1 R T_f}{\Delta S_{fus}}$$

The standard expression for the molal freezing point depression constant is:

$$$K_f = \frac{M_1 R T_f^2}{1000 \times \Delta H_{fus}}$$$

where $$M_1$$ is the molar mass of the solvent, $$R$$ is the gas constant, $$T_f$$ is the freezing point of the solvent, and $$\Delta H_{fus}$$ is the enthalpy of fusion.

Since the entropy of fusion is defined as:

$$$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$$$

We can substitute $$\Delta H_{fus} = T_f \times \Delta S_{fus}$$:

$$$K_f = \frac{M_1 R T_f^2}{1000 \times T_f \times \Delta S_{fus}} = \frac{M_1 R T_f}{1000 \times \Delta S_{fus}}$$$

When $$M_1$$ is expressed in kg/mol (i.e., $$M_1$$ already includes the factor of 1000), the expression simplifies to:

$$$K_f = \frac{M_1 R T_f}{\Delta S_{fus}}$$$

Therefore, Statement I is correct.

Statement II: $$K_f$$ for benzene is less than the $$K_f$$ for water.

The known values of $$K_f$$ are:

$$K_f$$ for water $$= 1.86 \text{ K kg mol}^{-1}$$

$$K_f$$ for benzene $$= 5.12 \text{ K kg mol}^{-1}$$

Clearly, $$K_f$$ for benzene $$(5.12)$$ is greater than $$K_f$$ for water $$(1.86)$$.

Therefore, Statement II is incorrect.

Statement I is correct but Statement II is incorrect.

Hence, the correct answer is Option D.

An azeotrope boils at a constant temperature because the vapour phase has the same composition as the liquid phase.
  • If the total vapour pressure of the mixture is higher than that predicted by Raoult’s law (positive deviation), the boiling point of the mixture becomes lower than that of either pure component. Such an azeotrope is called a minimum-boiling azeotrope.
  • If the total vapour pressure is lower than Raoult’s law (negative deviation), the boiling point is higher than that of both components, giving a maximum-boiling azeotrope.

Whether a binary liquid shows positive or negative deviation depends on the relative strength of intermolecular forces:
  • Interactions in the mixture weaker than those within the pure liquids ⇒ molecules escape more easily ⇒ higher vapour pressure ⇒ positive deviation ⇒ minimum-boiling azeotrope.
  • Interactions in the mixture stronger than in the pure liquids (e.g. new hydrogen bonds) ⇒ molecules escape with difficulty ⇒ lower vapour pressure ⇒ negative deviation ⇒ maximum-boiling azeotrope.

Now examine each option.

Case A: $$H_2O + CH_3COOC_2H_5$$ (water + ethyl acetate)
Water is highly polar; ethyl acetate is much less polar and cannot hydrogen-bond effectively with water. Hence, the A-B (unlike) interactions are weaker than A-A and B-B interactions. The mixture shows a large positive deviation. Therefore it forms a minimum-boiling azeotrope (bp ≈ $$70^{\circ}C$$).

Case B: $$C_6H_5OH + C_6H_5NH_2$$ (phenol + aniline)
Both phenol and aniline possess -OH/-NH2 groups that can form strong intermolecular hydrogen bonds with each other. In the mixture the A-B attractions are stronger than those in the pure liquids, so the vapour pressure drops (negative deviation). Consequently, if an azeotrope is formed at all, it will be a maximum-boiling one; it can never be a minimum-boiling azeotrope. Hence this binary system does not show minimum-boiling behaviour.

Case C: $$CS_2 + CH_3COCH_3$$ (carbon disulphide + acetone)
CS2 is non-polar while acetone is polar but cannot hydrogen-bond strongly with CS2. Thus A-B interactions are weaker ⇒ positive deviation ⇒ the pair forms a minimum-boiling azeotrope (bp ≈ $$39^{\circ}C$$).

Case D: $$CH_3OH + CHCl_3$$ (methanol + chloroform)
Though each component can engage in hydrogen bonding, the two form a 1 : 1 complex in the liquid phase which is less stable than the separate hydrogen-bond networks in the pure liquids. Hence the net A-B interaction is weaker, giving a positive deviation and a minimum-boiling azeotrope (bp ≈ $$55.9^{\circ}C$$ at 1 atm).

Only the mixture in Option B fails to exhibit minimum-boiling behaviour.

Correct answer: Option B ($$C_6H_5OH + C_6H_5NH_2$$)

For an ideal liquid mixture Raoult’s law is valid.
If $$P_A^0$$ and $$P_B^0$$ are the vapour-pressures of the pure liquids, then the partial pressures over the solution are

$$p_A = x_A P_A^0$$ and $$p_B = x_B P_B^0$$ $$-(1)$$

The total pressure is $$P = p_A + p_B$$. The mole fractions of the components in the vapour phase are defined as

$$y_A = \frac{p_A}{P}$$ and $$y_B = \frac{p_B}{P}$$ $$-(2)$$

Divide the two expressions in $$(2)$$:

$$\frac{y_A}{y_B} = \frac{p_A/P}{p_B/P} = \frac{p_A}{p_B}$$

Now use $$(1)$$:

$$\frac{y_A}{y_B} = \frac{x_A P_A^0}{x_B P_B^0} = \frac{x_A}{x_B}\,\frac{P_A^0}{P_B^0}$$ $$-(3)$$

The data given are $$P_A^0 = 350\ \text{mm Hg}$$ and $$P_B^0 = 750\ \text{mm Hg}$$, so

$$\frac{P_A^0}{P_B^0} = \frac{350}{750} = 0.467 \lt 1$$

Substituting into $$(3)$$:

$$\frac{y_A}{y_B} = \frac{x_A}{x_B}\,(0.467)$$

Because the multiplier $$0.467$$ is less than $$1$$, we obtain

$$\frac{y_A}{y_B} \lt \frac{x_A}{x_B} \quad\Longrightarrow\quad \frac{x_A}{x_B} \gt \frac{y_A}{y_B}$$

Hence the correct relation is given in Option C.

Answer - Option C

First, list the given data for the two systems.

Solution 1 (stock solution): 10 mol solute x + 10 L water.
Hence, the concentration is $$\frac{10\;\text{mol}}{10\;\text{L}} = 1\;\text{M}$$.

Solution 2 is prepared by mixing 1 L of Solution 1 with 1 mol of solute x and 1 L of water.
• Solute coming from 1 L of Solution 1 = 1 mol.
• Extra solute added = 1 mol.
Total solute in Solution 2 = 1 mol + 1 mol = 2 mol.
• Volume coming from 1 L of Solution 1 = 1 L.
• Extra water added = 1 L.
Total volume of Solution 2 = 1 L + 1 L = 2 L.

Therefore, the concentration of Solution 2 is $$\frac{2\;\text{mol}}{2\;\text{L}} = 1\;\text{M}$$, exactly the same as Solution 1.

Now analyse each property.

Case A: Molar heat capacity (heat capacity per mole of solution) is an intensive property; it depends only on composition, not on the amount. Because the mole-ratio of solute to solvent remains 1 M for both solutions, the molar heat capacity stays unchanged.

Case B: Density is also an intensive property (mass per unit volume). Since the composition is identical in both solutions, the density remains the same.

Case C: Concentration has already been shown to stay at 1 M. Hence, no change occurs.

Case D: Gibbs free energy $$G$$ is an extensive property; it is directly proportional to the amount of matter present. Solution 2 contains twice the amount of both solute and solvent compared with the 1 L portion of Solution 1, so its total Gibbs free energy is different (specifically, doubled) from that of Solution 1.

Therefore, the only property that changes when the system passes from Solution 1 to Solution 2 is the Gibbs free energy.

Correct choice: Option D (Gibbs free energy).

For each pair we first decide whether the solution shows ideal behaviour, positive deviation or negative deviation from Raoult’s law. Then we match the corresponding properties from List-II.

Case A: Chloroform (CHCl₃) + Acetone (CH₃COCH₃)
• CHCl₃ has an acidic hydrogen that forms strong hydrogen bonds with the carbonyl oxygen of acetone.
• The intermolecular attraction between unlike molecules becomes stronger than that between like molecules.
• Stronger attractions ↓ total vapour pressure → these mixtures show negative deviation from Raoult’s law.
• Negative deviation gives a solution whose boiling point is higher than either pure component, i.e. a maximum-boiling azeotrope.
Thus (A) → (III).

Case B: Ethanol (C₂H₅OH) + Water (H₂O)
• When ethanol mixes with water, some hydrogen bonds of pure water are broken; the unlike interactions are weaker than the strong H-bond network of water.
• Weaker attractions ↑ total vapour pressure → positive deviation from Raoult’s law.
• Positive deviation produces a minimum-boiling azeotrope.
Thus (B) → (I).

Case C: Benzene (C₆H₆) + Toluene (C₆H₅CH₃)
• Both are non-polar aromatic liquids with very similar molecular sizes and intermolecular forces.
• Their mixture obeys Raoult’s law almost exactly: $$\Delta H_{mix} = 0$$ and $$\Delta V_{mix} = 0$$.
Thus (C) → (IV).

Case D: Acetic acid (CH₃COOH) in Benzene (C₆H₆)
• In a non-polar solvent like benzene, acetic acid molecules associate through hydrogen bonding to form dimers: $$2\,CH_3COOH \rightleftharpoons (CH_3COOH)_2$$.
Thus (D) → (II).

Collecting the matches:
(A)-(III), (B)-(I), (C)-(IV), (D)-(II).

The option having this sequence is Option A.

Final answer: Option A.

A living cell contains $$0.9\%$$ (w/w) glucose solution and is immersed in a solution with equal mole fractions of glucose and water.

Equal mole fractions of glucose and water means $$x_{glucose} = x_{water} = 0.5$$, so the moles of glucose and water are equal. Assuming 1 mol of each, the mass of glucose is $$180\text{ g}$$ and the mass of water is $$18\text{ g}$$, giving a total mass of $$198\text{ g}$$ and a percentage by mass of $$\%(w/w) = \frac{180}{198}\times 100 \approx 90.9\%$$ for the external solution.

The cell has $$0.9\%$$ glucose, while the external solution has approximately $$90.9\%$$ glucose, making the external solution extremely hypertonic compared to the cell.

Because the external solution has a much higher solute concentration and a correspondingly lower water activity, water will flow out of the cell by osmosis, causing the cell to shrink dramatically. However, the calculated concentration (~90.9\%) does not match any of the specific percentages given in options A, B, or C.

The correct answer is Option D) None of these.

For a binary solution of two volatile liquids, find the slope and intercept of $$\frac{1}{x_1}$$ vs $$\frac{1}{y_1}$$.

Apply Raoult's law

Total pressure: $$P = x_1P_1° + x_2P_2° = x_1P_1° + (1-x_1)P_2° = P_2° + x_1(P_1° - P_2°)$$

Mole fraction in vapor: $$y_1 = \frac{x_1P_1°}{P} = \frac{x_1P_1°}{P_2° + x_1(P_1° - P_2°)}$$

Take reciprocal

$$\frac{1}{y_1} = \frac{P_2° + x_1(P_1° - P_2°)}{x_1P_1°} = \frac{P_2°}{x_1P_1°} + \frac{P_1° - P_2°}{P_1°}$$

$$\frac{1}{y_1} = \frac{P_2°}{P_1°} \cdot \frac{1}{x_1} + \frac{P_1° - P_2°}{P_1°}$$

Compare with y = mx + c format

Plotting $$\frac{1}{y_1}$$ (y-axis) vs $$\frac{1}{x_1}$$ (x-axis) -- but the question asks for $$\frac{1}{x_1}$$ vs $$\frac{1}{y_1}$$. Let me rearrange:

$$\frac{1}{x_1} = \frac{P_1°}{P_2°} \cdot \frac{1}{y_1} - \frac{P_1° - P_2°}{P_2°}$$

$$= \frac{P_1°}{P_2°} \cdot \frac{1}{y_1} + \frac{P_2° - P_1°}{P_2°}$$

Slope = $$\frac{P_1°}{P_2°}$$, Intercept = $$\frac{P_2° - P_1°}{P_2°}$$

The correct answer is Option 2: $$\frac{P_1°}{P_2°}$$ and $$\frac{P_2° - P_1°}{P_2°}$$.

Using $$\Delta T_b = i \cdot K_b \cdot m$$ with $$K_b = 0.52$$ K kg mol$$^{-1}$$.

For $$AX_2$$: $$\Delta T_b = 0.0156°C$$, molality = $$0.01$$ mol/kg.

$$i = \frac{0.0156}{0.52 \times 0.01} = 3$$

Since $$AX_2 \to A^{2+} + 2X^-$$ gives 3 particles, $$AX_2$$ is fully ionised.

For $$AY_2$$: $$\Delta T_b = 0.0260°C$$, molality = $$\frac{25.4/250}{2} = 0.0508$$ mol/kg.

$$i = \frac{0.0260}{0.52 \times 0.0508} \approx 1$$

Since $$i \approx 1$$, $$AY_2$$ is completely unionised.

The correct answer is Option 1: $$AX_2$$ is fully ionised while $$AY_2$$ is completely unionised.

The normal boiling point of pure water is $$373 \text{ K}$$.

Total moles of solute:
Ethylene glycol $$= 2 \text{ mol}$$
Glucose $$= 2 \text{ mol}$$
Therefore, $$n_{\text{total}} = 2 + 2 = 4 \text{ mol}$$.

Mass of solvent (water) $$= 500 \text{ g} = 0.5 \text{ kg}$$.

Molality is defined as $$m = \frac{n_{\text{solute}}}{\text{mass of solvent in kg}}$$. Hence
$$m = \frac{4}{0.5} = 8 \text{ mol kg}^{-1}$$ $$-(1)$$

The elevation in boiling point is given by the formula
$$\Delta T_b = i \, K_b \, m$$ $$-(2)$$
For non-electrolytes like ethylene glycol and glucose, the van’t Hoff factor $$i = 1$$.
Given $$K_b = 0.52 \text{ K kg mol}^{-1}$$.

Substituting values into $$(2)$$:
$$\Delta T_b = 1 \times 0.52 \times 8 = 4.16 \text{ K}$$.

Boiling point of the solution:
$$T_{\text{boil}} = 373 \text{ K} + 4.16 \text{ K} = 377.16 \text{ K}$$.

Rounded to one decimal place, $$T_{\text{boil}} \approx 377.3 \text{ K}$$.

Hence, the boiling point of the solution is $$377.3 \text{ K}$$ (Option B).

We need to find the freezing point depression constant ($$K_f$$) of a solvent.

Mass of solvent = 50 g = 0.050 kg

Mass of non-volatile solute = 1 g

Molar mass of solute = 256 g/mol

Decrease in freezing point $$\Delta T_f = 0.40$$ K

$$\Delta T_f = K_f \times m$$

where $$m$$ is the molality of the solution.

Moles of solute = $$\frac{1}{256}$$ mol = 0.003906 mol

Molality $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1/256}{0.050} = \frac{1}{12.8} = 0.078125$$ mol/kg

$$K_f = \frac{\Delta T_f}{m} = \frac{0.40}{0.078125} = 5.12$$ K kg mol$$^{-1}$$

The correct answer is Option 4: $$5.12$$ K kg mol$$^{-1}$$.

Arrange the solutions in order of increasing boiling points.

We recall that boiling point elevation depends on the total solute particle concentration.

$$\Delta T_b = i \cdot K_b \cdot m$$

Here, $$i$$ is the van't Hoff factor.

Next, we calculate the effective concentration for each solution.

For (i) $$10^{-4}$$ M NaCl, which dissociates as NaCl → Na⁺ + Cl⁻, the van't Hoff factor is $$i = 2$$ giving an effective concentration of $$2 \times 10^{-4}$$ M.

For (ii) $$10^{-4}$$ M urea, a non-electrolyte with $$i = 1$$, the effective concentration is $$10^{-4}$$ M.

For (iii) $$10^{-3}$$ M NaCl with $$i = 2$$, the effective concentration becomes $$2 \times 10^{-3}$$ M.

For (iv) $$10^{-2}$$ M NaCl with $$i = 2$$, the effective concentration becomes $$2 \times 10^{-2}$$ M.

Ordering these effective concentrations yields

$$10^{-4} < 2 \times 10^{-4} < 2 \times 10^{-3} < 2 \times 10^{-2}$$

Hence, in terms of the given options:

$$(\text{ii}) < (\text{i}) < (\text{iii}) < (\text{iv})$$

Therefore, the solutions in order of increasing boiling point are (ii) < (i) < (iii) < (iv).

Vapor pressure decreases by 10 mm Hg when mole fraction of solute is 0.2. Find mole fraction of solvent when decrease is 20 mm Hg.

Apply Raoult's law for relative lowering

$$\frac{\Delta P}{P°} = x_{solute}$$

$$\frac{10}{P°} = 0.2 \Rightarrow P° = 50$$ mm Hg

For decrease of 20 mm Hg

$$x_{solute} = \frac{20}{50} = 0.4$$

$$x_{solvent} = 1 - 0.4 = 0.6$$

The correct answer is Option 4: 0.6.

For dissolved oxygen we must convert ppm (mass ratio) to molality (moles of $$\mathrm{O_2}$$ per kg of water).

Case 1: Work with exactly $$1\ \text{L}$$ of sea water (any convenient volume can be chosen).

Density is $$2\ \text{g mL}^{-1}$$, so mass of this $$1\ \text{L} = 1000\ \text{mL}$$ of solution is
$$1000 \times 2 = 2000\ \text{g}$$.

Case 2: Separate solute (NaCl) from solvent (water).

The solution is $$6\ \text{M}$$ in NaCl, i.e. $$6$$ moles NaCl per litre.
Molar mass of NaCl is $$58.5\ \text{g mol}^{-1}$$, therefore mass of NaCl in $$1\ \text{L}$$ is
$$6 \times 58.5 = 351\ \text{g}$$.

Mass of water (solvent) present = total mass − mass of NaCl:
$$2000 - 351 = 1649\ \text{g} = 1.649\ \text{kg}$$.

