Chapter 1: Some Basic Concepts of Chemistry

The molecular mass of a substance equals the sum of the atomic masses of all the atoms present in one molecule of the substance.

Glucose has the formula $$\mathrm{C_6H_{12}O_6}$$, so one molecule of glucose contains $$6$$ carbon atoms, $$12$$ hydrogen atoms and $$6$$ oxygen atoms.

Using the standard atomic masses $$\mathrm{C} = 12 \, \mathrm{u}$$, $$\mathrm{H} = 1 \, \mathrm{u}$$ and $$\mathrm{O} = 16 \, \mathrm{u}$$:

$$\text{Molecular mass of glucose} = 6 \times (12 \, \mathrm{u}) + 12 \times (1 \, \mathrm{u}) + 6 \times (16 \, \mathrm{u})$$

$$= 72 \, \mathrm{u} + 12 \, \mathrm{u} + 96 \, \mathrm{u} = 180 \, \mathrm{u}$$

Step 1: Convert mass percentages to moles. Take $$100 \, \mathrm{g}$$ of the compound so that the percentages directly give the masses of each element. Divide each by its atomic mass to get moles.

$$n_{\mathrm{H}} = \dfrac{4.07}{1.008} = 4.04 \; \mathrm{mol}$$

$$n_{\mathrm{C}} = \dfrac{24.27}{12.01} = 2.021 \; \mathrm{mol}$$

$$n_{\mathrm{Cl}} = \dfrac{71.65}{35.45} = 2.021 \; \mathrm{mol}$$

Step 2: Find the simplest whole-number ratio. Divide every mole value by the smallest (here, $$2.021$$):

$$\mathrm{H} : \mathrm{C} : \mathrm{Cl} = \dfrac{4.04}{2.021} : \dfrac{2.021}{2.021} : \dfrac{2.021}{2.021} = 2 : 1 : 1$$

So the empirical formula is $$\mathrm{CH_2Cl}$$.

Step 3: Determine the molecular formula. Empirical formula mass:

$$M_{\text{emp}} = 12.01 + 2(1.008) + 35.45 = 49.48 \, \mathrm{g\,mol^{-1}}$$

The integer multiple $$n$$ relating empirical mass to molar mass is

$$n = \dfrac{M_{\text{molar}}}{M_{\text{emp}}} = \dfrac{98.96}{49.48} = 2$$

Therefore the molecular formula is $$(\mathrm{CH_2Cl})_2 = \mathrm{C_2H_4Cl_2}$$.

The balanced combustion equation for methane is

$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$$

From the stoichiometry, $$1 \, \mathrm{mol}$$ of $$\mathrm{CH_4}$$ produces $$2 \, \mathrm{mol}$$ of $$\mathrm{H_2O}$$.

Step 1: Moles of methane. Molar mass of $$\mathrm{CH_4} = 12 + 4 \times 1 = 16 \, \mathrm{g\,mol^{-1}}$$.

$$n_{\mathrm{CH_4}} = \dfrac{16 \, \mathrm{g}}{16 \, \mathrm{g\,mol^{-1}}} = 1 \, \mathrm{mol}$$

Step 2: Moles of water produced.

$$n_{\mathrm{H_2O}} = 2 \times n_{\mathrm{CH_4}} = 2 \times 1 = 2 \, \mathrm{mol}$$

Step 3: Mass of water produced. Molar mass of $$\mathrm{H_2O} = 18 \, \mathrm{g\,mol^{-1}}$$.

$$m_{\mathrm{H_2O}} = 2 \, \mathrm{mol} \times 18 \, \mathrm{g\,mol^{-1}} = 36 \, \mathrm{g}$$

The combustion reaction is

$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$$

From the balanced equation, $$1 \, \mathrm{mol}$$ of $$\mathrm{CH_4}$$ produces $$1 \, \mathrm{mol}$$ of $$\mathrm{CO_2}$$.

