Physical Chemistry Questions and Answers

Answer: C

8 g of NaOH is needed.

H₂SO₄ (molar mass 98 g mol⁻¹) is dibasic, so each mole needs two moles of NaOH:

\[ \begin{aligned} n(\mathrm{H_{2}SO_{4}}) &= \frac{9.8}{98} = 0.1 \text{ mol} \\ n(\mathrm{NaOH}) &= 2 \times 0.1 = 0.2 \text{ mol} \\ m &= 0.2 \times 40 = 8 \text{ g} \end{aligned} \]

0.2 g is the number of moles of NaOH written as grams. Treating the acid as monobasic (a 1 : 1 ratio) would give 4 g, which is not offered, and 0.4 g, 1.2 g and 1.6 g do not come from the stoichiometry.

Answer: A

The amount of methane needed is 0.25 mol.

Combustion of methane gives one CO₂ per CH₄: CH₄ + 2O₂ → CO₂ + 2H₂O.

\[ \begin{aligned} n(\mathrm{CO_{2}}) &= \frac{11}{44} = 0.25\ \text{mol} \\ &= n(\mathrm{CH_{4}}) \end{aligned} \]

0.5 mol would give 22 g of CO₂, and the larger values more still. The answer equals the moles of CO₂ because the mole ratio is 1 : 1.

Answer: B

The mass of CO₂ is 3.96 g.

Only 90% of the 10 g is CaCO₃ (molar mass 100 g mol⁻¹), and each mole gives one mole of CO₂ (44 g mol⁻¹):

\[ \begin{aligned} \mathrm{CaCO_{3} + 2HCl} &\rightarrow \mathrm{CaCl_{2} + H_{2}O + CO_{2}} \\ n &= \frac{9}{100} = 0.09\ \text{mol} \\ m(\mathrm{CO_{2}}) &= 0.09 \times 44 = 3.96\ \text{g} \end{aligned} \]

4.4 g ignores the impurity, and 0.44 g and 0.22 g misplace the decimal.

Answer: B

The highest molar mass belongs to Lead nitrate, Pb(NO₃)₂.

Each nitrate group, NO₃, weighs 14 + 3 × 16 = 62. Lead is divalent, so its nitrate carries two of them:

\[ \begin{aligned} \mathrm{Pb(NO_{3})_{2}} &= 208 + 2 \times 62 = 332 \\ \mathrm{AgNO_{3}} &= 108 + 62 = 170 \end{aligned} \]

The others are lighter still: potassium nitrate 39 + 62 = 101, sodium nitrate 23 + 62 = 85, and ammonium nitrate NH₄NO₃ = 18 + 62 = 80. Silver nitrate is the trap, since silver is heavy, but it has only one nitrate group.

Answer: B

The mass of ammonia formed is 3.4 g.

\(\text{N}_{2} + 3\text{H}_{2} \rightarrow 2\text{NH}_{3}\). The moles are \(\frac{2.8}{28} = 0.1\) of nitrogen and \(\frac{0.8}{2} = 0.4\) of hydrogen.

0.1 mol of nitrogen needs only 0.3 mol of hydrogen, so nitrogen is the limiting reagent.

\[ \begin{aligned} n(\text{NH}_{3}) &= 2 \times 0.1 = 0.2\ \text{mol} \\ m &= 0.2 \times 17 = 3.4\ \text{g} \end{aligned} \]

3.6 g would be the total mass of the reactants, which is more than can react because 0.1 mol of hydrogen is left over.

Answer: B

3 g of ethane contains \(6.023 \times 10^{22}\) molecules.

Find the number of moles, then multiply by Avogadro's number.

\[ \begin{aligned} n &= \frac{3}{30} = 0.1 \text{ mol} \\ N &= 0.1 \times 6.023 \times 10^{23} \\ &= 6.023 \times 10^{22} \end{aligned} \]

\(6.023 \times 10^{23}\) would be one whole mole, 30 g. Each of the smaller options is a further factor of ten too small, the result of slipping a decimal place in the mole count.

Answer: C

The empirical formula is M₂O₃.

Divide each percentage by the atomic mass to get the mole ratio:

\[ \begin{aligned} \mathrm{M} &: \frac{54}{27} = 2 \\ \mathrm{O} &: \frac{46}{16} = 2.875 \end{aligned} \]

The ratio M : O is 2 : 2.875, or 1 : 1.44, which is closest to 1 : 1.5, that is 2 : 3. (A metal of atomic mass 27 is aluminium, and Al₂O₃ contains about 53 % metal, matching the data.)

M₂O and MO₂ give ratios of 1 : 0.5 and 1 : 2, and M₃O₂ and M₂O₅ are further from 1 : 1.44.

Answer: D

The molarity is 0.4 M.

Molarity is moles of solute per litre of solution (taking the solution volume as 250 mL).

\[ \begin{aligned} n &= \frac{3.0}{30} = 0.1\ \text{mol} \\ M &= \frac{0.1}{0.25} = 0.4\ \text{M} \end{aligned} \]

0.1 M forgets to divide by the volume in litres, and 0.3 M or 0.5 M come from using the wrong volume. Molarity uses litres of solution, while molality would use kilograms of solvent; with 250 mL of water both give about 0.4 here.