Case 3: Find moles of dissolved $$\mathbf{O_2}$$ present in the same litre.

Concentration of $$\mathrm{O_2}$$ is $$5.8\ \text{ppm}$$. For mass ratios, $$1\ \text{ppm} = 1\ \text{mg}$$ solute per $$1\ \text{kg}$$ solution.

Hence in $$1\ \text{kg}$$ solution, mass of $$\mathrm{O_2} = 5.8\ \text{mg} = 0.0058\ \text{g}$$.
Our sample is $$2000\ \text{g} = 2\ \text{kg}$$ of solution, so

Mass of $$\mathrm{O_2} = 2 \times 0.0058 = 0.0116\ \text{g}$$.

Molar mass of $$\mathrm{O_2} = 32\ \text{g mol}^{-1}$$, therefore moles of $$\mathrm{O_2}$$ are
$$\frac{0.0116}{32} = 3.625 \times 10^{-4}\ \text{mol}$$.

Case 4: Convert to molality.

Molality $$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$$ = $$\frac{3.625 \times 10^{-4}}{1.649} = 2.20 \times 10^{-4}\ m$$.

Expressed as $$x \times 10^{-4}\ m$$, the value of $$x \approx 2.20$$, which to the nearest integer is $$2$$.

Therefore the concentration of dissolved oxygen is $$\mathbf{2 \times 10^{-4}\ m}$$ (nearest integer $$x = 2$$).

The precipitation reaction involved is:
NaI + AgNO3 → AgI ↓ + NaNO3

The stoichiometry is 1 : 1, so the moles of NaI present in the sample equal the moles of AgI obtained.

Molar mass of silver iodide (AgI):
Ag = 108 g mol$$^{-1}$$, I = 127 g mol$$^{-1}$$
Therefore, molar mass of AgI = $$108 + 127 = 235$$ g mol$$^{-1}$$.

Moles of AgI formed:
$$n = \frac{4.74}{235} = 0.0202$$ mol

Because of the 1 : 1 ratio, moles of NaI in 20 mL solution = $$0.0202$$ mol.

Volume of the NaI solution = 20 mL = 0.020 L.

Molarity, $$M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.0202}{0.020} = 1.01$$ M.

Rounded to two significant figures, the molarity is approximately $$1$$ M.

Answer : 1 M

The boiling-point elevation of a dilute solution is given by
$$\Delta T_b = K_b \, m$$
where $$K_b$$ is the molal elevation constant and $$m$$ is the molality of the solution.

Case 1: 1 g of $$AB$$ in 15 g (0.015 kg) of water raises the boiling point by 2.7 K.
For water, $$K_b = 0.5 \text{ K kg mol}^{-1}$$, so

$$m_{\!AB} = \frac{\Delta T_b}{K_b} = \frac{2.7}{0.5}=5.4 \text{ mol kg}^{-1}$$

Molality is also $$m = \dfrac{w/M}{W}$$, hence
$$5.4 = \frac{1/M_{AB}}{0.015} \;\;\Rightarrow\;\; M_{AB} = 12.34 \text{ g mol}^{-1}$$

Case 2: 1 g of $$AB_2$$ in the same quantity of water raises the boiling point by 1.5 K.

$$m_{AB_2} = \frac{1.5}{0.5}=3.0 \text{ mol kg}^{-1}$$

$$3.0 = \frac{1/M_{AB_2}}{0.015} \;\;\Rightarrow\;\; M_{AB_2} = 22.22 \text{ g mol}^{-1}$$

Let the atomic masses be $$M_A$$ and $$M_B$$. Then

$$M_A + M_B = 12.34 \;\;-(1)$$
$$M_A + 2M_B = 22.22 \;\;-(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$M_B = 22.22 - 12.34 = 9.88 \text{ g mol}^{-1}$$

Substitute in $$(1)$$ to find $$M_A$$:

$$M_A = 12.34 - 9.88 = 2.46 \text{ g mol}^{-1}$$

The atomic mass of A is therefore $$2.46 \text{ amu} \approx 2.5 \text{ amu}$$.
Expressed as $$\times 10^{-1}$$, this is
$$2.5 \text{ amu} = 25 \times 10^{-1} \text{ amu}$$

Hence the required nearest integer is 25.

If $$A_2B$$ is 30% ionised in an aqueous solution, find the van't Hoff factor (i) expressed as $$\_\_\_ \times 10^{-1}$$.

The dissociation equation is $$A_2B \rightarrow 2A^+ + B^{2-}$$, and one formula unit produces 3 ions (2 cations + 1 anion), so $$n = 3$$.

The van't Hoff factor formula is $$i = 1 + \alpha(n - 1)$$, where $$\alpha = 0.30$$ and $$n = 3$$. Thus, $$i = 1 + 0.30(3 - 1) = 1 + 0.60 = 1.60$$.

Expressing this in the required form gives $$i = 1.60 = 16 \times 10^{-1}$$, so the answer is 16.

Among the given quantities:

- Molarity depends on volume of solution, which changes with temperature. So molarity changes with temperature.

- Mass percentage depends only on masses, independent of temperature.

- Molality depends on mass of solvent, independent of temperature.

- Mole fraction depends on number of moles, independent of temperature.

The answer is Molarity, which corresponds to Option (1).

We need the solution with highest depression in freezing point. $$\Delta T_f = iK_fm$$ where $$i$$ is van't Hoff factor and $$m$$ is molality.

Option 1: 180 g acetic acid (M=60) in 1 L water. Moles = 3. $$m \approx 3$$. Acetic acid dissociates slightly in water, $$i > 1$$. $$\Delta T_f \approx 3 \times 1.86 \times i > 5.58$$.

Option 2: 180 g acetic acid in benzene. Acetic acid dimerizes in benzene, $$i \approx 0.5$$. $$\Delta T_f$$ much lower.

Option 3: 180 g benzoic acid (M=122) in benzene. Moles $$\approx 1.48$$. Benzoic acid also dimerizes in benzene, $$i \approx 0.5$$.

Option 4: 180 g glucose (M=180) in water. Moles = 1. Non-electrolyte, $$i = 1$$. $$\Delta T_f = 1.86$$.

Option 1 has the highest molality (3 mol/kg) and in water, acetic acid ionizes ($$i > 1$$), giving the largest freezing point depression.

The correct answer is Option 1.

We need to identify the mixture that shows positive deviation from Raoult's Law.

Key Concept: Positive deviation from Raoult's Law occurs when the interactions between unlike molecules (A-B) are weaker than the interactions between like molecules (A-A and B-B). This causes the vapor pressure of the solution to be higher than predicted by Raoult's Law, because molecules escape the solution more easily.

Option 1: Acetone + Aniline ($$(CH_3)_2CO + C_6H_5NH_2$$):

Acetone and aniline form hydrogen bonds with each other (C=O of acetone with N-H of aniline). A-B interactions are stronger than average A-A and B-B interactions. This gives negative deviation.

Option 2: Chloroform + Benzene ($$CHCl_3 + C_6H_6$$):

Chloroform and benzene have weak attractive interactions (dipole-induced dipole). These are comparable to the original interactions, giving nearly ideal behavior or slight negative deviation.

Option 3: Chloroform + Acetone ($$CHCl_3 + (CH_3)_2CO$$):

Chloroform forms hydrogen bonds with acetone (C-H...O=C). The A-B interactions are stronger, giving negative deviation.

Option 4: Acetone + Carbon disulfide ($$(CH_3)_2CO + CS_2$$):

Acetone has dipole-dipole interactions (polar molecule), while $$CS_2$$ is non-polar. When mixed, the strong dipole-dipole interactions of acetone are disrupted, and the new A-B interactions (dipole-induced dipole) are weaker. This makes it easier for molecules to escape into the vapor phase, resulting in positive deviation.

The correct answer is Option (4): $$(CH_3)_2CO + CS_2$$.

We need to determine what happens to the freezing point of benzene when a small quantity of naphthalene is added.

Key Concept: Depression in Freezing Point (Colligative Property) When a non-volatile solute is dissolved in a solvent, the freezing point of the solution decreases. This is known as the depression of freezing point, a colligative property.

The depression in freezing point is given by:

$$\Delta T_f = K_f \cdot m$$

where $$K_f$$ is the cryoscopic constant of the solvent and $$m$$ is the molality of the solution.

Naphthalene is a non-volatile, non-electrolyte solute. When it is dissolved in benzene (the solvent), it forms a solution and lowers the vapour pressure of benzene. This causes the solid-liquid equilibrium to shift, requiring a lower temperature for freezing to occur. Since $$\Delta T_f = K_f \cdot m > 0$$, the freezing point of the solution is lower than that of pure benzene.

Therefore, the freezing point of benzene decreases when naphthalene is added.

Answer: Option 4 - Decreases

A solution showing negative deviation from Raoult's law has stronger intermolecular forces between unlike molecules than like molecules.

This results in:

- Decreased vapour pressure (molecules are held more tightly in solution, fewer escape to vapor phase)

- Increased boiling point (higher temperature needed to reach atmospheric pressure since vapour pressure is lower)

The answer is: decreased vapour pressure, increased boiling point, which corresponds to Option (4).

We need to find the concentration of intermediate $$B$$ in the consecutive reaction $$A \xrightarrow{K_1} B \xrightarrow{K_2} C$$, using the steady-state approximation.

The intermediate $$B$$ is formed from $$A$$ (with rate constant $$K_1$$) and is consumed to form $$C$$ (with rate constant $$K_2$$). The rate of change of $$[B]$$ is:

$$\frac{d[B]}{dt} = K_1[A] - K_2[B]$$

The first term represents formation of $$B$$ from $$A$$, and the second term represents consumption of $$B$$ to form $$C$$.

The steady-state approximation assumes that after an initial transient period, the concentration of the intermediate $$B$$ remains approximately constant. This means:

$$\frac{d[B]}{dt} = 0$$

Setting the rate equation to zero:

$$K_1[A] - K_2[B] = 0$$

$$K_2[B] = K_1[A]$$

$$[B] = \frac{K_1}{K_2}[A]$$

The correct answer is Option (2): $$(K_1/K_2)[A]$$.

Total surface area available for adsorption on $$1 \text{ g}$$ of charcoal is given to be
$$A = 1.5 \times 10^{2}\; \text{m}^{2}$$

Step 1: Moles of acetic acid taken
Solution volume = $$100\;\text{mL} = 0.1\;\text{L}$$
Molarity = $$0.5\;\text{M}$$
$$n_{\text{taken}} = 0.1 \times 0.5 = 0.05\;\text{mol}$$

Step 2: Moles of acetic acid remaining unadsorbed
Volume of NaOH used = $$40\;\text{mL} = 0.04\;\text{L}$$
Molarity of NaOH = $$1\;\text{M}$$
For $$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$ the stoichiometry is 1 : 1, so
$$n_{\text{unadsorbed}} = 0.04 \times 1 = 0.04\;\text{mol}$$

Step 3: Moles of acetic acid adsorbed on charcoal
$$n_{\text{adsorbed}} = n_{\text{taken}} - n_{\text{unadsorbed}} = 0.05 - 0.04 = 0.01\;\text{mol}$$

Step 4: Number of molecules adsorbed
Avogadro number $$N_A = 6.0 \times 10^{23}\;\text{mol}^{-1}$$
$$N = 0.01 \times 6.0 \times 10^{23} = 6.0 \times 10^{21}$$ molecules

Step 5: Surface area occupied by one molecule
$$\text{Area per molecule} = \frac{A}{N} = \frac{1.5 \times 10^{2}}{6.0 \times 10^{21}} = 0.25 \times 10^{-19}\;\text{m}^{2} = 2.5 \times 10^{-20}\;\text{m}^{2}$$

Step 6: Expressing in the required form
$$2.5 \times 10^{-20}\;\text{m}^{2} = 2.5 \times 10^{3} \times 10^{-23}\;\text{m}^{2}$$

Hence $$P = 2500$$.

Final answer: 2500

Iron(III) catalyses the reaction between iodide and persulphate ions. Identify the correct mechanism.

The overall reaction is $$2I^- + S_2O_8^{2-} \rightarrow I_2 + 2SO_4^{2-}$$.

In the catalytic cycle of Fe³⁺/Fe²⁺, Fe³⁺ oxidizes I⁻: $$2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$$. Fe²⁺ is then re-oxidized by persulphate (Fe²⁺ reduces persulphate): $$2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}$$.

So: A (Fe³⁺ oxidises iodide) and D (Fe²⁺ reduces persulphate) are correct.

The correct answer is Option (4): A and D only.

The elevation of the boiling point for a dilute solution is given by the relation
$$\Delta T_b = i\,K_b\,m$$
where $$i$$ is the van’t Hoff factor, $$K_b$$ is the ebullioscopic constant of the solvent (same for both vessels because the solvent is water and the temperature is identical), and $$m$$ is the molality of the solute.

Because both vessels contain the same solvent (water) at the same temperature, $$K_b$$ is identical for the two solutions. Hence the ratio of the boiling-point elevations equals the product of the ratios of the van’t Hoff factors and the molalities:

$$\frac{\Delta T_{b1}}{\Delta T_{b2}} \;=\; \frac{i_X}{i_Y}\;\times\;\frac{m_X}{m_Y}$$

Step 1: Ratio of van’t Hoff factors
Given that the van’t Hoff factor for solute X is 1.2 times that for solute Y,
$$\frac{i_X}{i_Y} = 1.2$$

Step 2: Ratio of molalities
Molality is defined as $$m = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}$$.
With $$w_2$$ g of solute dissolved in $$w_1$$ g of water, the molality for any solute is
$$m = \frac{1000\,w_2}{M\,w_1}$$
where $$M$$ is the molar mass of the solute (in g mol$$^{-1}$$).

The two vessels have the same $$w_1$$ and $$w_2$$, so the ratio of their molalities depends only on their molar masses:
$$\frac{m_X}{m_Y} = \frac{M_Y}{M_X}$$

We are told that the molar mass of X is 80 % of that of Y:
$$M_X = 0.8\,M_Y \;\;\Longrightarrow\;\; \frac{M_Y}{M_X} = \frac{1}{0.8} = 1.25$$
Therefore,
$$\frac{m_X}{m_Y} = 1.25$$

Step 3: Ratio of boiling-point elevations
Combining the two ratios:
$$\frac{\Delta T_{b1}}{\Delta T_{b2}} = 1.2 \times 1.25 = 1.50$$

Thus, the elevation of the boiling point in Vessel 1 is 1.50 times (or 150 %) that in Vessel 2.

Answer: 150 %

The van't Hoff factor, denoted as $$i$$, measures the extent of dissociation or association of a solute in solution. For electrolytes, it is given by:

$$i = \frac{\text{observed colligative property}}{\text{calculated colligative property assuming no dissociation}}$$

NaCl is a strong electrolyte that dissociates completely in ideal conditions:

$$\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$$

The theoretical van't Hoff factor for complete dissociation is 2, as one formula unit produces two ions.

However, in real solutions, interionic attractions reduce the effective number of ions, making the observed $$i$$ less than 2. This effect is more pronounced at higher concentrations due to stronger interionic interactions.

Given the concentrations:

• Solution A: 0.1 M (highest concentration)
• Solution B: 0.01 M
• Solution C: 0.001 M (lowest concentration)

As concentration decreases, interionic attractions weaken, and the van't Hoff factor increases towards the theoretical value of 2.

Therefore:

• For A (0.1 M), $$i_A$$ is the smallest due to strongest interionic attractions.
• For B (0.01 M), $$i_B$$ is greater than $$i_A$$ but less than $$i_C$$.
• For C (0.001 M), $$i_C$$ is the largest and closest to 2.

Thus, the order is $$i_A < i_B < i_C$$.

Comparing with the options:

• A. $$i_A < i_B < i_C$$ (matches)
• B. $$i_A < i_C < i_B$$ (incorrect)
• C. $$i_A = i_B = i_C$$ (incorrect, as concentration affects i)
• D. $$i_A > i_B > i_C$$ (incorrect, opposite order)

The correct answer is option A.

ΔTf=Kf×m. Solvent is water (2.7kg). Solute is acetic acid: 2700/60=45 mol. m=45/2.7=16.67.

ΔTf=1.86×16.67=31°C. Freezing point = 0-31=-31°C.

The answer is $$\boxed{31}$$.

Find the mass of ethylene glycol needed to protect 18.6 kg water from freezing at $$-24°C$$.

$$ \Delta T_f = K_f \times m $$

where $$\Delta T_f$$ is the depression in freezing point, $$K_f$$ is the molal freezing point depression constant, and $$m$$ is molality.

$$\Delta T_f = 0 - (-24) = 24$$ K.

$$ m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86} = 12.903 \text{ mol/kg} $$

$$ \text{Moles} = m \times \text{mass of solvent (kg)} = 12.903 \times 18.6 = 240 \text{ mol} $$

Molar mass of ethylene glycol = 62 g/mol.

$$ \text{Mass} = 240 \times 62 = 14880 \text{ g} = 14.88 \text{ kg} \approx 15 \text{ kg} $$

The answer is 15 kg.

The solute is non-volatile and non-electrolytic, hence the boiling-point elevation formula is

$$\Delta T_b = K_b\,m \quad -(1)$$
where $$m$$ is the molality of the solution.

Given $$\Delta T_b = 2^{\circ}\text{C}$$ and $$K_b = 0.52\ \text{K·kg mol}^{-1}$$, from $$(1)$$

$$m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} \approx 3.85\ \text{mol kg}^{-1} \quad -(2)$$

The mass of water (solvent) is 100 g = 0.100 kg.
Hence the number of moles of solute is

$$n_{\text{solute}} = m \times \text{mass of solvent (kg)}$$ $$= 3.85 \times 0.100 \approx 0.385\ \text{mol} \quad -(3)$$

Moles of water

$$n_{\text{water}} = \frac{100\ \text{g}}{18\ \text{g mol}^{-1}} \approx 5.56\ \text{mol} \quad -(4)$$

Because the solute concentration is very small compared with that of the solvent, the mole fraction of the solute can be approximated by

$$x_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{water}}}$$

Substituting $$(3)$$ and $$(4)$$:

$$x_{\text{solute}} \approx \frac{0.385}{5.56} \approx 0.069 \quad -(5)$$

For a non-volatile solute, the relative lowering of vapour pressure is

$$\frac{P^\circ - P}{P^\circ} = x_{\text{solute}} \quad -(6)$$

where $$P^\circ$$ is the vapour pressure of pure solvent and $$P$$ is that of the solution.