Step 1: Moles of $$\mathrm{CO_2}$$ formed. Molar mass of $$\mathrm{CO_2} = 12 + 2 \times 16 = 44 \, \mathrm{g\,mol^{-1}}$$.

$$n_{\mathrm{CO_2}} = \dfrac{22 \, \mathrm{g}}{44 \, \mathrm{g\,mol^{-1}}} = 0.5 \, \mathrm{mol}$$

Step 2: Moles of $$\mathrm{CH_4}$$ needed. By stoichiometry,

$$n_{\mathrm{CH_4}} = n_{\mathrm{CO_2}} = 0.5 \, \mathrm{mol}$$

The balanced equation for the Haber process is

$$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$$

Step 1: Compute moles of each reactant.

$$n_{\mathrm{N_2}} = \dfrac{50.0 \times 10^3 \, \mathrm{g}}{28 \, \mathrm{g\,mol^{-1}}} = 1.785 \times 10^3 \, \mathrm{mol}$$

$$n_{\mathrm{H_2}} = \dfrac{10.0 \times 10^3 \, \mathrm{g}}{2 \, \mathrm{g\,mol^{-1}}} = 5.00 \times 10^3 \, \mathrm{mol}$$

Step 2: Identify the limiting reagent. The reaction requires $$3 \, \mathrm{mol}$$ of $$\mathrm{H_2}$$ per mole of $$\mathrm{N_2}$$. The $$\mathrm{H_2}$$ needed to consume all of the $$\mathrm{N_2}$$ would be

$$3 \times 1.785 \times 10^3 = 5.355 \times 10^3 \, \mathrm{mol}$$

which is greater than the $$5.00 \times 10^3 \, \mathrm{mol}$$ of $$\mathrm{H_2}$$ available. Hence $$\mathrm{H_2}$$ is the limiting reagent and $$\mathrm{N_2}$$ is present in excess.

Step 3: Moles and mass of $$\mathrm{NH_3}$$ formed. From stoichiometry, $$3 \, \mathrm{mol} \; \mathrm{H_2}$$ gives $$2 \, \mathrm{mol} \; \mathrm{NH_3}$$.

$$n_{\mathrm{NH_3}} = \dfrac{2}{3} \times n_{\mathrm{H_2}} = \dfrac{2}{3} \times 5.00 \times 10^3 = 3.33 \times 10^3 \, \mathrm{mol}$$

Molar mass of $$\mathrm{NH_3} = 14 + 3 \times 1 = 17 \, \mathrm{g\,mol^{-1}}$$.

$$m_{\mathrm{NH_3}} = 3.33 \times 10^3 \, \mathrm{mol} \times 17 \, \mathrm{g\,mol^{-1}} = 5.66 \times 10^4 \, \mathrm{g} = 56.6 \, \mathrm{kg}$$

The mass percent of solute is defined as

$$\text{Mass \% of A} = \dfrac{\text{mass of A}}{\text{mass of solution}} \times 100$$

Mass of solution = mass of solute + mass of solvent = $$2 + 18 = 20 \, \mathrm{g}$$.

$$\text{Mass \% of A} = \dfrac{2 \, \mathrm{g}}{20 \, \mathrm{g}} \times 100 = 10\%$$

Molarity is defined as moles of solute per litre of solution:

$$M = \dfrac{\text{moles of solute}}{\text{volume of solution in L}}$$

Step 1: Moles of NaOH. Molar mass of $$\mathrm{NaOH} = 23 + 16 + 1 = 40 \, \mathrm{g\,mol^{-1}}$$.

$$n_{\mathrm{NaOH}} = \dfrac{4 \, \mathrm{g}}{40 \, \mathrm{g\,mol^{-1}}} = 0.1 \, \mathrm{mol}$$

Step 2: Convert volume to litres.

$$V = 250 \, \mathrm{mL} = 0.250 \, \mathrm{L}$$

Step 3: Molarity.

$$M = \dfrac{0.1 \, \mathrm{mol}}{0.250 \, \mathrm{L}} = 0.4 \, \mathrm{mol\,L^{-1}} = 0.4 \, \mathrm{M}$$

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

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

Step 1: Choose a convenient sample. Take $$1 \, \mathrm{L}$$ (i.e. $$1000 \, \mathrm{mL}$$) of the $$3 \, \mathrm{M}$$ solution. This contains exactly $$3 \, \mathrm{mol}$$ of NaCl.

Step 2: Mass of NaCl present. Molar mass of $$\mathrm{NaCl} = 23 + 35.5 = 58.5 \, \mathrm{g\,mol^{-1}}$$.

$$m_{\mathrm{NaCl}} = 3 \, \mathrm{mol} \times 58.5 \, \mathrm{g\,mol^{-1}} = 175.5 \, \mathrm{g}$$

Step 3: Mass of the solution. Using the given density,

$$m_{\text{soln}} = V \times d = 1000 \, \mathrm{mL} \times 1.25 \, \mathrm{g\,mL^{-1}} = 1250 \, \mathrm{g}$$

Step 4: Mass of water (solvent).

$$m_{\text{water}} = 1250 - 175.5 = 1074.5 \, \mathrm{g} = 1.0745 \, \mathrm{kg}$$

Step 5: Molality.

$$m = \dfrac{3 \, \mathrm{mol}}{1.0745 \, \mathrm{kg}} = 2.79 \, \mathrm{m}$$

Step 1: Compute the molar mass of $$\mathrm{Na_2SO_4}$$.

Using atomic masses $$\mathrm{Na} = 23$$, $$\mathrm{S} = 32$$ and $$\mathrm{O} = 16 \, \mathrm{g\,mol^{-1}}$$:

$$M(\mathrm{Na_2SO_4}) = 2 \times 23 + 1 \times 32 + 4 \times 16 = 46 + 32 + 64 = 142 \, \mathrm{g\,mol^{-1}}$$

Step 2: Mass percent of each element. The mass percent is

$$\%\mathrm{element} = \dfrac{\text{total mass of element in } 1 \text{ mole}}{\text{molar mass of compound}} \times 100$$

$$\%\mathrm{Na} = \dfrac{2 \times 23}{142} \times 100 = \dfrac{46}{142} \times 100 = 32.39\%$$

$$\%\mathrm{S} = \dfrac{32}{142} \times 100 = 22.54\%$$

$$\%\mathrm{O} = \dfrac{4 \times 16}{142} \times 100 = \dfrac{64}{142} \times 100 = 45.07\%$$

The percentages add to $$32.39 + 22.54 + 45.07 = 100.00\%$$, as required.

Take a $$100 \, \mathrm{g}$$ sample so that the percentages directly give masses in grams: $$69.9 \, \mathrm{g}$$ of Fe and $$30.1 \, \mathrm{g}$$ of O.

Step 1: Convert masses to moles. Using $$M(\mathrm{Fe}) = 55.85 \, \mathrm{g\,mol^{-1}}$$ and $$M(\mathrm{O}) = 16 \, \mathrm{g\,mol^{-1}}$$:

$$n_{\mathrm{Fe}} = \dfrac{69.9}{55.85} = 1.251 \, \mathrm{mol}$$

$$n_{\mathrm{O}} = \dfrac{30.1}{16} = 1.881 \, \mathrm{mol}$$

Step 2: Divide by the smallest mole value.

$$\dfrac{n_{\mathrm{Fe}}}{1.251} = 1.000 \qquad \dfrac{n_{\mathrm{O}}}{1.251} = 1.504$$

Step 3: Multiply by a small integer to obtain whole numbers. Multiplying both by $$2$$ gives

$$\mathrm{Fe} : \mathrm{O} = 2 : 3$$

Therefore the empirical formula is $$\mathrm{Fe_2O_3}$$.

From the definition of molarity, the number of moles of solute is

$$n = M \times V$$

where $$V$$ is the volume of solution in litres.