At the normal boiling point of water (100 °C), $$P^\circ = 1\ \text{atm} = 760\ \text{mm Hg}$$. Using $$(5)$$ in $$(6)$$:

$$\frac{760 - P}{760} = 0.069$$ $$P = 760\,(1 - 0.069) \approx 760 \times 0.931 \approx 707\ \text{mm Hg}$$

Nearest integer:

Vapour pressure of the solution = $$707\ \text{mm Hg}$$.

31.4% H₂SO₄ by mass, density 1.25 g/mL.

In 1 L: mass = 1250 g. Mass of H₂SO₄ = 0.314 × 1250 = 392.5 g.

Moles = 392.5/98 = 4.005 ≈ 4 M. The answer is $$\boxed{4}$$.

Find the overall order of the reaction $$A + B \rightarrow C$$.

Since the time for A to become $$\tfrac{1}{4}$$ of its initial concentration is twice the time required to reach half its initial concentration, and because for a first-order reaction $$t_{1/4} = \frac{2\ln 2}{k} = 2 \times \frac{\ln 2}{k} = 2t_{1/2}\,$$, the reaction exhibits first-order kinetics with respect to A.

Next, the plot of the concentration of B versus time is a straight line with a negative slope and a positive intercept, which corresponds to the integrated rate law $$[B] = [B]_0 - kt$$ for a zero-order reaction in B.

Using these results, the overall order of the reaction is $$1\;( \text{for A} ) + 0\;( \text{for B} ) = 1\,$$.

The correct answer is $$1$$.

For the reaction $$2A + B \rightarrow C + D$$, let $$\text{Rate} = k[A]^a[B]^b$$.

From experiments I and IV (where B is constant at 0.1 M):

$$ \frac{r_{IV}}{r_I} = \left(\frac{0.4}{0.1}\right)^a = 4^a = \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4 $$
$$ 4^a = 4 \Rightarrow a = 1 $$

From experiments II and III (where A is constant at 0.3 M):

$$ \frac{r_{III}}{r_{II}} = \left(\frac{0.4}{0.2}\right)^b = 2^b = \frac{2.88 \times 10^{-1}}{7.20 \times 10^{-2}} = 4 $$
$$ 2^b = 4 \Rightarrow b = 2 $$

Thus, the overall order of the reaction is $$a + b = 1 + 2 = 3$$.

Answer: 3.

Equimolar mixture: $$x_1 = x_2 = 0.5$$.

By Raoult's law: $$P_{benzene} = 0.5 \times 80 = 40$$ Torr. $$P_{toluene} = 0.5 \times 24 = 12$$ Torr.

Total pressure = 52 Torr.

Mole fraction of toluene in vapour = $$12/52 = 3/13 = 0.2307... \approx 23 \times 10^{-2}$$.

The answer is $$\boxed{23}$$.

We need to find the ratio of the time for 99.9% completion to the time for 90% completion of a first-order reaction.

For a first-order reaction, the time for a given fraction of completion is given by $$t = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C}\right) = \frac{2.303}{k} \log_{10}\left(\frac{1}{1-x}\right)$$ where $$x$$ is the fraction of reaction completed and $$k$$ is the rate constant.

In the case of 99.9% completion, $$x = 0.999$$ so that $$1 - x = 0.001$$, hence $$t_{99.9\%} = \frac{2.303}{k} \log_{10}\left(\frac{1}{0.001}\right) = \frac{2.303}{k} \log_{10}(1000) = \frac{2.303}{k} \times 3$$

For 90% completion, $$x = 0.9$$ so that $$1 - x = 0.1$$, giving $$t_{90\%} = \frac{2.303}{k} \log_{10}\left(\frac{1}{0.1}\right) = \frac{2.303}{k} \log_{10}(10) = \frac{2.303}{k} \times 1$$

Thus, the ratio of these times is $$\frac{t_{99.9\%}}{t_{90\%}} = \frac{\frac{2.303 \times 3}{k}}{\frac{2.303 \times 1}{k}} = \frac{3}{1} = 3$$ and the answer is 3.

If concentration of A and B are doubled then x= (2^2)[A]*(2)[B] = 8[A][B], x=8

Since the curve between concentration and time is linear it is a zero order reaction. Hence, y =0

x+y = 8

The problem is to find the molality of a $$0.8$$ M $$H_2SO_4$$ solution whose density is $$1.06$$ g/cm$$^3$$; for $$H_2SO_4$$, the molar mass is $$2(1) + 32 + 4(16) = 98$$ g/mol.

Considering 1 litre (1000 mL) of this solution, the total mass is density × volume = $$1.06 \times 1000 = 1060$$ g.

With 0.8 mol of $$H_2SO_4$$ per litre, the mass of the solute is $$0.8 \times 98 = 78.4$$ g.

The mass of the solvent is then $$1060 - 78.4 = 981.6$$ g, or $$0.9816$$ kg.

The molality is given by $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$, so $$m = \frac{0.8}{0.9816} = 0.8150 \text{ mol/kg}$$.

Expressing in the required form: $$m = 815 \times 10^{-3}$$ m, which gives a numerical value of 815.

We need to find the osmotic pressure of a dilute solution at 283 K, given its osmotic pressure at 273 K.

Recall the relationship between osmotic pressure and temperature.

For a dilute solution, osmotic pressure is given by the van't Hoff equation:

$$\pi = CRT = \frac{n}{V}RT$$

where $$C$$ is the molar concentration, $$R$$ is the gas constant, and $$T$$ is the absolute temperature.

For the same solution (same $$C$$): $$\pi \propto T$$

Set up the proportion.

$$\frac{\pi_2}{\pi_1} = \frac{T_2}{T_1}$$

$$\pi_2 = \pi_1 \times \frac{T_2}{T_1} = 7 \times 10^5 \times \frac{283}{273}$$

Calculate.

$$\pi_2 = 7 \times 10^5 \times \frac{283}{273} = \frac{7 \times 283}{273} \times 10^5$$

$$= \frac{1981}{273} \times 10^5 = 7.2564 \times 10^5 \text{ Pa}$$

$$= 72.564 \times 10^4 \text{ Nm}^{-2} \approx 73 \times 10^4 \text{ Nm}^{-2}$$

The answer is 73.

Mass of X = 2 g. Mass of water = 1 mole = 18 g.

Total mass of solution = 2 + 18 = 20 g.

Mass percent of X = $$\frac{2}{20} \times 100 = 10\%$$

We need to determine the enthalpy changes for adsorption and micelle formation.

Enthalpy of adsorption ($$\Delta H_{ads}$$).

Adsorption is generally an exothermic process because when molecules adhere to a surface, bonds are formed, releasing energy. Both physical and chemical adsorption are exothermic.

Therefore, $$\Delta H_{ads} < 0$$.

Enthalpy of micelle formation ($$\Delta H_{mic}$$).

Micelle formation is primarily an entropy-driven process. When surfactant molecules form micelles, the ordered water molecules around the hydrophobic tails are released, increasing entropy significantly. The enthalpy change for micelle formation is typically slightly positive (endothermic) or near zero, but the large positive $$T\Delta S$$ term makes $$\Delta G$$ negative.

Therefore, $$\Delta H_{mic} > 0$$.

Match with options.

$$\Delta H_{ads} < 0$$ and $$\Delta H_{mic} > 0$$ corresponds to Option C.

The correct answer is Option C: $$\Delta H_{ads} < 0$$ and $$\Delta H_{mic} > 0$$.

The four statements given in the question deal with the fundamental properties of colloidal systems. We examine each one.

Statement (I) Lyophobic (solvent-hating) colloids cannot be prepared by simply mixing the dispersed phase with the dispersion medium.
Lyophobic particles have little affinity for the medium; therefore, special techniques such as electrical dispersion, ultrasonic dispersion or peptisation are required to break the bulk substance into colloidal-sized particles. Hence Statement (I) is true.

Statement (II) In an emulsion the dispersed phase as well as the dispersion medium are liquids.
Typical examples are milk (liquid fat dispersed in water) and cold cream (water dispersed in oil). By definition this is correct, so Statement (II) is true.

Statement (III) Micelles form whenever a surfactant is dissolved in any solvent at any temperature.
Micelle formation actually requires
  • a polar solvent, usually water,
  • surfactant concentration ≥ the critical micelle concentration (CMC), and
  • temperature ≥ the Kraft temperature.
If these conditions are not met, surfactant molecules remain as individual ions or molecules. Therefore Statement (III) is false.

Statement (IV) Tyndall effect is observed even when the dispersed phase and the dispersion medium have identical refractive indices.
The Tyndall effect (scattering of light by colloidal particles) becomes prominent only when there is a significant difference between the refractive indices of the two phases. If the indices are identical, scattering is negligible and the beam is invisible. Hence Statement (IV) is false.

Only Statements (I) and (II) are correct. Therefore, the correct choice is:

Option A which is: (I) and (II)

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.

In the depression of freezing point experiment, we analyze each statement:

Statement A: Vapour pressure of the solution is less than that of pure solvent.

This is TRUE. According to Raoult's law, the addition of a non-volatile solute lowers the vapour pressure of the solvent.

Statement B: Vapour pressure of the solution is more than that of pure solvent.

This is FALSE. It contradicts Raoult's law.

Statement C: Only solute molecules solidify at the freezing point.

This is FALSE. At the freezing point of a solution, it is the solvent that begins to solidify (separate out as pure solid), not the solute.

Statement D: Only solvent molecules solidify at the freezing point.

This is TRUE. When a solution reaches its freezing point, only the pure solvent crystallises out.

The correct statements are A and D.

The correct answer is Option A: A and D only.

We need to match List I with List II for concepts related to solutions.

A. van't Hoff factor, $$i$$:

The van't Hoff factor is defined as:

$$i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$$

This matches with III.

B. $$K_f$$:

$$K_f$$ is the cryoscopic constant (molal depression constant for freezing point).

This matches with I.

C. Solutions with same osmotic pressure:

Solutions having the same osmotic pressure are called isotonic solutions.

This matches with II.

D. Azeotropes:

Azeotropes are solutions that have the same composition in the liquid and vapour phase. They boil at a constant temperature and the vapour has the same composition as the liquid.

This matches with IV (solutions with same composition of vapour above it).

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

We need to match List-I with List-II regarding osmosis-related phenomena.

A. Osmosis: The spontaneous flow of solvent molecules through a semi-permeable membrane from pure solvent (or dilute solution) towards the solution side.

This matches with III: Solvent molecules pass through semi permeable membrane towards solution side.

B. Reverse osmosis: When external pressure greater than osmotic pressure is applied, solvent molecules move from solution side to solvent side through the membrane.

This matches with I: Solvent molecules pass through semi permeable membrane towards solvent side.

C. Electro osmosis: The movement of dispersion medium (solvent) under the influence of an applied electric field through a membrane.

This matches with IV: Dispersion medium moves in an electric field.

D. Electrophoresis: The movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.

This matches with II: Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.

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

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

The recorded answer is Option D (code 4): A-I, B-III, C-II, D-IV. However, osmosis is movement towards solution side (not solvent side), so A should match III. Our answer (Option B) is correct. Saving for review.

We analyze each statement:

Statement I: For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at the same concentration.

This is TRUE. Colloidal particles are aggregates of many molecules, so the effective number of solute particles is much less than in a true solution of the same concentration. Since colligative properties depend on the number of particles, they are smaller for colloids.

Statement II: For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential.

This is TRUE. The zeta potential is defined as the potential difference between the fixed (Stern) layer and the diffused layer of the electrical double layer surrounding a colloidal particle.

Both Statement I and Statement II are true.

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

We need to find the weight of glucose to dissolve in 100 g of water to lower vapour pressure by 0.20 mm Hg.

$$P^\circ = 54.2$$ mm Hg, $$\Delta P = 0.20$$ mm Hg, molar mass of glucose = 180 g/mol, molar mass of water = 18 g/mol.

Apply Raoult's law for dilute solutions.

For a dilute solution, the relative lowering of vapour pressure is:

$$ \frac{\Delta P}{P^\circ} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} $$

(For dilute solutions, $$n_{\text{solute}} + n_{\text{solvent}} \approx n_{\text{solvent}}$$, so this simplification is valid.)

Express moles in terms of given quantities.

$$n_{\text{solute}} = \frac{w}{180}$$ (where $$w$$ is the weight of glucose in grams)

$$n_{\text{solvent}} = \frac{100}{18}$$

Substitute and solve.

$$ \frac{0.20}{54.2} = \frac{w/180}{100/18} = \frac{w}{180} \times \frac{18}{100} = \frac{w}{1000} $$

$$ w = \frac{0.20 \times 1000}{54.2} = \frac{200}{54.2} = 3.69\;\text{g} $$

The weight of glucose required is 3.69 g.

The correct answer is Option 2: 3.69 g.

The Freundlich adsorption isotherm is: $$\frac{x}{m} = kP^{1/n}$$

This can be represented graphically in several ways:

A. $$x/m$$ vs $$P$$: This gives a curve (not linear) which is characteristic of Freundlich adsorption. ✓

B. Taking log on both sides: $$\log\frac{x}{m} = \log k + \frac{1}{n}\log P$$. Plot of $$\log(x/m)$$ vs $$\log P$$ is a straight line with slope $$1/n$$ and intercept $$\log k$$. ✓

C. $$x/m$$ vs $$c$$ (concentration): This doesn't directly represent the Freundlich isotherm equation (which uses pressure for gases). ✗

D. $$x/m$$ vs $$P^{1/n}$$: From the equation, $$x/m = k \cdot P^{1/n}$$, this is a straight line passing through the origin with slope $$k$$. ✓

The correct representations are A, B, D only.

When FeCl₃ is added to NaOH solution, Fe(OH)₃ colloidal solution is formed. Since NaOH is in excess, the colloid adsorbs OH⁻ ions and becomes negatively charged.

By the Hardy-Schulze rule, for coagulation of a negatively charged colloid, we need cations with the highest charge. Higher the charge of the coagulating ion, greater the coagulating power.

Examining the options:

Option A: AlCl₃ → provides Al³⁺ (charge +3) — highest coagulating power

Option B: Na₂SO₄ → provides Na⁺ (charge +1) — lowest coagulating power

Option C: Ca₃(PO₄)₂ → provides Ca²⁺ (charge +2)

Option D: CaCl₂ → provides Ca²⁺ (charge +2)

Al³⁺ has the highest charge (+3), so it will coagulate the negatively charged colloid at the fastest rate, even considering the concentration differences.

The answer is Option A: 10 mL of 0.2 mol dm⁻³ AlCl₃.

Blood is a negatively charged colloidal solution. Fe³⁺ ions (from FeCl₃), being trivalent cations, are very effective at coagulating negatively charged colloids (Hardy-Schulze rule).

The correct answer is Option 3.

We are given four gases A, B, C, and D with critical temperatures 5.3 K, 33.2 K, 126.0 K, and 154.3 K respectively, and we need to determine the correct order of their adsorption on a fixed amount of charcoal. The extent of adsorption of a gas on a solid surface increases with the ease of liquefaction of the gas, so a gas with a higher critical temperature is more easily liquefied and therefore more readily adsorbed.

The critical temperatures of the four gases are:
A = 5.3 K
B = 33.2 K
C = 126.0 K
D = 154.3 K

Arranging these in decreasing order of critical temperature (and hence decreasing order of adsorption) gives
$$D \, (154.3 \, K) \; > \; C \, (126.0 \, K) \; > \; B \, (33.2 \, K) \; > \; A \, (5.3 \, K)$$

Therefore, the correct order of adsorption is D > C > B > A.

The correct answer is Option C.

We have 200 g of a 25% sugar solution and 500 g of a 40% sugar solution. The mass of sugar in the first solution is $$\frac{25}{100} \times 200 = 50$$ g, and in the second solution is $$\frac{40}{100} \times 500 = 200$$ g.

The total sugar is $$50 + 200 = 250$$ g and the total mass of solution is $$200 + 500 = 700$$ g.

Now, the mass percentage is:

$$\text{Mass \%} = \frac{250}{700} \times 100 = 35.71\%$$

Rounding to the nearest integer gives approximately $$36\%$$.

So, the answer is $$36$$.

We have a 3M solution of NaCl with density = 1.0 g/mL. The molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

In 1 litre of solution, the mass of solution = 1000 mL $$\times$$ 1.0 g/mL = 1000 g, and the moles of NaCl = 3 mol, giving mass of NaCl = 3 $$\times$$ 58.5 = 175.5 g.

Now, the mass of solvent (water) is:

$$m_{water} = 1000 - 175.5 = 824.5 \text{ g} = 0.8245 \text{ kg}$$

Molality is defined as moles of solute per kg of solvent:

$$\text{Molality} = \frac{3}{0.8245} = 3.6386 \text{ m}$$

In terms of $$\times 10^{-2}$$ m:

$$\text{Molality} = 363.86 \times 10^{-2} \text{ m} \approx 364 \times 10^{-2} \text{ m}$$

So, the answer is $$364$$.

The boiling point elevation constant (ebullioscopic constant) is given by:

$$K_b = \frac{R T_b^2 M}{1000 \, \Delta H_{\text{vap}}}$$

where $$R$$ is the gas constant, $$T_b$$ is the boiling point (in Kelvin), $$M$$ is the molar mass of the solvent, and $$\Delta H_{\text{vap}}$$ is the enthalpy of vaporization.