Step 1: Convert volume.

$$V = 500 \, \mathrm{mL} = 0.500 \, \mathrm{L}$$

Step 2: Moles of sodium acetate required.

$$n = 0.375 \, \mathrm{mol\,L^{-1}} \times 0.500 \, \mathrm{L} = 0.1875 \, \mathrm{mol}$$

Step 3: Mass required.

$$m = n \times M_m = 0.1875 \, \mathrm{mol} \times 82.0245 \, \mathrm{g\,mol^{-1}} = 15.38 \, \mathrm{g}$$

Take $$1 \, \mathrm{L} = 1000 \, \mathrm{mL}$$ of the solution as a convenient sample.

Step 1: Mass of the solution.

$$m_{\text{soln}} = V \times d = 1000 \, \mathrm{mL} \times 1.41 \, \mathrm{g\,mL^{-1}} = 1410 \, \mathrm{g}$$

Step 2: Mass of $$\mathrm{HNO_3}$$. $$69\%$$ of the solution mass is the acid:

$$m_{\mathrm{HNO_3}} = \dfrac{69}{100} \times 1410 = 972.9 \, \mathrm{g}$$

Step 3: Moles of $$\mathrm{HNO_3}$$. Molar mass of $$\mathrm{HNO_3} = 1 + 14 + 3 \times 16 = 63 \, \mathrm{g\,mol^{-1}}$$.

$$n = \dfrac{972.9 \, \mathrm{g}}{63 \, \mathrm{g\,mol^{-1}}} = 15.44 \, \mathrm{mol}$$

Step 4: Concentration. Since this is the amount contained in exactly $$1 \, \mathrm{L}$$,

$$M = \dfrac{n}{V} = \dfrac{15.44 \, \mathrm{mol}}{1 \, \mathrm{L}} = 15.44 \, \mathrm{mol\,L^{-1}}$$

Every mole of $$\mathrm{CuSO_4}$$ contains exactly one mole of Cu, so the mass of Cu obtainable is given by

$$m_{\mathrm{Cu}} = m_{\mathrm{CuSO_4}} \times \dfrac{M(\mathrm{Cu})}{M(\mathrm{CuSO_4})}$$

Step 1: Molar mass of $$\mathrm{CuSO_4}$$. Using $$\mathrm{Cu} = 63.5$$, $$\mathrm{S} = 32$$, $$\mathrm{O} = 16 \, \mathrm{g\,mol^{-1}}$$:

$$M(\mathrm{CuSO_4}) = 63.5 + 32 + 4 \times 16 = 63.5 + 32 + 64 = 159.5 \, \mathrm{g\,mol^{-1}}$$

Step 2: Mass fraction of Cu.

$$\dfrac{M(\mathrm{Cu})}{M(\mathrm{CuSO_4})} = \dfrac{63.5}{159.5} = 0.3981$$

Step 3: Mass of Cu in $$100 \, \mathrm{g}$$ of $$\mathrm{CuSO_4}$$.

$$m_{\mathrm{Cu}} = 100 \, \mathrm{g} \times 0.3981 = 39.81 \, \mathrm{g}$$

The composition is the same as in Exercise 1.3.

Step 1: Empirical formula. For a $$100 \, \mathrm{g}$$ sample,

$$n_{\mathrm{Fe}} = \dfrac{69.9}{55.85} = 1.251 \, \mathrm{mol}, \qquad n_{\mathrm{O}} = \dfrac{30.1}{16} = 1.881 \, \mathrm{mol}$$

Dividing both by the smaller:

$$\mathrm{Fe} : \mathrm{O} = 1 : 1.504$$

Multiplying by $$2$$: $$\mathrm{Fe} : \mathrm{O} = 2 : 3$$, so the empirical formula is $$\mathrm{Fe_2O_3}$$.

Step 2: Empirical formula mass.

$$M_{\text{emp}} = 2 \times 55.85 + 3 \times 16 = 111.7 + 48 = 159.7 \, \mathrm{g\,mol^{-1}}$$

Step 3: Molecular formula. $$\mathrm{Fe_2O_3}$$ (haematite) is a stable, well-known iron oxide; its actual molar mass coincides with the empirical formula mass, so

$$n = \dfrac{M_{\text{mol}}}{M_{\text{emp}}} = 1$$

Therefore the molecular formula is the same as the empirical formula: $$\mathrm{Fe_2O_3}$$.