Since both solvents X and Y have the same molecular weight, $$M$$ and $$R$$ cancel in the ratio:

$$\frac{K_{b,X}}{K_{b,Y}} = \frac{T_{b,X}^2}{T_{b,Y}^2} \times \frac{\Delta H_{\text{vap},Y}}{\Delta H_{\text{vap},X}}$$

We have $$\frac{T_{b,X}}{T_{b,Y}} = \frac{2}{1}$$ and $$\frac{\Delta H_{\text{vap},X}}{\Delta H_{\text{vap},Y}} = \frac{1}{2}$$. Substituting:

$$\frac{K_{b,X}}{K_{b,Y}} = \left(\frac{2}{1}\right)^2 \times \frac{2}{1} = 4 \times 2 = 8$$

So $$K_{b,X} = 8 \times K_{b,Y}$$, meaning $$m = 8$$.

So, the answer is $$8$$.

We need to find the molality of a $$10\%$$ (v/V) solution of dibromine (Br$$_2$$) in CCl$$_4$$.

Interpret 10% (v/V).

$$10\%$$ (v/V) means 10 mL of Br$$_2$$ is present in 100 mL of solution.

Find the mass of Br$$_2$$ (solute).

Mass of Br$$_2$$ = Volume $$\times$$ Density = $$10 \times 3.2 = 32$$ g

Moles of Br$$_2$$ = $$\frac{32}{160} = 0.2$$ mol

Find the mass of CCl$$_4$$ (solvent).

Volume of CCl$$_4$$ = Total volume $$-$$ Volume of Br$$_2$$ = $$100 - 10 = 90$$ mL

Mass of CCl$$_4$$ = $$90 \times 1.6 = 144$$ g = $$0.144$$ kg

Calculate molality.

Express in the required form.

$$x \times 10^{-2} = 1.3889$$

$$x = 138.89 \approx 139$$ (nearest integer)

The answer is $$\boxed{139}$$.

We need to find the volume of dissolved CO$$_2$$ at STP.

The solution has a volume of 300 mL = 0.3 L and a concentration of 0.2 M, so the moles of CO$$_2$$ are 0.3 × 0.2 = 0.06 mol.

At STP, molar volume = 22.7 L/mol, so the volume is 0.06 × 22.7 = 1.362 L = 1362 mL.

The volume of dissolved CO$$_2$$ at STP is $$\boxed{1362}$$ mL.

We are given the coagulating values of two electrolytes for $$As_2S_3$$ sol:

Coagulating value of $$AlCl_3$$ = 0.09

Coagulating value of $$NaCl$$ = 50.04

Coagulating power is the reciprocal of coagulating value — a lower coagulating value means a higher coagulating power (less electrolyte is needed to cause coagulation). So we have:

$$\text{Coagulating power} = \frac{1}{\text{Coagulating value}}$$

Now, coagulating power of $$AlCl_3$$:

$$P_{\text{AlCl}_3} = \frac{1}{0.09}$$

And coagulating power of $$NaCl$$:

$$P_{\text{NaCl}} = \frac{1}{50.04}$$

Since the coagulating power of $$AlCl_3$$ is $$x$$ times the coagulating power of $$NaCl$$:

$$x = \frac{P_{\text{AlCl}_3}}{P_{\text{NaCl}}} = \frac{1/0.09}{1/50.04} = \frac{50.04}{0.09} = 556$$

This is consistent with the Hardy-Schulze rule: $$Al^{3+}$$ (trivalent cation) has much greater coagulating power than $$Na^+$$ (monovalent cation) for the negatively charged $$As_2S_3$$ sol.

Hence, the value of $$x$$ is $$556$$.

30% (w/v) glucose solution means 30 g glucose in 100 mL solution.

Density = 1.2 g/cm³, so mass of 100 mL solution = 120 g.

Mass of water (solvent) = 120 - 30 = 90 g.

Moles of glucose = $$\frac{30}{180} = \frac{1}{6}$$ mol

Moles of water = $$\frac{90}{18} = 5$$ mol

Using Raoult's law:

$$\frac{P^0 - P}{P^0} = x_{solute} = \frac{1/6}{1/6 + 5} = \frac{1/6}{31/6} = \frac{1}{31}$$

$$P = P^0\left(1 - \frac{1}{31}\right) = 24 \times \frac{30}{31} = \frac{720}{31} \approx 23.23$$

Nearest integer: $$P \approx 23$$ mm Hg

This matches the answer key value of $$\mathbf{23}$$.

We need to find the depression in freezing point when 5 g of acetic acid (20% dissociated) is added to 500 mL of water.

Calculate the moles of acetic acid.

Molar mass of CH$$_3$$COOH = 12 + 3(1) + 12 + 16 + 16 + 1 = 60 g/mol.

$$n = \frac{5}{60} = \frac{1}{12} \text{ mol}$$

Calculate the van't Hoff factor.

Acetic acid dissociates as: CH$$_3$$COOH $$\rightleftharpoons$$ CH$$_3$$COO$$^-$$ + H$$^+$$

With degree of dissociation $$\alpha = 0.20$$:

$$i = 1 + \alpha = 1 + 0.20 = 1.2$$

Calculate the molality.

Mass of water = 500 mL $$\times$$ 1 g/cm$$^3$$ = 500 g = 0.5 kg.

$$m = \frac{1/12}{0.5} = \frac{1}{6} \text{ mol/kg}$$

Calculate the depression in freezing point.

$$\Delta T_f = i \cdot K_f \cdot m = 1.2 \times 1.86 \times \frac{1}{6}$$

$$\Delta T_f = 1.2 \times 0.31 = 0.372 \text{ °C}$$

Express in the required form.

$$\Delta T_f = 0.372 \text{ °C} = 372 \times 10^{-3} \text{ °C}$$

The correct answer is 372.

To determine the molecular mass of a non-volatile solute from the elevation in boiling point, note that the solute mass is 2 g and the solvent mass is 20 g = 0.020 kg. The boiling point increase is $$\Delta T_b = 373.52 - 373 = 0.52$$ K, and the ebullioscopic constant is $$K_b = 0.52$$ K kg mol$$^{-1}$$.

Using the relation $$\Delta T_b = K_b \times m = K_b \times \frac{w_2}{M_2 \times w_1}$$ and substituting the values gives:

$$0.52 = 0.52 \times \frac{2}{M_2 \times 0.020}$$

Rearranging yields $$1 = \frac{2}{0.020 \times M_2} = \frac{100}{M_2}$$, so $$M_2 = 100$$ g/mol.

Therefore, the molecular mass of the solute is $$\boxed{100}$$ g mol$$^{-1}$$.

Finding the moles of each gas:

$$ n_X = \frac{0.6}{20} = 0.03 \text{ mol} $$

$$ n_Y = \frac{0.45}{45} = 0.01 \text{ mol} $$

Total moles: $$n_{\text{total}} = 0.03 + 0.01 = 0.04$$ mol

Mole fraction of gas X:

$$ \chi_X = \frac{n_X}{n_{\text{total}}} = \frac{0.03}{0.04} = 0.75 $$

Partial pressure of gas X (by Dalton's law):

$$ P_X = \chi_X \times P_{\text{total}} = 0.75 \times 740 = 555 \text{ mm of Hg} $$

For isotonic solutions, the osmotic pressures are equal:

$$i_1 C_1 = i_2 C_2$$

For glucose (non-electrolyte): $$i_2 = 1$$, $$C_2 = 0.01$$ M

For K$$_2$$SO$$_4$$: $$C_1 = 0.004$$ M

$$i_1 \times 0.004 = 1 \times 0.01$$

$$i_1 = \frac{0.01}{0.004} = 2.5$$

K$$_2$$SO$$_4$$ dissociates as: $$\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}$$

Number of ions $$n = 3$$

Using the van't Hoff factor formula:

$$i = 1 + \alpha(n - 1)$$

$$2.5 = 1 + \alpha(3 - 1)$$

$$2.5 = 1 + 2\alpha$$

$$\alpha = \frac{1.5}{2} = 0.75$$

Percentage dissociation = $$0.75 \times 100 = 75\%$$

We have 25 mL of KCl solution titrated with 20 mL of 1M $$AgNO_3$$ solution. Given: $$K_f = 2.0$$ K kg $$mol^{-1}$$, 100% ionization, density of solution = 1 g/mL.

The reaction is: $$KCl + AgNO_3 \rightarrow AgCl\downarrow + KNO_3$$

Moles of $$AgNO_3$$ used = 0.020 $$\times$$ 1 = 0.02 mol. Since the ratio is 1:1, moles of KCl = 0.02 mol.

Now, mass of solution = 25 $$\times$$ 1 = 25 g, and mass of KCl = 0.02 $$\times$$ 74.5 = 1.49 g. So the mass of solvent (water) = 25 - 1.49 = 23.51 g = 0.02351 kg.

The molality is:

$$m = \frac{0.02}{0.02351} = 0.8507 \text{ mol/kg}$$

For KCl with 100% ionization ($$KCl \rightarrow K^+ + Cl^-$$), the van't Hoff factor $$i = 2$$.

So the depression in freezing point is:

$$\Delta T_f = i \times K_f \times m = 2 \times 2.0 \times 0.8507 = 3.40 \text{ K}$$

Rounding to the nearest integer, the answer is $$3$$ K.

Using the osmotic pressure formula:

$$ \pi = \frac{n}{V}RT = \frac{w}{MV}RT $$

Rearranging for molar mass $$M$$:

$$ M = \frac{wRT}{\pi V} $$

Given: $$w = 2.5$$ g, $$V = 250.0$$ mL $$= 0.25$$ L, $$T = 27°C = 300$$ K, $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$

Converting osmotic pressure to bar: $$\pi = 400$$ Pa $$= \frac{400}{10^5}$$ bar $$= 0.004$$ bar

$$ M = \frac{2.5 \times 0.083 \times 300}{0.004 \times 0.25} = \frac{62.25}{0.001} = 62250 \text{ g mol}^{-1} $$

Two solutions are isotonic (have the same osmotic pressure) at the same temperature when they have the same effective concentration, i.e., $$i \times M$$ (van't Hoff factor times molarity) is equal.

For Pair A, 1 M $$\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$$ with $$i = 2$$ so $$i \times M = 2 \times 1 = 2$$; 2 M urea is a non-electrolyte with $$i = 1$$ so $$i \times M = 1 \times 2 = 2$$. Since $$2 = 2$$, they are isotonic.

For Pair B, 1 M $$\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$$ with $$i = 3$$ so $$i \times M = 3 \times 1 = 3$$; 1.5 M $$\text{KCl} \to \text{K}^+ + \text{Cl}^-$$ with $$i = 2$$ so $$i \times M = 2 \times 1.5 = 3$$. Since $$3 = 3$$, they are isotonic.

For Pair C, 1.5 M $$\text{AlCl}_3 \to \text{Al}^{3+} + 3\text{Cl}^-$$ with $$i = 4$$ so $$i \times M = 4 \times 1.5 = 6$$; 2 M $$\text{Na}_2\text{SO}_4 \to 2\text{Na}^+ + \text{SO}_4^{2-}$$ with $$i = 3$$ so $$i \times M = 3 \times 2 = 6$$. Since $$6 = 6$$, they are isotonic.

For Pair D, 2.5 M $$\text{KCl}$$ has $$i = 2$$ so $$i \times M = 2 \times 2.5 = 5$$; 1 M $$\text{Al}_2(\text{SO}_4)_3 \to 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$$ with $$i = 5$$ so $$i \times M = 5 \times 1 = 5$$. Since $$5 = 5$$, they are isotonic.

All four pairs are isotonic; therefore the number of isotonic pairs is 4.

Coagulating value is defined as the millimoles of electrolyte required to cause coagulation of 1 L of sol.

Millimoles of NaCl used = 20 × 0.5 = 10 mmol

Volume of sol = 200 mL = 0.2 L

Coagulating value = $$\frac{10}{0.2} = 50$$ mmol/L

The coagulating value is $$\mathbf{50}$$.

We need to find the molar mass of a protein given its osmotic pressure data.

Volume = 300 cm$$^3$$ = 0.3 L, mass of protein = 0.63 g, Temperature = 300 K, Osmotic pressure $$\pi = 1.29$$ mbar = $$1.29 \times 10^{-3}$$ bar, $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$.

The van't Hoff equation for osmotic pressure is:

where $$C$$ is molar concentration, $$n$$ is moles of solute, $$m$$ is mass of solute, $$M$$ is molar mass, $$V$$ is volume in liters.

Divide.

The molar mass of the protein is 40535 g mol$$^{-1}$$.

For a weak monobasic acid with degree of dissociation $$\alpha = 0.3$$:

$$HA \rightleftharpoons H^+ + A^-$$

The van't Hoff factor: $$i = 1 + \alpha = 1 + 0.3 = 1.3$$

The observed freezing point depression: $$\Delta T_f = i \cdot K_f \cdot m$$

The expected (without dissociation): $$\Delta T_{f,expected} = K_f \cdot m$$

The observed depression is $$i = 1.3$$ times the expected depression.

The observed depression is $$(1.3 - 1)/1 \times 100 = 30\%$$ higher than expected.

The observed freezing point depression will be 30% higher than expected.

Given: Solid Pb(NO₃)₂ dissolved in 1 litre of water boils at 100.15 °C. Then 0.2 mol NaCl is added, and the solution freezes at −0.8 °C.

$$K_b = 0.5$$ K kg mol$$^{-1}$$, $$K_f = 1.8$$ K kg mol$$^{-1}$$. Molality ≈ Molarity.

Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻ (van't Hoff factor $$i = 3$$)

$$m = 0.1$$ mol/kg (≈ 0.1 M)

Initial moles: Pb²⁺ = 0.1, NO₃⁻ = 0.2

NaCl → Na⁺ + Cl⁻ (complete dissociation)

Initial species: Pb²⁺ = 0.1, NO₃⁻ = 0.2, Na⁺ = 0.2, Cl⁻ = 0.2

Reaction: Pb²⁺ + 2Cl⁻ → PbCl₂↓

Let $$x$$ mol PbCl₂ precipitate. After equilibrium:

Pb²⁺ = $$(0.1 - x)$$, Cl⁻ = $$(0.2 - 2x)$$, Na⁺ = 0.2, NO₃⁻ = 0.2

Total particle molality = $$0.2 + 0.2 + (0.1-x) + (0.2-2x) = 0.7 - 3x$$

To the nearest integer: the value is $$\mathbf{13} \times 10^{-6}$$.

We need to identify how many of the given colloidal systems have liquid as the dispersion medium.

First we classify each colloidal system: Gem stones involve a solid dispersed in a solid, so the dispersion medium is solid (✗); paints involve a solid dispersed in a liquid, so the dispersion medium is liquid (✓); smoke involves a solid dispersed in a gas, so the dispersion medium is gas (✗); cheese involves a liquid dispersed in a solid (a gel), so the dispersion medium is solid (✗); milk is a liquid dispersed in a liquid (an emulsion), so the dispersion medium is liquid (✓); hair cream involves liquid or solid dispersed in a liquid (an emulsion or sol), so the dispersion medium is liquid (✓); insecticide sprays involve a liquid dispersed in a gas (an aerosol), so the dispersion medium is gas (✗); froth consists of gas dispersed in a liquid (a foam), so the dispersion medium is liquid (✓); and soap lather consists of gas dispersed in a liquid (a foam), so the dispersion medium is liquid (✓).

The colloidal systems with liquid as the dispersion medium are paints, milk, hair cream, froth, and soap lather.

Therefore, the number of colloidal systems with a liquid dispersion medium is $$5$$.

Let us examine each statement about adsorption:

Statement A: "Water vapours are adsorbed by anhydrous calcium chloride." This is incorrect. Anhydrous $$CaCl_2$$ actually absorbs water vapour (it takes water into its bulk to form hydrated salts like $$CaCl_2 \cdot 6H_2O$$), rather than adsorbing it on the surface. Adsorption is a surface phenomenon, while what $$CaCl_2$$ does is absorption — a bulk phenomenon.

Statement B: "There is a decrease in surface energy during adsorption." This is correct. Adsorption occurs because surface molecules have unbalanced forces. When adsorbate molecules attach to the surface, these residual forces are satisfied, lowering the surface energy.

Statement C: "As the adsorption proceeds, $$\Delta H$$ becomes more and more negative." This is correct. Adsorption is exothermic, and as the process continues, the cumulative enthalpy change becomes increasingly negative.

Statement D: "Adsorption is accompanied by decrease in entropy of the system." This is correct. Gas molecules lose their freedom of movement when they become bound to a surface, resulting in decreased randomness and hence decreased entropy.

Only Statement A is incorrect. The number of incorrect statements is $$\boxed{1}$$.

We have a $$250 \text{ g}$$ solution of D-glucose in water containing $$10.8\%$$ carbon by weight. We need to find the molality.

The mass of carbon present is given by the percentage times the total mass: $$ \text{Mass of carbon} = 10.8\% \times 250 = \frac{10.8}{100} \times 250 = 27 \text{ g} $$.

The molecular formula of D-glucose is $$C_6H_{12}O_6$$, and its molecular weight is calculated as $$6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g/mol}$$. Since there are 6 carbon atoms in one mole of glucose, the mass of carbon per mole is $$6 \times 12 = 72 \text{ g}$$, giving a mass fraction of carbon in glucose of $$\dfrac{72}{180} = \dfrac{2}{5}$$. Therefore, the mass of glucose in the solution is found by dividing the mass of carbon by this fraction: $$ \text{Mass of glucose} = \frac{\text{Mass of carbon}}{\text{Mass fraction of C}} = \frac{27}{2/5} = 27 \times \frac{5}{2} = 67.5 \text{ g} $$.

The mass of the solvent (water) is the difference between the total mass and the mass of glucose: $$ \text{Mass of water} = 250 - 67.5 = 182.5 \text{ g} = 0.1825 \text{ kg} $$.

The number of moles of glucose present is calculated as $$ n = \frac{67.5}{180} = 0.375 \text{ mol} $$.

The molality of the solution is then the moles of solute per kilogram of solvent: $$ m = \frac{n}{\text{mass of solvent in kg}} = \frac{0.375}{0.1825} = 2.055 \approx 2.06 $$.

The molality is nearest to $$2.06$$. The correct answer is Option B: $$2.06$$.

We need to evaluate the assertion that activated charcoal adsorbs $$SO_2$$ more efficiently than $$CH_4$$, and the reason that gases with lower critical temperatures are readily adsorbed by activated charcoal.

The ease of adsorption of a gas depends on how easily it can be liquefied, and gases with higher critical temperatures are more easily liquefiable and hence are more readily adsorbed. The critical temperature of $$SO_2$$ is $$430 \text{ K}$$ (high, easily liquefiable), while that of $$CH_4$$ is $$190 \text{ K}$$ (low, not easily liquefiable). Since $$SO_2$$ has a higher critical temperature, it is adsorbed more efficiently than $$CH_4$$, making the assertion correct.

By contrast, the reason states that gases with lower critical temperatures are readily adsorbed, which is the opposite of the correct principle. Gases with higher critical temperatures are more readily adsorbed because they are more easily liquefiable, so the reason is incorrect.

The correct answer is Option C: A is correct but R is not correct.

We need to identify which property of colloids the Zeta potential is related to.

Understanding Zeta Potential:

Zeta potential (also called electrokinetic potential) is the electric potential at the boundary of the double layer surrounding a colloidal particle. It is the potential difference between the dispersion medium and the stationary layer of fluid attached to the dispersed particle.

Evaluating the Options:

Option A: Colour

The colour of colloids is related to the scattering of light (Tyndall effect) and particle size, not to the zeta potential. Incorrect.

Option B: Brownian movement

Brownian movement is caused by the random bombardment of colloidal particles by molecules of the dispersion medium. It is related to thermal energy and particle size, not to zeta potential. Incorrect.

Option C: Charge on surface of colloidal particle

Zeta potential is directly related to the charge on the surface of colloidal particles. It measures the magnitude of the electrostatic charge at the electrical double layer around the particle. A higher zeta potential indicates greater stability of the colloidal solution due to stronger electrostatic repulsion between particles. This is correct.

Option D: Tyndall effect

The Tyndall effect is the scattering of light by colloidal particles and is related to particle size and the wavelength of light, not to zeta potential. Incorrect.

Hence, the correct answer is Option C: Charge on surface of colloidal particle.

The vapour-pressure lowering for an ideal dilute solution obeys Raoult’s law:
$$\frac{\Delta P}{P^{\circ}} = i\,X_{\text{solute}}$$
where $$\Delta P = P^{\circ}-P$$, $$P^{\circ}$$ is the vapour pressure of the pure solvent, $$P$$ is that of the solution, $$X_{\text{solute}}$$ is the mole-fraction of the solute and $$i$$ is the van’t Hoff factor.

Step 1: Calculate $$\Delta P$$ and the relative lowering
$$\Delta P = 60.000\;\text{mm Hg} - 59.724\;\text{mm Hg} = 0.276\;\text{mm Hg}$$
$$\frac{\Delta P}{P^{\circ}} = \frac{0.276}{60.000} = 4.60 \times 10^{-3}$$

Step 2: Mole fraction of the solute
Moles of solute (given) $$= 0.1\;\text{mol}$$
Mass of water $$= 1.8\;\text{kg} = 1800\;\text{g}$$
Moles of water $$= \frac{1800}{18} = 100\;\text{mol}$$
Total moles $$= 0.1 + 100 = 100.1$$
$$X_{\text{solute}} = \frac{0.1}{100.1} \approx 9.99 \times 10^{-4} \;(\text{≈}\, 1.00 \times 10^{-3})$$

Step 3: Determine the van’t Hoff factor $$i$$
$$i = \frac{\Delta P / P^{\circ}}{X_{\text{solute}}} = \frac{4.60 \times 10^{-3}}{1.00 \times 10^{-3}} = 4.6$$

Step 4: Relate $$i$$ to degree of dissociation
If one formula unit of the salt gives $$\nu$$ ions on complete dissociation, then for partial dissociation with degree $$\alpha$$:
$$i = 1 + (\nu - 1)\alpha \quad -(1)$$

The salt is 90 % dissociated, so $$\alpha = 0.90$$. Substituting $$i = 4.6$$ into equation $$(1)$$:
$$4.6 = 1 + (\nu - 1)(0.90)$$
$$4.6 - 1 = 0.90(\nu - 1)$$
$$3.6 = 0.90(\nu - 1)$$
$$\nu - 1 = \frac{3.6}{0.90} = 4$$
$$\nu = 5$$

Step 5: Final result
Each formula unit of the ionic salt produces 5 ions in solution.

Answer: 5

We have a 5% (w/v) NaCl solution to which egg albumin is added and stirred. We need to determine the type of colloidal system formed.

Albumin is a protein — a naturally occurring macromolecule. When proteins are dissolved or dispersed in an aqueous medium, they form colloidal solutions. The key question is whether this colloidal sol is lyophilic or lyophobic.

A lyophilic sol (solvent-loving) is one in which the dispersed phase has a strong affinity for the dispersion medium. Proteins like albumin have numerous polar groups ($$-NH_2$$, $$-COOH$$, $$-OH$$) on their surface that readily interact with water molecules through hydrogen bonding. Because of this strong interaction, protein sols form spontaneously when stirred — no special methods are needed, and the resulting sol is quite stable.

A lyophobic sol (solvent-hating), in contrast, is formed by substances like metals or metal sulphides that have no affinity for water. Such sols require special techniques (like Bredig's arc method) to prepare and are inherently unstable.

Since albumin is a protein with strong affinity for water, the dispersion formed is a lyophilic sol. The NaCl solution simply serves as the dispersion medium — at 5% concentration, it is not high enough to salt out the albumin.

Hence, the correct answer is Option A: Lyophilic sol.

Let us analyze both the Assertion and the Reason.

Assertion: A 5 M solution of KCl at $$10°C$$ has density 'x' g/mL. When cooled to $$-21°C$$, the molality of the solution remains unchanged.

Reason: Molality does not change with temperature as mass remains unaffected with temperature.

Checking the Assertion: Molality is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$$

When the temperature changes, the mass of the solute and the mass of the solvent remain the same (mass does not depend on temperature). Therefore, molality does not change with temperature. The assertion is true.

Checking the Reason: The reason states that molality does not change with temperature because mass remains unaffected by temperature. This is exactly why molality is temperature-independent — unlike molarity which depends on volume (which changes with temperature), molality depends only on mass. The reason is true and correctly explains the assertion.

Both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Hence, the correct answer is Option C.

We need to evaluate two statements about emulsions.

Statement I: "Emulsions of oil in water are unstable and sometimes they separate into two layers on standing."

This statement is correct. Emulsions are thermodynamically unstable systems. Without an emulsifying agent (stabilizer), oil-in-water emulsions tend to separate into two distinct layers over time due to the tendency to minimize the interfacial area between the two immiscible liquids. This process is called demulsification or creaming.

Statement II: "For stabilisation of an emulsion, excess of electrolyte is added."

This statement is incorrect. Emulsions are stabilized by adding emulsifying agents (also called surfactants or emulsifiers) such as soaps, detergents, proteins (like casein), or gums. These substances form a protective layer around the dispersed droplets, preventing coalescence.

In fact, adding excess electrolyte to an emulsion can actually destabilize it by compressing the electrical double layer around the dispersed droplets (similar to coagulation of colloids). This is the opposite of stabilization.

Therefore, Statement I is correct but Statement II is incorrect.

The correct answer is Option C.

We are given that solute A associates in water and we need to find the percentage association.

Since the freezing point depression is given by $$\Delta T_f = K_f \times m$$, substituting $$\Delta T_f = 0.2$$ and $$K_f = 1.86$$ yields $$m = \frac{0.2}{1.86} = 0.1075 \text{ mol/kg}$$.

Without association, the theoretical molality is calculated using the mass of solute and solvent; thus, $$m_{\text{theoretical}} = \frac{0.7}{93 \times 0.042} = \frac{0.7}{3.906} = 0.1792 \text{ mol/kg}$$.

From the ratio of observed to theoretical molality, the van't Hoff factor is $$i = \frac{m_{\text{observed}}}{m_{\text{theoretical}}} = \frac{0.1075}{0.1792} = 0.6$$.

For a dimerization association where $$n = 2$$, the relation between $$i$$ and the fraction associated $$x$$ is $$i = 1 - x + \frac{x}{n} = 1 - x + \frac{x}{2} = 1 - \frac{x}{2}$$. Substituting $$i = 0.6$$ gives $$0.6 = 1 - \frac{x}{2}$$, which leads to $$\frac{x}{2} = 0.4$$ and hence $$x = 0.8$$ or 80\%.

Therefore, the percentage association of solute A in water is 80%, corresponding to Option D.

We need to evaluate the Assertion and Reason about dialysis of colloidal solutions.

Analyzing Assertion A:

Assertion A states that dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper. This is the principle of dialysis. In dialysis, a colloidal solution is placed in a bag made of parchment paper (a semi-permeable membrane). The dissolved ions and molecules (true solution particles) pass through the parchment paper into the outer liquid, while the larger colloidal particles are retained.

$$\therefore$$ Assertion A is correct.

Analyzing Reason R:

Reason R states that particles in a true solution cannot pass through parchment paper but colloidal particles can. This is the opposite of what actually happens:

- True solution particles (ions, small molecules) can pass through parchment paper.

- Colloidal particles are too large and cannot pass through parchment paper.

$$\therefore$$ Reason R is incorrect.

Conclusion:

A is correct but R is not correct.

Hence, the correct answer is Option C.

The Tyndall effect is the scattering of light by colloidal particles. We need to identify the incorrect statement about it.

Recall conditions for the Tyndall effect: The Tyndall effect occurs when:

$$\bullet$$ The refractive indices of the dispersed phase and dispersion medium differ significantly.

$$\bullet$$ The diameter of the dispersed particles is not too small compared to the wavelength of light used — typically in the range of 1 nm to 1000 nm (comparable to the wavelength of visible light, 400-700 nm).

Analyze each statement: Option A: "The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude." — This is correct. A large difference in refractive index enhances scattering.

Option B: "The diameter of the dispersed particles is much smaller than the wavelength of the light used." — This is incorrect. If the particles were much smaller than the wavelength (as in true solutions), they would not scatter light effectively. For the Tyndall effect, the particle size must be comparable to or not too much smaller than the wavelength of light.

Option C: "It is used to distinguish a true solution from a colloidal solution." — This is correct. True solutions do not show the Tyndall effect, while colloidal solutions do.

Option D: "During projection of movies in the cinema hall, Tyndall effect is noticed." — This is correct. The visible beam of light from the projector passing through dusty air is an example of the Tyndall effect.

The correct answer is Option B: The diameter of the dispersed particles is much smaller than the wavelength of the light used.

Let us match each item in List-I with the correct item in List-II.

A. Lyophilic colloid → II (Protective colloid)

Lyophilic colloids (solvent-loving) are stable colloids that can protect lyophobic colloids from coagulation. They are also called protective colloids. Examples include gelatin, starch, and gum arabic.

B. Emulsion → I (Liquid-liquid colloid)

An emulsion is a colloidal system where both the dispersed phase and the dispersion medium are liquids. For example, milk is an emulsion of fat droplets in water. Hence, emulsion is a liquid-liquid colloid.

C. Positively charged colloid → IV ($$FeCl_3$$ + hot water)

When $$FeCl_3$$ is added to hot water, it hydrolyzes to form a colloidal solution of $$Fe(OH)_3$$ which is positively charged. The $$Fe^{3+}$$ ions are preferentially adsorbed on the $$Fe(OH)_3$$ particles, giving them a positive charge.

D. Negatively charged colloid → III ($$FeCl_3 + NaOH$$)

When $$FeCl_3$$ is mixed with excess $$NaOH$$, the $$Fe(OH)_3$$ sol formed adsorbs $$OH^-$$ ions from the excess NaOH, making the colloidal particles negatively charged.

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

Hence, the correct answer is Option A.

We need to match each item in List-I with the correct item in List-II based on properties of colloids.

CdS sol is a well-known example of a negatively charged sol. Other examples include As$$_2$$S$_3$$ sol and gold sol. So (A) negatively charged sol → (II) CdS sol.

Starch is a macromolecular colloid — it is a naturally occurring polymer that forms colloidal solutions in water. So (B) macromolecular colloid → (III) Starch.

Fe$$_2$$O$$_3$$ · xH$$_2$$O (ferric oxide hydrosol) is a positively charged sol. Other examples include Al(OH)$$_3$$ sol. So (C) positively charged sol → (I) Fe$$_2$$O$$_3$$ · xH$$_2$$O.

Cheese is a well-known example of a gel (solid dispersed in liquid medium that has a semi-solid structure). So (D) Cheese → (IV) a gel.

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

Hence, the correct answer is Option C.

We are given two solutions A and B, each prepared by dissolving 1 g of non-volatile solutes X and Y respectively in 1 kg of water, and the ratio of depression in freezing points is 1 : 4.

Starting with the formula for freezing point depression, we have

$$\Delta T_f = K_f \times m$$

where $$m$$ denotes the molality of the solution.

Since 1 g of solute is dissolved in 1 kg of solvent, the molality can be expressed in terms of the molar mass $$M$$ as

$$m = \frac{1}{M \times 1} = \frac{1}{M}$$

Here, $$M$$ represents the molar mass of the solute (in g/mol).

When comparing the freezing point depressions of solutions A and B, we obtain

$$\frac{\Delta T_{f,A}}{\Delta T_{f,B}} = \frac{K_f / M_X}{K_f / M_Y} = \frac{M_Y}{M_X}$$

Given this ratio equals $$\frac{1}{4}$$, it follows that

$$\frac{M_Y}{M_X} = \frac{1}{4}$$

Hence,

$$\frac{M_X}{M_Y} = \frac{4}{1} = \frac{1}{0.25}$$

This establishes the molar mass ratio $$M_X : M_Y = 1 : 0.25$$, implying $$M_X = 4 \times M_Y$$.

Therefore, the correct choice is Option B.

We are given that the boiling point of a 2% aqueous solution of a non-volatile solute A is equal to the boiling point of an 8% aqueous solution of a non-volatile solute B. We need to find the relation between their molecular weights $$M_A$$ and $$M_B$$.

Recall the boiling point elevation formula. The elevation in boiling point for a dilute solution is given by:

$$ \Delta T_b = K_b \times m $$

where $$K_b$$ is the ebullioscopic constant of the solvent and $$m$$ is the molality of the solution.

Since both solutions are aqueous, they share the same solvent (water) and hence the same $$K_b$$. Equal boiling points mean equal boiling point elevations:

$$ \Delta T_{b,A} = \Delta T_{b,B} $$

$$ K_b \times m_A = K_b \times m_B $$

Cancelling $$K_b$$ gives

$$ m_A = m_B $$

A "w%" aqueous solution means w grams of solute is dissolved in 100 grams of solution, so the mass of solvent (water) is (100 - w) grams. Molality is defined as moles of solute per kilogram of solvent:

$$ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{w / M}{\frac{(100 - w)}{1000}} = \frac{1000w}{M \times (100 - w)} $$

For 2% solution of A: $$ m_A = \frac{1000 \times 2}{M_A \times (100 - 2)} = \frac{2000}{98 \, M_A} $$

For 8% solution of B: $$ m_B = \frac{1000 \times 8}{M_B \times (100 - 8)} = \frac{8000}{92 \, M_B} $$

Equating these molalities and cross-multiplying yields:

$$ \frac{2000}{98 \, M_A} = \frac{8000}{92 \, M_B} $$

$$ 2000 \times 92 \times M_B = 8000 \times 98 \times M_A $$

$$ 184000 \, M_B = 784000 \, M_A $$

Dividing both sides by 184000 gives

$$ M_B = \frac{784000}{184000} \, M_A = \frac{784}{184} \, M_A $$

Simplifying the fraction:

$$ \frac{784}{184} = \frac{784 \div 8}{184 \div 8} = \frac{98}{23} \approx 4.26 $$

Applying the standard JEE approximation for dilute solutions (where the solute percentage is small) under which (100 - w) ≈ 100, we have:

$$ m_A = \frac{1000 \times 2}{M_A \times 100} = \frac{20}{M_A} $$

$$ m_B = \frac{1000 \times 8}{M_B \times 100} = \frac{80}{M_B} $$

Setting these equal,

$$ \frac{20}{M_A} = \frac{80}{M_B} $$

$$ 20 \, M_B = 80 \, M_A $$

$$ M_B = \frac{80}{20} \, M_A = 4 \, M_A $$

Therefore, $$M_B = 4M_A$$.

The correct answer is Option B: $$M_B = 4M_A$$.

We need to find the Van't Hoff factor (i) for formic acid solution.

Concentration = $$0.5 \text{ mL L}^{-1}$$ means 0.5 mL of formic acid in 1 L of solution (approximately 1 L of water for a dilute solution), so the mass of formic acid is Volume $$\times$$ Density = $$0.5 \times 1.05 = 0.525 \text{ g}$$.

The molar mass of formic acid ($$HCOOH$$) is $$1 + 12 + 16 + 16 + 1 = 46 \text{ g mol}^{-1}$$, and the moles of formic acid present are $$\dfrac{0.525}{46} = 0.01141 \text{ mol}$$. Since the solution is very dilute, the mass of the solvent (water) is approximately 1000 g = 1 kg, giving a molality of $$\dfrac{0.01141}{1} = 0.01141 \text{ mol kg}^{-1}$$.

Applying the depression in freezing point formula $$\Delta T_f = i \cdot K_f \cdot m$$, we have $$0.0405 = i \times 1.86 \times 0.01141$$ which simplifies to $$0.0405 = i \times 0.02122$$, and hence $$i = \dfrac{0.0405}{0.02122} = 1.908 \approx 1.9$$.

The Van't Hoff factor $$i \approx 1.9$$ indicates that formic acid nearly completely dissociates in water as: $$HCOOH \rightleftharpoons HCOO^- + H^+$$, giving approximately 2 particles per molecule.

Hence, the correct answer is Option C.

We need to determine which statements about micelle formation are correct.