The average atomic mass is the weighted mean of the isotopic masses, with the natural abundances as weights:

$$\bar M = \sum_i f_i M_i$$

where $$f_i$$ is the fractional abundance (percent abundance $$\div\, 100$$) and $$M_i$$ the isotopic molar mass.

Step 1: Fractional abundances.

$$f(^{35}\mathrm{Cl}) = \dfrac{75.77}{100} = 0.7577$$

$$f(^{37}\mathrm{Cl}) = \dfrac{24.23}{100} = 0.2423$$

Step 2: Weighted average.

$$\bar M = (0.7577)(34.9689) + (0.2423)(36.9659)$$

$$= 26.4959 + 8.9568 = 35.4527 \, \mathrm{u}$$

Step 1: Molar mass of sucrose. Using $$\mathrm{C} = 12$$, $$\mathrm{H} = 1$$, $$\mathrm{O} = 16$$:

$$M(\mathrm{C_{12}H_{22}O_{11}}) = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 \, \mathrm{g\,mol^{-1}}$$

Step 2: Moles of sugar.

$$n = \dfrac{20 \, \mathrm{g}}{342 \, \mathrm{g\,mol^{-1}}} = 0.0585 \, \mathrm{mol}$$

Step 3: Molarity.

$$M = \dfrac{n}{V} = \dfrac{0.0585 \, \mathrm{mol}}{2 \, \mathrm{L}} = 0.0292 \, \mathrm{mol\,L^{-1}}$$

Step 1: Moles of methanol needed.

$$n = M \times V = 0.25 \, \mathrm{mol\,L^{-1}} \times 2.5 \, \mathrm{L} = 0.625 \, \mathrm{mol}$$

Step 2: Mass of methanol. Molar mass of $$\mathrm{CH_3OH} = 12 + 4(1) + 16 = 32 \, \mathrm{g\,mol^{-1}}$$.

$$m = n \times M = 0.625 \, \mathrm{mol} \times 32 \, \mathrm{g\,mol^{-1}} = 20 \, \mathrm{g} = 0.020 \, \mathrm{kg}$$

Step 3: Volume of methanol. Using $$V = m/d$$ with $$d = 0.793 \, \mathrm{kg\,L^{-1}}$$:

$$V = \dfrac{0.020 \, \mathrm{kg}}{0.793 \, \mathrm{kg\,L^{-1}}} = 0.02522 \, \mathrm{L} = 25.22 \, \mathrm{mL}$$

The pressure exerted by a column of air at the surface is

$$P = \dfrac{F}{A} = \dfrac{m \, g}{A}$$

The given quantity $$m/A = 1034 \, \mathrm{g\,cm^{-2}}$$ already has the right structure; multiply by $$g$$ after converting to SI.

Step 1: Convert $$m/A$$ to SI units. $$1 \, \mathrm{g} = 10^{-3} \, \mathrm{kg}$$ and $$1 \, \mathrm{cm^{-2}} = 10^{4} \, \mathrm{m^{-2}}$$:

$$\dfrac{m}{A} = 1034 \, \mathrm{g\,cm^{-2}} = 1034 \times 10^{-3} \, \mathrm{kg} \times 10^{4} \, \mathrm{m^{-2}} = 1.034 \times 10^{4} \, \mathrm{kg\,m^{-2}}$$

Step 2: Multiply by $$g = 9.8 \, \mathrm{m\,s^{-2}}$$.

$$P = \dfrac{m}{A} \, g = 1.034 \times 10^{4} \, \mathrm{kg\,m^{-2}} \times 9.8 \, \mathrm{m\,s^{-2}}$$

$$= 1.013 \times 10^{5} \, \mathrm{kg\,m^{-1}\,s^{-2}} = 1.013 \times 10^{5} \, \mathrm{Pa}$$

This is consistent with the well-known sea-level atmospheric pressure of about $$1.013 \times 10^{5} \, \mathrm{Pa}$$ ($$1 \, \mathrm{atm}$$).