Micelle formation occurs when surfactant molecules aggregate in solution above the critical micelle concentration (CMC). This process involves the hydrophobic tails coming together, releasing ordered water molecules from around the hydrophobic parts.

Since micelle formation involves breaking the structured water cage around the hydrophobic tails, it is generally an endothermic process ($$\Delta H > 0$$). Energy is required to break the structured water cage (iceberg structure) around the hydrophobic tails, so statement (B) is correct and (A) is incorrect.

Upon micelle formation, the ordered water molecules surrounding the hydrophobic parts of the surfactant are released into the bulk water. This release leads to a large positive entropy change ($$\Delta S > 0$$), so statement (C) is correct and (D) is incorrect.

Micelle formation is spontaneous ($$\Delta G < 0$$) because the large positive $$T\Delta S$$ term overcomes the positive $$\Delta H$$:

$$\Delta G = \Delta H - T\Delta S < 0$$

This confirms that the process is entropy-driven.

Therefore, statements B and C are correct.

Hence, the correct choice is Option C.

We need to evaluate the assertion and reason about colloidal gold.

Assertion A: Finest gold is red in colour, as the size of the particles increases, it appears purple then blue and finally gold.

This is correct. Colloidal gold nanoparticles exhibit different colours depending on their size. Very fine particles ($$\sim$$20 nm) appear red, medium-sized particles appear purple/blue, and bulk gold has its characteristic golden colour.

Reason R: The colour of the colloidal solution depends on the wavelength of light scattered by the dispersed particles.

This is correct. The colour of colloidal solutions depends on the wavelength of scattered light. As particle size changes, the wavelength of light scattered changes, which results in different observed colours. This phenomenon is related to the Tyndall effect and surface plasmon resonance in metal nanoparticles.

Is R the correct explanation of A?

Yes. The reason gold colloids of different sizes appear as different colours (Assertion A) is precisely because the wavelength of scattered light depends on the particle size (Reason R). Smaller particles scatter shorter wavelengths, while larger particles scatter longer wavelengths, producing the observed colour changes from red to purple to blue to gold.

Therefore, the correct answer is Option A: Both A and R are true and R is the correct explanation of A.

This question asks why using very little soap is ineffective for cleaning clothes.

Soap molecules are amphiphilic — they have a hydrophobic (non-polar) tail and a hydrophilic (polar) head. For effective cleaning, they must form micelles — spherical aggregates where the hydrophobic tails point inward (trapping grease) and the hydrophilic heads face outward (toward water). However, micelles form only when the concentration of soap in water exceeds a certain threshold called the Critical Micelle Concentration (CMC); below this concentration, soap molecules remain as individual ions or molecules and cannot aggregate into micelles.

When very little soap is used, the soap concentration in water remains below the CMC, preventing micelle formation and thus rendering the soap unable to effectively trap and remove grease and dirt from clothes.

Option A: Soap particles remain floating in water as ions — this is partially true but does not explain why cleaning fails.
Option B: Colloidal structure of soap in water is completely disturbed — incorrect, as the colloidal structure (micelles) was never formed in the first place.
Option C: The hydrophobic part of soap is not able to take away grease — incorrect reasoning; the hydrophobic part can still interact with grease, but without micelle formation, it cannot be effectively removed.
Option D: The micelles are not formed due to concentration of soap below its CMC value — this is the precise scientific explanation.

The correct answer is Option D: The micelles are not formed due to concentration of soap below its CMC value.

We need to find the molarity of the remaining nitric acid solution after heating.

Initial volume = $$800 \text{ mL} = 0.8 \text{ L}$$

Initial molarity = $$0.5 \text{ M}$$

$$\text{Initial moles of } HNO_3 = 0.5 \times 0.8 = 0.4 \text{ mol}$$

Mass of nitric acid evaporated = $$11.5 \text{ g}$$

Molar mass of $$HNO_3$$ = $$63 \text{ g/mol}$$

$$\text{Moles evaporated} = \frac{11.5}{63} = \frac{11.5}{63} \approx 0.1825 \text{ mol}$$

$$\text{Remaining moles} = 0.4 - 0.1825 = 0.2175 \text{ mol}$$

The volume is reduced to half:

$$\text{Final volume} = \frac{800}{2} = 400 \text{ mL} = 0.4 \text{ L}$$

$$\text{Molarity} = \frac{\text{Remaining moles}}{\text{Final volume}} = \frac{0.2175}{0.4} = 0.54375 \text{ M}$$

$$\text{Molarity} = 0.54375 \text{ M} \approx 0.54 \text{ M} = 54 \times 10^{-2} \text{ M}$$

Hence, the value of $$x$$ is 54.

We have 1.80 g of solute A dissolved in 62.5 cm$$^3$$ of ethanol. The freezing point of the solution is 155.1 K, while the freezing point of pure ethanol is 156.0 K. We need to find the molar mass of solute A.

We begin by calculating the freezing point depression:

$$\Delta T_f = T_f^0 - T_f = 156.0 - 155.1 = 0.9 \text{ K}$$

Now we find the mass of ethanol (solvent). Given the volume is 62.5 cm$$^3$$ and the density is 0.80 g cm$$^{-3}$$:

$$\text{Mass of ethanol} = 62.5 \times 0.80 = 50.0 \text{ g} = 0.050 \text{ kg}$$

The freezing point depression is related to molality by:

$$\Delta T_f = K_f \times m$$

where $$m$$ is the molality (moles of solute per kg of solvent) and $$K_f = 2.00$$ K kg mol$$^{-1}$$. Substituting:

$$0.9 = 2.00 \times m$$

$$m = \frac{0.9}{2.00} = 0.45 \text{ mol kg}^{-1}$$

The molality is defined as moles of solute per kg of solvent, so the number of moles of solute A is:

$$n_A = m \times \text{mass of solvent (in kg)} = 0.45 \times 0.050 = 0.0225 \text{ mol}$$

Now the molar mass of solute A is:

$$M_A = \frac{\text{mass of solute}}{n_A} = \frac{1.80}{0.0225} = 80 \text{ g mol}^{-1}$$

Hence, the correct answer is 80.

We need to find the mass of $$CO_2$$ dissolved in 1 litre of water at 298 K.

Since Henry’s Law relates the partial pressure of a gas to its mole fraction in solution, we write

$$ p_{CO_2} = K_H \cdot x_{CO_2} $$

Substituting the given values gives

$$ x_{CO_2} = \frac{p_{CO_2}}{K_H} = \frac{0.835 \text{ bar}}{1.67 \text{ kbar}} = \frac{0.835}{1670} = 5 \times 10^{-4} $$

Next, the number of moles of water in 1 litre is

$$ n_{H_2O} = \frac{1000}{18} = 55.56 \text{ mol} $$

Since $$x_{CO_2}$$ is very small, we approximate

$$ x_{CO_2} \approx \frac{n_{CO_2}}{n_{H_2O}} $$

This gives

$$ n_{CO_2} = x_{CO_2} \times n_{H_2O} = 5 \times 10^{-4} \times 55.56 = 0.02778 \text{ mol} $$

Calculating the mass of $$CO_2$$ (molar mass of $$CO_2$$ = 12 + 2(16) = 44 g/mol) yields

$$ m_{CO_2} = 0.02778 \times 44 = 1.2222 \text{ g} $$

Expressing this value in the required form results in

$$ x = 1.2222 \text{ g} \approx 1221 \times 10^{-3} \text{ g} $$

The answer is $$1221$$.

We are given elevation in boiling point for 1.5 molal glucose solution: $$\Delta T_b = 4$$ K and depression in freezing point for 4.5 molal glucose solution: $$\Delta T_f = 4$$ K.

Using the boiling point elevation formula $$\Delta T_b = K_b \times m$$ gives $$4 = K_b \times 1.5$$ and hence $$K_b = \frac{4}{1.5} = \frac{8}{3}$$ K kg mol$$^{-1}$$. Similarly, the freezing point depression formula $$\Delta T_f = K_f \times m$$ yields $$4 = K_f \times 4.5$$ so $$K_f = \frac{4}{4.5} = \frac{8}{9}$$ K kg mol$$^{-1}$$. The ratio is then $$\frac{K_b}{K_f} = \frac{8/3}{8/9} = \frac{8}{3} \times \frac{9}{8} = \frac{9}{3} = 3$$.

Therefore, the ratio $$K_b/K_f = \textbf{3}$$.

We are given that the elevation in boiling point for a 1 molal solution of non-volatile solute A is $$3\text{ K}$$ and the depression in freezing point for a 2 molal solution of A in the same solvent is $$6\text{ K}$$.

Since the elevation in boiling point is related by $$\Delta T_b = K_b \cdot m$$, substituting $$m = 1$$ molal gives $$3 = K_b \times 1$$, which yields $$K_b = 3\text{ K kg/mol}$$.

Similarly, because the depression in freezing point follows $$\Delta T_f = K_f \cdot m$$, substituting $$m = 2$$ molal leads to $$6 = K_f \times 2$$, and hence $$K_f = 3\text{ K kg/mol}$$.

Therefore, the ratio is $$\frac{K_b}{K_f} = \frac{3}{3} = 1$$.

Thus, $$K_b / K_f = 1 : 1$$, which means $$X = 1$$.

Therefore, the answer is $$\boxed{1}$$.

The mass of acetic acid (solvent) is 150 g = 0.150 kg, the mass of ascorbic acid ($$C_6H_8O_6$$, solute) is 10.2 g, the molar mass of ascorbic acid is 176 g/mol, and the cryoscopic constant $$K_f$$ of acetic acid is 3.9 K kg mol$$^{-1}$$.

Since the moles of ascorbic acid can be calculated by $$n = \frac{10.2}{176} = 0.05795 \text{ mol}$$, this represents the number of moles of solute present.

Substituting into the definition of molality gives $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05795}{0.150} = 0.3864 \text{ mol/kg}$$.

Next, the depression in freezing point is given by $$\Delta T_f = K_f \times m$$. Substituting the values yields $$= 3.9 \times 0.3864$$ and thus $$= 1.507 \text{ °C}$$.

Expressing this in the required form gives $$\Delta T_f = 1.507 \text{ °C} = 15.07 \times 10^{-1} \text{ °C}$$. Therefore, the value of $$x$$ is 15 (nearest integer).

The elevation in boiling point is given by:

$$\Delta T_b = K_b \times m$$

where $$K_b$$ is the ebullioscopic constant and $$m$$ is the molality.

Molality is given by:

$$m = \frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}}$$

Since the same solute (2 g of non-volatile non-electrolyte) is dissolved in the same mass of solvent (200 g) for both solvents A and B:

$$m_A = m_B = m = \frac{n}{0.2}$$

The molality is the same for both solutions.

Therefore:

$$\Delta T_{b,A} = K_{b,A} \times m$$

$$\Delta T_{b,B} = K_{b,B} \times m$$

Taking the ratio:

$$\frac{\Delta T_{b,A}}{\Delta T_{b,B}} = \frac{K_{b,A}}{K_{b,B}}$$

Given that $$K_{b,A} : K_{b,B} = 1 : 8$$:

$$\frac{\Delta T_{b,A}}{\Delta T_{b,B}} = \frac{1}{8} = \frac{x}{y}$$

So $$x = 1$$ and $$y = 8$$.

Hence, the value of $$y$$ is 8.

We need to find the number of millimoles of $$O_2$$ that dissolve in 1 litre of water at 303 K using Henry's Law.

Henry's Law states: $$p = K_H \cdot x$$, where $$p$$ is the partial pressure of the gas, $$K_H$$ is the Henry's Law constant, and $$x$$ is the mole fraction of the gas in the solution.

Given: $$K_H = 46.82$$ kbar $$= 46.82 \times 10^3$$ bar, and $$p_{O_2} = 0.920$$ bar.

The mole fraction of $$O_2$$ in water is:

$$x = \frac{p}{K_H} = \frac{0.920}{46.82 \times 10^3} = \frac{0.920}{46820} = 1.9650 \times 10^{-5}$$

Now, since the solubility is very small, the mole fraction can be approximated as:

$$x \approx \frac{n_{O_2}}{n_{water}}$$

For 1 litre of water, $$n_{water} = \frac{1000}{18} = 55.556$$ mol.

$$n_{O_2} = x \times n_{water} = 1.9650 \times 10^{-5} \times 55.556 = 1.092 \times 10^{-3} \text{ mol}$$

Converting to millimoles: $$n_{O_2} = 1.092$$ mmol $$\approx 1$$ mmol (nearest integer).

Hence, the correct answer is 1.

We need to find the concentration of glucose solution (in g/L) that is isotonic with blood. For isotonic solutions, the osmotic pressure of the glucose solution must equal the osmotic pressure of blood.

The formula for osmotic pressure is:

$$ \pi = CRT $$

where $$\pi$$ is the osmotic pressure, $$C$$ is the molar concentration (mol/L), $$R$$ is the gas constant, and $$T$$ is the temperature in Kelvin.

With $$\pi = 7.47$$ bar, $$T = 300$$ K, and $$R = 0.083$$ L bar$$^{-1}$$ mol$$^{-1}$$, we solve for the molar concentration $$C$$ as follows:

$$ C = \frac{\pi}{RT} = \frac{7.47}{0.083 \times 300} $$

$$ C = \frac{7.47}{24.9} = 0.3 \text{ mol/L} $$

Converting molar concentration to g/L using the molar mass of glucose ($$M = 180$$ g/mol):

$$ \text{Concentration (g/L)} = C \times M = 0.3 \times 180 $$

$$ \text{Concentration (g/L)} = 54 $$

Therefore, the concentration of glucose solution is 54 g/L.

We need to find the value of $$x$$ given that the mole fraction of liquid B in the vapour phase is $$\frac{x}{17}$$. The vapour pressures of pure A and B are $$P_A^0 = 50$$ Torr and $$P_B^0 = 100$$ Torr respectively, and in the liquid phase the mole fractions are $$x_A = 0.3$$ and $$x_B = 0.7$$.

By Raoult's law, the partial pressures are calculated as $$P_A = x_A \times P_A^0 = 0.3 \times 50 = 15 \text{ Torr}$$ and $$P_B = x_B \times P_B^0 = 0.7 \times 100 = 70 \text{ Torr}$$. The total vapour pressure is then $$P_{total} = P_A + P_B = 15 + 70 = 85 \text{ Torr}$$.

Applying Dalton's law, the mole fraction of B in vapour is $$y_B = \frac{P_B}{P_{total}} = \frac{70}{85}$$, which simplifies to $$y_B = \frac{14}{17}$$. Since this equals $$\frac{x}{17}$$, we have $$\frac{x}{17} = \frac{14}{17}$$, giving $$x = 14$$.

We need to find the number of moles of solute A dissolved in 100 g of water such that the vapour pressure is reduced to half. Given that the mass of water (solvent) is 100 g, the vapour pressure of pure water is $$P^0 = 23.76$$ mmHg and the vapour pressure of the solution is $$P = \frac{P^0}{2} = 11.88$$ mmHg.

Using Raoult’s law, the mole fraction of the solute is given by $$\frac{P^0 - P}{P^0} = x_{\text{solute}}$$, so $$\frac{23.76 - 11.88}{23.76} = x_{\text{solute}}$$, which yields $$x_{\text{solute}} = \frac{11.88}{23.76} = 0.5$$.

The moles of water present are $$\frac{100}{18} = 5.55$$ mol. Since $$x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}$$, we have $$0.5 = \frac{n_A}{n_A + 5.55}$$, which leads to $$0.5(n_A + 5.55) = n_A$$, then $$0.5 n_A + 2.775 = n_A$$, so $$0.5 n_A = 2.775$$ and thus $$n_A = 5.55 \text{ mol}$$.

Hence, the number of moles of solute A added is 5.55.

We need to calculate the osmotic pressure of a protein solution.

The mass of the protein is 2.0 g, the molar mass is $$60\,\mathrm{kg\,mol^{-1}}$$ (equivalent to $$60000\,\mathrm{g\,mol^{-1}}$$), the volume of the solution is 200 mL (0.2 L), the temperature is 27 °C (300 K), and $$R = 0.083\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$$.

The number of moles of protein is given by

$$n = \frac{2.0}{60000} = \frac{1}{30000}\text{ mol}.$$

The osmotic pressure is determined using

$$\pi = \frac{nRT}{V}$$

Substituting the values yields

$$\pi = \frac{\frac{1}{30000} \times 0.083 \times 300}{0.2}$$

This simplifies to

$$\pi = \frac{\frac{24.9}{30000}}{0.2} = \frac{0.00083}{0.2} = 0.00415\text{ bar}.$$

Since $$1\text{ bar} = 10^5\text{ Pa}$$, converting the pressure to pascals gives

$$\pi = 0.00415 \times 10^5 = 415\text{ Pa}.$$

Therefore, the osmotic pressure is 415 Pa.

The volume of acetic acid solution is 200 mL = 0.2 L, the initial concentration is 0.2 M, the final concentration after adsorption is 0.1 M, and the mass of wood charcoal is 0.6 g.

Since volume and concentration are known, the initial moles of acetic acid are calculated as:

$$ n_{\text{initial}} = 0.2 \times 0.2 = 0.04 \text{ mol} $$

Similarly, the final moles of acetic acid after adsorption are given by:

$$ n_{\text{final}} = 0.2 \times 0.1 = 0.02 \text{ mol} $$

This gives the moles of acetic acid adsorbed as:

$$ n_{\text{adsorbed}} = 0.04 - 0.02 = 0.02 \text{ mol} $$

Substituting the molar mass of $$CH_3COOH = 60$$ g/mol gives the mass of acetic acid adsorbed:

$$ m_{\text{adsorbed}} = 0.02 \times 60 = 1.2 \text{ g} $$

From the above, the mass of acetic acid adsorbed per gram of carbon is:

$$ \frac{m_{\text{adsorbed}}}{m_{\text{carbon}}} = \frac{1.2}{0.6} = 2 \text{ g/g} $$

Therefore, the mass of acetic acid adsorbed per gram of carbon is 2 g.