The SI unit of mass is the kilogram, symbol $$\mathrm{kg}$$.

Since May 2019, the kilogram has been defined by fixing the numerical value of the Planck constant, $$h$$, to be exactly

$$h = 6.62607015 \times 10^{-34} \, \mathrm{J\,s} = 6.62607015 \times 10^{-34} \, \mathrm{kg\,m^2\,s^{-1}}$$

with the metre and the second defined by the speed of light and the caesium hyperfine transition, respectively.

(Historically — and as stated in the NCERT textbook — the kilogram was defined as the mass of a particular cylinder of platinum-iridium alloy kept at the International Bureau of Weights and Measures in Sèvres, France.)

The SI prefixes denote the following powers of ten:

Pairing each prefix with the matching entry from the second column gives the answer below.

The significant figures of a measured quantity are all the digits known with certainty plus the first uncertain (estimated) digit. They convey the precision of a measurement.

The standard rules used to count significant figures are:

1. All non-zero digits are significant. E.g. $$285 \, \mathrm{cm}$$ has $$3$$ significant figures and $$0.25 \, \mathrm{mL}$$ has $$2$$.
2. Zeros preceding the first non-zero digit are not significant; they only fix the decimal point. E.g. $$0.03$$ has $$1$$ significant figure and $$0.0052$$ has $$2$$.
3. Zeros between two non-zero digits are significant. E.g. $$2.005$$ has $$4$$ significant figures.
4. Zeros at the end of a number with a decimal point are significant. E.g. $$0.200 \, \mathrm{g}$$ has $$3$$ significant figures.
5. For numbers without a decimal point, terminal zeros are ambiguous; expressing the value in scientific notation removes the ambiguity. For example, $$100$$ may be written $$1 \times 10^{2}$$ (one s.f.) or $$1.00 \times 10^{2}$$ (three s.f.).
6. Exact numbers (e.g. counts, defined quantities such as $$60 \, \mathrm{s}/\mathrm{min}$$) have an infinite number of significant figures.

Use $$d = v \, t$$ after converting the time to SI seconds.

Step 1: Convert time. $$1 \, \mathrm{ns} = 10^{-9} \, \mathrm{s}$$, so

$$t = 2.00 \, \mathrm{ns} = 2.00 \times 10^{-9} \, \mathrm{s}$$

Step 2: Compute distance.

$$d = v \, t = (3.0 \times 10^{8} \, \mathrm{m\,s^{-1}})(2.00 \times 10^{-9} \, \mathrm{s})$$

$$= 6.0 \times 10^{-1} \, \mathrm{m} = 0.60 \, \mathrm{m}$$

The two quantities measure different things.

• $$0.50 \, \mathrm{mol \; Na_2CO_3}$$ specifies an amount of the substance — namely $$0.5$$ times Avogadro's number ($$0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$$) of formula units of $$\mathrm{Na_2CO_3}$$, which corresponds to a mass $$0.50 \times 106 = 53 \, \mathrm{g}$$. It says nothing about volume or solvent.
• $$0.50 \, \mathrm{M \; Na_2CO_3}$$ specifies a concentration in solution — $$0.50 \, \mathrm{mol}$$ of $$\mathrm{Na_2CO_3}$$ dissolved in enough water to give $$1 \, \mathrm{L}$$ of solution. Different volumes of this solution contain different amounts of solute.

For example, $$1 \, \mathrm{L}$$ of $$0.50 \, \mathrm{M \; Na_2CO_3}$$ contains $$0.50 \, \mathrm{mol}$$ of $$\mathrm{Na_2CO_3}$$, but $$500 \, \mathrm{mL}$$ of the same solution contains only $$0.25 \, \mathrm{mol}$$.

The balanced equation is

$$\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)}$$

According to Gay-Lussac's law of combining volumes (or, equivalently, Avogadro's law), the volume ratios of reactants and products of a gas-phase reaction (at the same temperature and pressure) are equal to the mole ratios.