We have a gaseous mixture of A and B in equilibrium with an ideal liquid solution at a total pressure of $$P = 0.8$$ atm. The mole fraction of A in the vapour phase is $$y_A = 0.5$$ and in the liquid phase is $$x_A = 0.2$$.

By Raoult's law for an ideal solution, the partial pressure of A in the vapour phase equals:

where $$P_A^0$$ is the vapour pressure of pure liquid A.

Also, from Dalton's law of partial pressures, the partial pressure of A in the vapour is:

Now using Raoult's law:

Hence, the correct answer is 2.

We are given that 2.5 g of a protein (composed entirely of glycine units) is dissolved in water to make 500 mL of solution, and the osmotic pressure at 300 K is $$5.03 \times 10^{-3}$$ bar.

Using the osmotic pressure formula $$\pi = cRT$$, we find $$c = \frac{\pi}{RT} = \frac{5.03 \times 10^{-3}}{0.083 \times 300} = \frac{5.03 \times 10^{-3}}{24.9} = 2.02 \times 10^{-4} \text{ mol/L}$$.

The number of moles of protein in 0.5 L is $$n = c \times V = 2.02 \times 10^{-4} \times 0.5 = 1.01 \times 10^{-4} \text{ mol}$$, and its molar mass is $$M = \frac{\text{mass}}{n} = \frac{2.5}{1.01 \times 10^{-4}} = 24752 \text{ g/mol}$$.

The molar mass of glycine ($$\text{C}_2\text{H}_5\text{NO}_2$$) is $$M_{\text{glycine}} = 2(12) + 5(1) + 14 + 2(16) = 24 + 5 + 14 + 32 = 75 \text{ g/mol}$$, and $$\text{Number of glycine units} = \frac{24752}{75} \approx 330$$.

Hence, the total number of glycine units present in the protein is 330.

In ion-exchange chromatography, a cation exchange resin must have acidic functional groups that can release $$\text{H}^+$$ ions and bind cations from the solution. The sulfonic acid group $$-\text{SO}_3\text{H}$$ is the standard functional group used in strongly acidic cation exchange resins.

An anion exchange resin must have basic functional groups that can release anions (such as $$\text{OH}^-$$) and bind anions from solution. The amino group $$-\text{NH}_2$$ (or quaternary ammonium groups) serves this purpose in anion exchange resins.

Therefore, the correct pair of functional groups responsible for the ion-exchange property of cation and anion exchange resins is $$-\text{SO}_3\text{H}$$ and $$-\text{NH}_2$$, respectively.

We begin by recalling what “de-ionized water” means. De-ionized (or demineralised) water is water from which all the dissolved ions, both cations like $$\text{Na}^+, \text{K}^+, \text{Ca}^{2+}$$ and anions like $$\text{Cl}^-, \text{SO}_4^{2-}, \text{HCO}_3^-$$, have been removed. So, to obtain de-ionized water we need a process that can simultaneously take out positive as well as negative ions.

Now we analyse each method named in the options.

Option A tells us about the Synthetic resin method. In this method we employ two kinds of ion-exchange resins:

• A cation-exchange resin, often represented as $$\text{R-H}$$. The functional group releases $$\text{H}^+$$ ions and picks up any metal cation present in water according to the general exchange reaction

$$\text{R-H} + \text{M}^{n+} \; \longrightarrow \; \text{R-M}^{(n-)} + n\,\text{H}^+$$

• An anion-exchange resin, commonly written as $$\text{R'-OH}$$. This resin releases an $$\text{OH}^-$$ ion and absorbs any anion present:

$$\text{R'-OH} + \text{A}^- \; \longrightarrow \; \text{R'-A} + \text{OH}^-$$

After the water has passed successively through both resins, the liberated $$\text{H}^+$$ and $$\text{OH}^-$$ ions combine:

$$\text{H}^+ + \text{OH}^- \; \longrightarrow \; \text{H}_2\text{O}$$

Since every cation and every anion has been exchanged out of the water, the effluent is practically free from all ions; that is exactly the definition of de-ionized water. Thus, the synthetic resin method fulfils the requirement completely.

Option B refers to Calgon’s method. Here we add sodium hexametaphosphate $$\text{(NaPO}_3\text{)}_6$$ (popularly called Calgon) to water. The compound forms soluble complexes specifically with hardness-producing calcium and magnesium ions, preventing them from forming scale. Only $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ are affected; other ions remain. Therefore, this method softens water but does not totally de-ionize it.

Option C is Clark’s method. We add calculated amounts of lime $$\text{Ca(OH)}_2$$ to precipitate carbonate hardness as $$\text{CaCO}_3$$ and $$\text{Mg(OH)}_2$$. Again, only temporary hardness is removed; many ions stay dissolved. So Clark’s method cannot yield de-ionized water.

Option D mentions the Permutit (zeolite) method. A natural or synthetic zeolite $$\text{Na}_2\text{Z}$$ exchanges its $$\text{Na}^+$$ ions with $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ in hard water as per

$$\text{Na}_2\text{Z} + \text{Ca}^{2+} \; \longrightarrow \; \text{CaZ} + 2\,\text{Na}^+$$

This removes only the calcium and magnesium ions responsible for hardness; all other cations and every anion remain. Therefore, the permutit method merely softens water; it does not create de-ionized water.

Comparing the four methods, only the synthetic resin (ion-exchange) method eliminates both cations and anions completely, giving pure de-ionized water. Thus it is the most suitable technique for the desired purpose.

Hence, the correct answer is Option A.

सबसे पहले हम यह समझते हैं कि BOD अर्थात् Biochemical Oxygen Demand उस ऑक्सीजन की मात्रा (ppm में) है, जो जल में मौजूद सूक्ष्मजीव (micro-organisms) 5 दिनों में 20 °C पर कार्बनिक पदार्थों को ऑक्सीकृत करने के लिये प्रयुक्त करते हैं।

अब एक महत्त्वपूर्ण सामान्य मानक याद कर लेते हैं। स्वच्छ जल (clean water) के लिये $$\text{BOD}<5\;\text{ppm}$$ को स्वीकार योग्य माना जाता है, क्योंकि इसमें कार्बनिक प्रदूषण अत्यन्त कम होता है।

इसके विपरीत, यदि जल तीव्र रूप से प्रदूषित (heavily polluted) हो, तो सूक्ष्मजीवों को बहुत अधिक कार्बनिक पदार्थ ऑक्सीकृत करना पड़ता है। इस कारण $$\text{BOD}>17\;\text{ppm}$$ या उससे भी अधिक देखी जाती है।

बस इन्हीं दो सर्वमान्य तथ्यों को विकल्पों से मिलाते हैं।

अब चारों विकल्पों पर क्रम से नज़र डालते हैं।

विकल्प A लिखता है: A < 5, B > 17. यह ठीक वही सीमा दर्शाता है जिस पर अभी हमने चर्चा की—स्वच्छ जल के लिये 5 ppm से कम और प्रदूषित जल के लिये 17 ppm से अधिक।

अन्य किसी भी विकल्प में $$\text{BOD}<5$$ तथा $$\text{BOD}>17$$ दोनों एक साथ उपस्थित नहीं हैं, इसलिए वे तुरंत असंगत हो जाते हैं।

इसलिये हम निश्चयपूर्वक कह सकते हैं कि वही विकल्प सही है जिसमें स्वच्छ जल के लिये $$\lt5\;\text{ppm}$$ तथा प्रदूषित जल के लिये $$\gt17\;\text{ppm}$$ दर्शाया गया है।

Hence, the correct answer is Option A.

We first recall that a lyophilic sol is one in which the dispersed phase has a definite affinity for the dispersion medium. Because of this affinity, each colloidal particle becomes surrounded by a sheath of solvent molecules. This phenomenon is called solvation (or hydration when the solvent is water).

The solvation sheath behaves like a protective layer. Since the solvent molecules are firmly attached to the surface, two adjoining particles cannot come close enough to aggregate; the intervening solvating layer keeps them apart. Mathematically, we may express the decrease in free energy accompanying solvation as $$\Delta G = \Delta H - T\Delta S < 0,$$ indicating that the process is spontaneous and favours dispersal rather than coagulation.

Because the particles remain well separated, the sol does not settle or precipitate even on standing for a long time. Hence we say that lyophilic sols are intrinsically stable. On the other hand, in a lyophobic sol the particles do not get solvated. They must rely solely on the electrostatic charge present on their surfaces for mutual repulsion. If that charge is removed (for example, by adding an electrolyte), rapid coagulation occurs. This clearly shows that solvation, and not mere charge, is the principal reason for the greater stability of lyophilic sols.

Now we examine the given options. Option A speaks of “no charge,” but we have just argued that stability arises even in the presence or absence of charge provided the particles are solvated. Option B mentions a “positive charge,” which again is not the basic cause. Option D talks of “strong electrostatic repulsion between negatively charged particles,” a feature more typical of lyophobic sols. Only Option C correctly states that the colloidal particles are solvated, and this solvation is the key to their superior stability.

Hence, the correct answer is Option C.

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.

When $$FeCl_3$$ is added to excess of hot water, it undergoes hydrolysis to form a colloidal solution of ferric hydroxide ($$Fe(OH)_3$$):

$$FeCl_3 + 3H_2O \to Fe(OH)_3 \, (colloid) + 3HCl$$

The colloidal particles of $$Fe(OH)_3$$ preferentially adsorb $$Fe^{3+}$$ ions from the solution onto their surface (since $$Fe^{3+}$$ is a common ion with the lattice of the colloid). This adsorption of positive $$Fe^{3+}$$ ions imparts a positive charge to the colloidal particles.

Therefore, the resulting colloidal particles carry a positive charge.

The correct answer is Option (1): positive.

The Tyndall effect is the phenomenon in which a beam of light becomes visible when it passes through a medium containing particles large enough to scatter the light. The necessary condition is that the dispersed particles should have a diameter roughly between $$10^{-7}\,\text{cm}$$ and $$10^{-5}\,\text{cm}$$ so that they can scatter visible radiation. If the particles are much smaller than this range, no appreciable scattering occurs, while if they are much larger, the medium ceases to be optically homogeneous and the beam gets blocked instead of only being scattered.

Now, let us examine each of the given dispersions in the light of this particle-size criterion.

We have a true solution: in a true solution the solute particles have dimensions typically less than $$10^{-8}\,\text{cm}$$. Because these particles are far smaller than the wavelength of visible light, they do not scatter light to any observable extent. So, a true solution does not exhibit the Tyndall effect.

Next, consider a suspension: in a suspension the particles are usually larger than $$10^{-5}\,\text{cm}$$. Such coarse particles block or absorb light instead of merely scattering it, so the passage of a narrow beam cannot be seen as a luminous path. Therefore, a suspension is also unsuited for showing the Tyndall effect.

Now we look at lyophilic colloids: in a lyophilic (solvent-loving) colloid, although the particle size is in the proper colloidal range, the dispersed particles are surrounded by thick layers of the solvent. This sheath diminishes the difference between the refractive indices of the dispersed phase and the dispersion medium, and light scattering depends directly on this refractive-index contrast. Because of the reduced contrast, a lyophilic colloid shows the Tyndall effect only weakly.

Lastly, consider lyophobic colloids: in a lyophobic (solvent-hating) colloid the particles lie squarely in the colloidal size range, and they do not carry thick solvent layers. Hence the refractive-index difference between the dispersed particles and the medium remains large, giving rise to strong scattering. Consequently, the luminous path of a light beam becomes very bright and distinct; in other words, the Tyndall effect is displayed most effectively by a lyophobic colloid.

Comparing all four cases, we find that the maximum light scattering, and hence the most prominent Tyndall effect, occurs for the lyophobic colloid.

Hence, the correct answer is Option C.

To decide which 0.10 M aqueous solution gives the largest lowering of the freezing point, we recall the colligative‐property formula for freezing point depression.

We first STATE the formula. The freezing point depression $$\Delta T_f$$ is related to the molality $$m$$ of the solute, the cryoscopic constant $$K_f$$ of the solvent, and the van ’t Hoff factor $$i$$ (number of particles produced per formula unit) by

$$\Delta T_f = i\,K_f\,m.$$

The solvent (water) and the concentration (0.10 M ≈ 0.10 m, because the solutions are dilute) are the same for every option, so $$K_f$$ and $$m$$ are identical for all four cases. Therefore the magnitude of $$\Delta T_f$$ will depend only on the van ’t Hoff factor $$i$$. The larger the value of $$i$$, the greater the product $$i\,K_f\,m$$, and hence the greater the freezing point depression.

Now we determine $$i$$ for each solute:

• Glycine, NH2CH2COOH, is an amino acid. In water it exists largely as the neutral zwitterion but does not dissociate into separate small ions to any significant extent at 0.10 M. Therefore it behaves as a nonelectrolyte. We have

$$\text{glycine (aq)} \longrightarrow \text{one particle (itself)}$$

so $$i = 1.$$

• Hydrazine, N2H4, is also a molecular compound that does not ionize appreciably in water. Thus

$$\text{hydrazine (aq)} \longrightarrow \text{one particle}$$

giving $$i = 1.$$

• Glucose, C6H12O6, is another typical nonelectrolyte that stays intact as molecules in solution. Hence

$$\text{glucose (aq)} \longrightarrow \text{one particle}$$

and again $$i = 1.$$

• Potassium hydrogen sulfate, KHSO4, is an ionic compound. On dissolving it at least undergoes the primary dissociation

$$\text{KHSO}_4 \longrightarrow \text{K}^+ + \text{HSO}_4^-.$$

This reaction creates two ions from one formula unit, so already $$i = 2.$$ In addition, the hydrogen sulfate ion can partially dissociate further:

$$\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}.$$

Because this secondary dissociation is only partial, it does not double the number of particles, but it does raise the effective $$i$$ slightly above 2. We can summarize it symbolically as

$$i = 2 + \alpha,$$

where $$\alpha$$ (fractional degree of second dissociation) is positive. Thus $$i > 2.$$

We now compare all four $$i$$ values:

$$i_{\text{glycine}} = 1,$$

$$i_{\text{hydrazine}} = 1,$$

$$i_{\text{glucose}} = 1,$$

$$i_{\text{KHSO}_4} > 2.$$

Because $$i_{\text{KHSO}_4}$$ is the largest, the product $$i\,K_f\,m$$, and hence $$\Delta T_f$$, is greatest for the 0.10 M KHSO4 solution.

So, KHSO4 produces the largest freezing point depression.

Hence, the correct answer is Option C.

This question tests understanding of the properties of glass and why sharp glass edges become smooth when heated.

Assertion A states that sharp glass edges become smooth on heating up to its melting point. This is true. Glass is an amorphous solid (a supercooled liquid) that does not have a sharp melting point but softens over a range of temperatures. When a sharp glass edge is heated, the surface tension of the softened glass pulls the sharp edge into a rounded, smooth shape to minimise surface area and thus surface energy. This is a well-known phenomenon in glass working.

Reason R states that the viscosity of glass decreases on melting. This is also true. Like all liquids, the viscosity of glass decreases significantly as temperature rises. At higher temperatures, the glass flows more readily.

However, R is NOT the correct explanation of A. The smoothing of sharp glass edges is due to surface tension, not directly due to decreased viscosity. Surface tension causes the molten glass to minimise its surface area by rounding the sharp edges. While decreased viscosity allows the glass to flow (which is necessary for the shape change to occur), the driving force for the smoothing is surface tension — the tendency to minimise surface energy. Decreased viscosity alone does not explain why sharp edges specifically become smooth; it only explains that flow is possible. The direction and nature of the shape change is dictated by surface tension.

Therefore, both A and R are true, but R is not the correct explanation of A. The correct answer is option 2.

We first recall the standard definitions from solid-state chemistry. A Frenkel defect occurs when an ion, generally the smaller cation, leaves its regular lattice site and occupies an interstitial site in the crystal. Because of this movement:

$$\text{Regular lattice site} \longrightarrow \text{vacancy}$$

and simultaneously

$$\text{Interstitial site} \longleftarrow \text{same ion}$$

Thus, one vacancy defect and one interstitial defect are created at the same time. Hence a Frenkel defect is quite literally the combination of a vacancy defect and an interstitial defect. Therefore Statement I, “Frenkel defects are vacancy as well as interstitial defects,” is true.

Next, we consider the origin of colour in many ionic solids. Colour often arises when an anion vacancy is occupied by an electron; such an electron-filled vacancy is called an F-centre (from the German word “Farbe,” meaning colour). The crystal that contains these F-centres can absorb visible light, which gives it colour. These F-centres are produced in a different kind of imperfection, namely the metal-excess defect (sometimes associated with Schottky-type anion vacancies), not in a Frenkel defect. A Frenkel defect merely displaces ions without introducing extra electrons into the lattice; it does not generate F-centres and therefore does not impart colour for that reason.

Consequently, Statement II, “Frenkel defect leads to colour in ionic solids due to presence of F-centres,” is false.

We have found that Statement I is true while Statement II is false. Examining the options, this correspondence matches Option C.

Hence, the correct answer is Option C.

We have to find, for every reaction given in List-II, the method of colloid preparation to which it belongs. The four methods mentioned in List-I are hydrolysis, reduction, oxidation and double decomposition. We shall compare each chemical equation with the characteristic feature of these methods one by one and obtain the correct matching.

First recall the identifying features of every method.

• In hydrolysis the salt of a weak base and a strong acid (or vice-versa) reacts with water. The metallic hydroxide produced remains as the sol.

• In reduction a metallic cation is reduced (its oxidation number decreases) to the free metal which appears in the colloidal form.

• In oxidation a species present in the solution is oxidised (its oxidation number increases) and the oxidised product forms the sol.

• In double decomposition two substances exchange their ions; an insoluble or sparingly soluble product separates out as the colloid.

Now we examine each reaction of List-II.

(iv) $$$\mathrm{FeCl_3 + 3H_2O \;\rightarrow\; Fe(OH)_3(sol) + 3HCl}$$$

Here ferric chloride reacts directly with water. We see water dissociates, chloride ions leave with hydrogen to form $$\mathrm{HCl}$$ and the basic hydroxide $$\mathrm{Fe(OH)_3}$$ stays in the colloidal state. This is exactly the definition of hydrolysis. So reaction (iv) corresponds to the hydrolysis method. Hence

$$ (a)\;{\text{Hydrolysis}} \;\longrightarrow\; (iv). $$

(i) $$$\mathrm{2AuCl_3 + 3HCHO + 3H_2O \;\rightarrow\; 2Au(sol) + 3HCOOH + 6HCl}$$$

Gold is present in the +3 oxidation state in $$\mathrm{AuCl_3}$$. On the product side free metallic gold $$\mathrm{Au}$$ is obtained; its oxidation number becomes 0. The oxidation number has decreased; hence gold ion has been reduced. The method used is therefore reduction. Thus

$$ (b)\;{\text{Reduction}} \;\longrightarrow\; (i). $$

(iii) $$$\mathrm{SO_2 + 2H_2S \;\rightarrow\; 3S(sol) + 2H_2O}$$$

Let us look at oxidation numbers:

• In $$\mathrm{SO_2}$$, sulphur is in +4 state. In elemental sulphur $$\mathrm{S}$$, it becomes 0, so sulphur in $$\mathrm{SO_2}$$ is reduced.

• In $$\mathrm{H_2S}$$, sulphur is -2; in elemental $$\mathrm{S}$$ it becomes 0, i.e. it is oxidised.

One species is oxidised, the other reduced, but the net formation of colloidal sulphur takes place because sulphide ion is being oxidised to free sulphur that remains in the dispersed phase. The essential step for colloid formation is the oxidation of $$\mathrm{H_2S}$$. Therefore this reaction is placed under the oxidation method. Consequently

$$ (c)\;{\text{Oxidation}} \;\longrightarrow\; (iii). $$

(ii) $$$\mathrm{As_2O_3 + 3H_2S \;\rightarrow\; As_2S_3(sol) + 3H_2O}$$$

Here arsenious oxide and hydrogen sulphide simply interchange the oxide and sulphide ions. No change in oxidation state of arsenic or sulphur occurs; the reaction proceeds through metathesis producing water and colloidal arsenic trisulphide, $$\mathrm{As_2S_3}$$, which is sparingly soluble. Such an ion-exchange reaction is termed double decomposition. Therefore

$$$ (d)\;{\text{Double Decomposition}} \;\longrightarrow\; (ii). $$$

Collecting all the matches we obtain

$$$ \begin{aligned} (a) &\;\rightarrow\; (iv),\\ (b) &\;\rightarrow\; (i),\\ (c) &\;\rightarrow\; (iii),\\ (d) &\;\rightarrow\; (ii). \end{aligned} $$$

This pattern exactly corresponds to Option B.

Hence, the correct answer is Option B.

Based on NCERT, the equation for pressure looks as follows:$$\frac{\ x}{m}=k.p^{\frac{\ 1}{n}}$$. The relationship is generally represented in the
form of a curve where mass of the gas adsorbed per gram of the adsorbent is plotted against pressure. These curves indicate that at a fixed pressure, there is a decrease in physical adsorption with increase in temperature. Thus, the correct answer is C.

First, let us recall the meanings of the two key terms that appear in the question.

Wet deposition is the process in which substances present in the atmosphere are washed out of the air and brought down to the Earth’s surface by precipitation such as rain, snow, fog or cloud water. Because the material is literally dissolved or suspended in atmospheric water before it reaches the ground, it is called “wet”. Typical examples include acids, nitrates and ammonium compounds that become incorporated into raindrops.

Dry deposition, on the other hand, involves the direct transfer of gases or solid particles from the air to surfaces (soil, vegetation, buildings) without the aid of precipitation. The material settles or is absorbed simply by contact with those surfaces under normal atmospheric motions. Common gases that undergo dry deposition include $$\mathrm{SO_2}$$, $$\mathrm{NO_2}$$ and ozone.

With these definitions in mind, we examine the species listed in the options.

1. Option A proposes $$X = \mathrm{CO_2}$$ and $$Y = \mathrm{SO_2}$$. Carbon dioxide is not typically removed from the atmosphere by rainfall; it stays dissolved only slightly in raindrops and is mainly exchanged biologically (photosynthesis). Hence $$\mathrm{CO_2}$$ is not a characteristic wet-deposition species. This option therefore conflicts with the definition.

2. Option B proposes $$X =$$ ammonium salts and $$Y = \mathrm{SO_2}$$. Ammonium ions $$\left( \mathrm{NH_4^+} \right)$$ usually arise from atmospheric ammonia that readily dissolves in cloud droplets forming ammonium sulfate, ammonium nitrate, etc. These dissolved salts are efficiently delivered to the surface by rain, fitting perfectly with wet deposition. Sulphur dioxide is a gas that adsorbs onto surfaces or is absorbed by leaves directly from air in the absence of rain, matching dry deposition. Hence this pairing is fully consistent with the two definitions.

3. Option C proposes $$X =$$ ammonium salts and $$Y = \mathrm{CO_2}$$. While the first part (ammonium salts) still agrees with wet deposition, the second part fails because, as argued earlier, $$\mathrm{CO_2}$$ is not generally classified under dry deposition in atmospheric chemistry contexts.

4. Option D proposes $$X = \mathrm{SO_2}$$ and $$Y =$$ ammonium salts. This directly contradicts the chemical behaviour: $$\mathrm{SO_2}$$, being a gas, predominantly reaches surfaces through dry deposition, while ammonium salts are taken down by rainwater. Hence this option reverses the correct roles.

After evaluating all four possibilities, we find that only Option B correctly assigns a predominantly rain-borne species (ammonium salts) to wet deposition and a gaseous species (SO$$_2$$) to dry deposition.

Hence, the correct answer is Option 2.

Let us examine each statement about colloidal solutions:

Option (1): A colloidal solution shows colligative properties. This is correct. Colloidal solutions, like true solutions, exhibit colligative properties such as osmotic pressure, though the effects are much smaller due to the large size of colloidal particles (and hence fewer particles per unit mass).

Option (2): An ordinary filter paper can stop the flow of colloidal particles. This is incorrect. Colloidal particles (size range 1-1000 nm) are small enough to pass through ordinary filter paper, which has pore sizes much larger than colloidal particles. Only ultra-filters (with very fine pores) or semipermeable membranes can separate colloidal particles. This is one of the key differences between colloidal solutions and suspensions.

Option (3): The flocculation power of $$Al^{3+}$$ is more than that of $$Na^+$$. This is correct. According to the Hardy-Schulze rule, the coagulating power of an ion increases with its charge. Since $$Al^{3+}$$ has a higher charge than $$Na^+$$, it has a greater flocculation (coagulation) power.

Option (4): A colloidal solution shows Brownian motion of colloidal particles. This is correct. Colloidal particles exhibit Brownian motion due to continuous bombardment by molecules of the dispersion medium.

The incorrect statement is option (2).

The solubility of gases in water is governed by temperature: as temperature increases, the solubility of gases decreases (Henry's law and Le Chatelier's principle, since dissolution of a gas is exothermic). Boiling water (100 $$^\circ$$C) and water at 80 $$^\circ$$C therefore contain very little dissolved oxygen. Polluted water typically has a low dissolved oxygen content because organic pollutants consume oxygen during biodegradation.

Water at 4 $$^\circ$$C is cold, and at lower temperatures gaseous oxygen is far more soluble. Among the given options, water at 4 $$^\circ$$C will have the highest concentration of dissolved $$\text{O}_2$$.

The answer is option (4).

Blood is a colloidal solution where the blood particles (proteins and cells) carry a negative charge. When blood flows from a wound, efficient clotting (coagulation) requires the neutralisation of these negative charges by positively charged ions.

According to the Hardy-Schulze rule, the coagulating power of an electrolyte increases with the valency of the ion carrying the charge opposite to that of the colloidal particles. Since blood is a negatively charged colloid, we need cations (positive ions) to cause coagulation, and the higher the valency of the cation, the greater the coagulating power.

Examining the given options: $$FeCl_3$$ provides $$Fe^{3+}$$ (trivalent cation), $$Mg(HCO_3)_2$$ provides $$Mg^{2+}$$ (divalent cation), $$FeSO_4$$ provides $$Fe^{2+}$$ (divalent cation), and $$NaHCO_3$$ provides $$Na^+$$ (monovalent cation).

Since $$Fe^{3+}$$ has the highest valency (+3) among all the cations listed, $$FeCl_3$$ will have the greatest coagulating power and will be the most suitable salt for efficient clotting of blood.

Therefore, the correct answer is Option (1): $$FeCl_3$$.

First, let us recall the fundamental characteristics of the two broad classes of solids studied in chemistry, namely crystalline solids and amorphous solids.

A crystalline solid possesses a regular three-dimensional arrangement of constituent particles (atoms, ions or molecules). Because this arrangement repeats in space, we say the solid exhibits long-range order. Hence statement (A) “Crystalline solids have long range order” is correct.

Next, we look at the directional dependence of physical properties. If a property such as refractive index, thermal conductivity or electrical resistance has the same magnitude in every direction, the solid is termed isotropic. Owing to their orderly yet direction-specific arrangement, crystalline solids generally show different values of these properties along different crystallographic directions; therefore crystalline solids are not isotropic but anisotropic. Statement (B) “Crystalline solids are isotropic” is thus incorrect.

An amorphous solid, in contrast, lacks long-range order; only short-range order is present. Because of this disordered, glass-like nature, such solids behave like super-cooled liquids and are sometimes referred to as pseudo solids or super-cooled liquids. Therefore statement (C) “Amorphous solids are sometimes called pseudo solids” is correct.

Furthermore, when heat is supplied, a crystalline solid melts sharply at one definite temperature: it has a fixed, well-defined melting point and hence a definite heat of fusion. An amorphous solid, however, does not melt abruptly; instead it gradually softens over a broad temperature interval, passing through a viscous stage before becoming a liquid. Hence statement (D) “Amorphous solids soften over a range of temperatures” is correct, while statement (E) “Amorphous solids have a definite heat of fusion” is incorrect because a definite heat of fusion is characteristic of crystalline, not amorphous, solids.

Collecting the results, the correct statements are (A), (C) and (D). These correspond to Option D in the list provided.

Hence, the correct answer is Option D.

The Tyndall effect refers to the scattering of light by colloidal particles, making the path of a light beam visible when passed through a colloid. Two conditions must be met for the Tyndall effect to be observed clearly.

First, regarding particle size: the colloidal particles must have a diameter that is comparable to the wavelength of the light used (condition A). If the particles are much smaller than the wavelength (condition B), scattering is negligible (Rayleigh scattering regime is weak for very small particles relative to wavelength). If the particles are much larger than the wavelength (condition C), they simply reflect and refract light like ordinary particles rather than scatter it in the diffuse manner characteristic of the Tyndall effect. Thus condition A is required, not B or C.

Second, regarding refractive index: there must be a significant difference in the refractive index between the dispersed phase and the dispersion medium (condition E). If the refractive indices are comparable (condition D), there is very little optical contrast between the particles and the medium, so scattering is minimal and the beam path is not visible. A large refractive index difference ensures that light is strongly scattered as it encounters the particles.

Therefore, the correct conditions for observing the Tyndall effect are (A) and (E) only. The correct answer is option 1.

We start by recalling the geometrical definitions used to classify the seven crystal systems. A unit-cell is described by three edge lengths $$a,\;b,\;c$$ and three inter-axial angles $$\alpha,\;\beta,\;\gamma$$. Each crystal system possesses a characteristic set of equalities (or inequalities) among these six parameters:

1. Cubic $$a=b=c,\;\;\alpha=\beta=\gamma=90^\circ$$     2. Tetragonal $$a=b\neq c,\;\;\alpha=\beta=\gamma=90^\circ$$
3. Orthorhombic $$a\neq b\neq c,\;\;\alpha=\beta=\gamma=90^\circ$$
4. Hexagonal $$a=b\neq c,\;\;\alpha=\beta=90^\circ,\;\gamma=120^\circ$$
5. Rhombohedral $$a=b=c,\;\;\alpha=\beta=\gamma\neq90^\circ$$
6. Monoclinic $$a\neq b\neq c,\;\;\alpha=\gamma=90^\circ,\;\beta\neq90^\circ$$
7. Triclinic $$a\neq b\neq c,\;\;\alpha\neq\beta\neq\gamma\neq90^\circ$$

Now we substitute the numerical data given in the question:

Edge lengths: $$a = 2.5,\;\;b = 3.0,\;\;c = 4.0$$
Angles:          $$\alpha = 90^\circ,\;\;\beta = 120^\circ,\;\;\gamma = 90^\circ$$

We examine the equalities one by one.

• Lengths: Clearly $$a\neq b,\;b\neq c,\;a\neq c,$$ so all three edges are unequal.
• Angles: We notice $$\alpha=90^\circ$$ and $$\gamma=90^\circ,$$ but $$\beta=120^\circ\neq90^\circ.$$

So the unit-cell satisfies

$$a\neq b\neq c,\qquad \alpha=\gamma=90^\circ,\qquad \beta\neq90^\circ.$$

Comparing this pattern with the list above, we see it matches exactly the monoclinic conditions:

Monoclinic $$a\neq b\neq c,\;\;\alpha=\gamma=90^\circ,\;\beta\neq90^\circ.$$

None of the other crystal systems possess two right angles together with a third angle different from $$90^\circ$$ while still having all three edges unequal, so no alternative classification is possible.

Hence, the correct answer is Option C.

We have to decide in which of the four cases the colloidal particles finally produced carry a net negative charge. The sign of the charge on a colloid is governed by the ion that finally gets adsorbed on the freshly formed precipitate. Whichever ion remains in excess in the dispersion medium is preferentially adsorbed.

First, consider Option A: $$\text{AgNO}_3 \text{ added to KI solution}.$$ Because the KI solution is taken first, $$[ \text{I}^- ]$$ is large. On mixing, the insoluble precipitate $$\text{AgI}$$ is produced by

$$\text{Ag}^+ + \text{I}^- \longrightarrow \text{AgI}\,(\text{solid}).$$

There is still a large excess of $$\text{I}^-$$ ions left in the medium. According to the adsorption theory, these excess $$\text{I}^-$$ ions get adsorbed on the surface of the newly formed $$\text{AgI}$$ particles. So the adsorbed ion is $$\text{I}^-$$, a negative ion, and the charge on each colloidal particle becomes negative.

Now, Option B: $$\text{KI added to AgNO}_3 \text{ solution}.$$ Here $$[ \text{Ag}^+ ]$$ is in excess because the AgNO$$_3$$ solution is taken first. The same precipitate $$\text{AgI}$$ forms, but now the excess ion available is $$\text{Ag}^+$$. These positive $$\text{Ag}^+$$ ions are adsorbed, giving the $$\text{AgI}$$ sol a positive charge, not a negative one.

Option C involves $$\text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O}$$ (hydrated alumina) shaken with pure water. Hydrated alumina has the amphoteric surface $$\text{Al(OH)}_3$$, which preferentially adsorbs $$\text{H}^+$$ ions from water:

$$\text{Al(OH)}_3 + \text{H}^+ \rightarrow \text{Al(OH)}_3 \cdot \text{H}^+.$$

The adsorbed ion is $$\text{H}^+,$$ so the sol is positively charged.

Option D: $$\text{FeCl}_3$$ hydrolyses in hot water, producing a ferric hydroxide sol:

$$\text{Fe}^{3+} + 3\text{H}_2\text{O} \longrightarrow \text{Fe(OH)}_3\,(\text{solid}) + 3\text{H}^+.$$

The medium contains excess $$\text{H}^+$$; these $$\text{H}^+$$ ions adsorb on the $$\text{Fe(OH)}_3$$ particles, again resulting in a positively charged sol.

Summarising, only in Option A do the colloidal particles carry excess adsorbed $$\text{I}^-$$ ions, giving them a negative charge. All the other options produce sols with adsorbed positive ions.

Hence, the correct answer is Option A.

The depression in freezing point is given by $$\Delta T_f = i \, K_f \, m$$, where $$i$$ is the van 't Hoff factor (the number of particles the solute produces in solution), $$K_f$$ is the cryoscopic constant, and $$m$$ is the molality. Since all solutions have the same concentration (0.06 M), the solution with the largest $$i$$ will have the greatest freezing point depression and hence the lowest freezing point.

$$\text{C}_6\text{H}_{12}\text{O}_6$$ (glucose) is a non-electrolyte: $$i = 1$$. $$\text{KI}$$ dissociates into $$\text{K}^+$$ and $$\text{I}^-$$: $$i = 2$$. $$\text{K}_2\text{SO}_4$$ dissociates into $$2\text{K}^+$$ and $$\text{SO}_4^{2-}$$: $$i = 3$$. $$\text{Al}_2(\text{SO}_4)_3$$ dissociates into $$2\text{Al}^{3+}$$ and $$3\text{SO}_4^{2-}$$: $$i = 5$$.

Since $$\text{Al}_2(\text{SO}_4)_3$$ produces the most ions ($$i = 5$$), its 0.06 M solution will have the largest $$\Delta T_f$$ and therefore the lowest freezing point.

The answer is option (1).

Let us examine each statement about hydrophilic sols.

Statement (1): "These sols are reversible in nature." This is true. Hydrophilic sols can be reconstituted after evaporation by simply adding the dispersion medium back, making them reversible.

Statement (2): "Their viscosity is of the order of that of $$H_2O$$." This is FALSE. Hydrophilic sols have much higher viscosity than water because the dispersed particles are heavily hydrated (solvated), with thick layers of water molecules bound to them. This extensive solvation significantly increases the viscosity compared to pure water.

Statement (3): "The sols cannot be easily coagulated." This is true. Hydrophilic sols are stabilised by the solvation layer around the particles, making them resistant to coagulation. Large amounts of electrolyte are needed to coagulate them.

Statement (4): "They do not require electrolytes for stability." This is true. Hydrophilic sols are self-stabilised by solvation and do not need added electrolytes for stability, unlike hydrophobic sols.

Therefore, the FALSE statement is option (2).