From the equation, $$2 \, \text{vol} \; \mathrm{H_2} : 1 \, \text{vol} \; \mathrm{O_2} : 2 \, \text{vol} \; \mathrm{H_2O}$$.

Here $$10 \, \text{vol} \; \mathrm{H_2}$$ require exactly $$5 \, \text{vol} \; \mathrm{O_2}$$ (which is what is supplied), so the reactants are in the stoichiometric ratio. They produce

$$\dfrac{2}{2} \times 10 = 10 \, \text{vol of water vapour}$$

Pick a convenient basis of total moles equal to $$1 \, \mathrm{mol}$$ of solution. Then

$$n_{\text{ethanol}} = x_{\text{ethanol}} \times 1 = 0.040 \, \mathrm{mol}$$

$$n_{\text{water}} = 1 - 0.040 = 0.960 \, \mathrm{mol}$$

Step 1: Mass of water. Molar mass of $$\mathrm{H_2O} = 18 \, \mathrm{g\,mol^{-1}}$$.

$$m_{\text{water}} = 0.960 \, \mathrm{mol} \times 18 \, \mathrm{g\,mol^{-1}} = 17.28 \, \mathrm{g}$$

Step 2: Volume of water (taken as the volume of the dilute solution). With $$d_{\text{water}} = 1 \, \mathrm{g\,mL^{-1}}$$:

$$V \approx \dfrac{m_{\text{water}}}{d_{\text{water}}} = \dfrac{17.28 \, \mathrm{g}}{1 \, \mathrm{g\,mL^{-1}}} = 17.28 \, \mathrm{mL} = 0.01728 \, \mathrm{L}$$

Step 3: Molarity.

$$M = \dfrac{n_{\text{ethanol}}}{V} = \dfrac{0.040 \, \mathrm{mol}}{0.01728 \, \mathrm{L}} = 2.315 \, \mathrm{mol\,L^{-1}}$$

By definition, $$1 \, \mathrm{mol}$$ of $$\mathrm{^{12}C}$$ atoms has a mass of exactly $$12 \, \mathrm{g}$$ and contains Avogadro's number $$N_A = 6.022 \times 10^{23}$$ atoms.

Therefore the mass of a single $$\mathrm{^{12}C}$$ atom is

$$m_{\mathrm{^{12}C}} = \dfrac{12 \, \mathrm{g}}{6.022 \times 10^{23}} = 1.993 \times 10^{-23} \, \mathrm{g}$$

The average molar mass is the weighted mean

$$\bar M = \sum_i \dfrac{a_i}{100} \, M_i$$

with $$a_i$$ the percent abundance and $$M_i$$ the isotopic molar mass.

Step 1: Compute each weighted contribution.

$$\dfrac{0.337}{100} \times 35.96755 = 0.12121 \, \mathrm{g\,mol^{-1}}$$

$$\dfrac{0.063}{100} \times 37.96272 = 0.02392 \, \mathrm{g\,mol^{-1}}$$

$$\dfrac{99.600}{100} \times 39.9624 = 39.80255 \, \mathrm{g\,mol^{-1}}$$

Step 2: Add the contributions.

$$\bar M = 0.12121 + 0.02392 + 39.80255 = 39.94768 \, \mathrm{g\,mol^{-1}}$$

Reported to a sensible number of figures, $$\bar M(\mathrm{Ar}) \approx 39.948 \, \mathrm{g\,mol^{-1}}$$.

All of the carbon in the gas ends up as $$\mathrm{CO_2}$$, and all the hydrogen ends up as $$\mathrm{H_2O}$$.

Step 1: Mass of C in the sample. Every $$44 \, \mathrm{g}$$ of $$\mathrm{CO_2}$$ contains $$12 \, \mathrm{g}$$ of C.

$$m_{\mathrm{C}} = 3.38 \times \dfrac{12}{44} = 0.9218 \, \mathrm{g}$$

Step 2: Mass of H in the sample. Every $$18 \, \mathrm{g}$$ of $$\mathrm{H_2O}$$ contains $$2 \, \mathrm{g}$$ of H.

$$m_{\mathrm{H}} = 0.690 \times \dfrac{2}{18} = 0.0767 \, \mathrm{g}$$

(Mass of C + Mass of H $$= 0.9985 \, \mathrm{g} \approx$$ mass of sample burnt, as expected.)

Step 3: Moles of each element.

$$n_{\mathrm{C}} = \dfrac{0.9218}{12} = 0.0768 \, \mathrm{mol}$$

$$n_{\mathrm{H}} = \dfrac{0.0767}{1} = 0.0767 \, \mathrm{mol}$$

Ratio $$n_{\mathrm{C}} : n_{\mathrm{H}} = 1 : 1$$, so the empirical formula is $$\mathrm{CH}$$ with empirical mass

$$M_{\text{emp}} = 12 + 1 = 13 \, \mathrm{g\,mol^{-1}}$$

Step 4: Molar mass from gas density at STP. Using the NCERT STP molar volume $$V_m = 22.4 \, \mathrm{L\,mol^{-1}}$$ (i.e. $$22.4 \, \mathrm{L}$$ of any gas weighs $$M$$ grams):

$$M = \dfrac{11.6 \, \mathrm{g}}{10.0 \, \mathrm{L}} \times 22.4 \, \mathrm{L\,mol^{-1}} = 25.98 \, \mathrm{g\,mol^{-1}} \approx 26 \, \mathrm{g\,mol^{-1}}$$

Step 5: Molecular formula.

$$n = \dfrac{M}{M_{\text{emp}}} = \dfrac{26}{13} = 2$$

Therefore the molecular formula is $$(\mathrm{CH})_2 = \mathrm{C_2H_2}$$ — ethyne (acetylene), which is indeed used as a welding fuel.

Step 1: Moles of HCl supplied.

$$n_{\mathrm{HCl}} = M \times V = 0.75 \, \mathrm{mol\,L^{-1}} \times 0.025 \, \mathrm{L} = 0.01875 \, \mathrm{mol}$$

Step 2: Moles of $$\mathrm{CaCO_3}$$ needed. From the equation, $$2 \, \mathrm{mol \; HCl}$$ react with $$1 \, \mathrm{mol \; CaCO_3}$$.

$$n_{\mathrm{CaCO_3}} = \dfrac{n_{\mathrm{HCl}}}{2} = \dfrac{0.01875}{2} = 9.375 \times 10^{-3} \, \mathrm{mol}$$

Step 3: Mass of $$\mathrm{CaCO_3}$$. Molar mass of $$\mathrm{CaCO_3} = 40 + 12 + 3 \times 16 = 100 \, \mathrm{g\,mol^{-1}}$$.

$$m_{\mathrm{CaCO_3}} = n \times M = 9.375 \times 10^{-3} \, \mathrm{mol} \times 100 \, \mathrm{g\,mol^{-1}} = 0.9375 \, \mathrm{g}$$

Step 1: Moles of $$\mathrm{MnO_2}$$. Using $$\mathrm{Mn} = 55$$, $$\mathrm{O} = 16$$:

$$M(\mathrm{MnO_2}) = 55 + 2 \times 16 = 87 \, \mathrm{g\,mol^{-1}}$$

$$n_{\mathrm{MnO_2}} = \dfrac{5.0 \, \mathrm{g}}{87 \, \mathrm{g\,mol^{-1}}} = 0.0575 \, \mathrm{mol}$$

Step 2: Moles of HCl required. From the balanced equation, $$1 \, \mathrm{mol \; MnO_2}$$ needs $$4 \, \mathrm{mol \; HCl}$$.

$$n_{\mathrm{HCl}} = 4 \times 0.0575 = 0.2299 \, \mathrm{mol}$$

Step 3: Mass of HCl. Molar mass of $$\mathrm{HCl} = 1 + 35.5 = 36.5 \, \mathrm{g\,mol^{-1}}$$.

$$m_{\mathrm{HCl}} = 0.2299 \, \mathrm{mol} \times 36.5 \, \mathrm{g\,mol^{-1}} = 8.39 \, \mathrm{g}$$