Solutions and Phase Equilibria Questions and Answers

Answer: C

The mass required is 5 g.

Moles needed = molarity \(\times\) volume in litres:

\[ \begin{aligned} n &= 0.5 \times 0.250 = 0.125\ \text{mol} \\ m &= 0.125 \times 40 = 5\ \text{g} \end{aligned} \]

20 g would make a full litre, 10 g ignores the 0.5 M, and 2 g, 4 g and 0.5 g do not follow from the data.

A quick check: 0.5 M means 20 g of X per litre (half of 40 g), and 250 mL is a quarter of a litre, giving 5 g.

Answer: B

The pair is carbon disulphide and acetone.

When X-Y attractions are weaker than X-X and Y-Y, molecules escape the mixture more easily, the vapour pressure exceeds Raoult's law, and the solution shows a positive deviation. Polar acetone mixed with non-polar CS₂ is the textbook example: the dipole-dipole attractions between acetone molecules are broken and replaced by weaker ones.

Chloroform with acetone and nitric acid with water form stronger X-Y attractions (hydrogen bonding), a negative deviation. Bromoethane with chloroethane and benzene with toluene are nearly ideal.

Answer: E

The mixture with positive deviation is Ethanol-Acetone.

Positive deviation occurs when the A-B attractions in the mixture are weaker than the A-A and B-B attractions in the pure liquids, so molecules escape more easily and the vapour pressure is higher than Raoult's law predicts. Pure ethanol is held together by hydrogen bonds; acetone molecules come between them and break many of those bonds, so the mixture is more volatile.

Chloroform-acetone and phenol-aniline show negative deviation, because new hydrogen bonds form between the components. Benzene-toluene and chloroethane-bromoethane are nearly ideal solutions of similar molecules.

Answer: C

The molar mass is 170 g mol⁻¹.

For a dilute solution, the relative lowering of vapour pressure equals the ratio of moles of solute to solvent:

\[ \begin{aligned} \frac{0.85 - 0.845}{0.85} &= \frac{0.5/M}{39/78} \\ \frac{0.005}{0.85} &= \frac{1}{M} \Rightarrow M = 170 \end{aligned} \]

The other molar masses do not satisfy this relation; 85 g mol⁻¹, for example, is what you get by taking the moles of benzene as 1 instead of \(\frac{39}{78} = 0.5\).

Strictly, Raoult's law gives \(\frac{n_{B}}{n_{A} + n_{B}}\), but for a dilute solution \(n_{B}\) is negligible beside \(n_{A}\), which is the approximation used here.

Answer: E

The osmotic pressure is 0.246 atm.

Urea does not ionise, so van 't Hoff's equation applies directly, \(\pi = CRT\), with R in L atm K^-1 mol^-1:

\[ \pi = 0.01 \times 0.082 \times 300 = 0.246\ \text{atm} \]

The value of R is printed with its kelvin unit missing, but 0.082 L atm K^-1 mol^-1 is the standard value. 2.46 and 24.6 atm are ten and a hundred times too large, as if the solution were 0.1 M or 1 M, and 0.0082 and 0.082 atm do not include the factor of 300 K.

Answer: D

The molarity is 0.2M.

Molarity is moles of solute per litre of solution:

\[ \begin{aligned} n &= \frac{149}{74.5} = 2\ \text{mol} \\ M &= \frac{2\ \text{mol}}{10\ \text{L}} = 0.2\ \text{M} \end{aligned} \]

2 M forgets to divide by the 10 litres, and 0.002 M divides by a further 100. 0.1M takes the solute as 1 mol instead of 2, and 1M makes both mistakes at once.

Answer: A

The molar mass is 500 g/mol.

Freezing point depression: \(\Delta T_{f} = K_{f}m\), where the molality \(m = \frac{w_{2}/M_{2}}{w_{1}/1000}\):

\[ \begin{aligned} M_{2} &= \frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}} \\ &= \frac{5 \times 1 \times 1000}{0.20 \times 50} = 500 \end{aligned} \]

The common slips are forgetting the 1000 that converts grams of benzene to kilograms, which gives 0.5, or using the masses the wrong way round. The other options do not satisfy the equation with these data.

Answer: A

The mass percentage of the solute is 8%.

Mass percentage is the mass of solute divided by the mass of the whole solution, not of the solvent alone:

\[ \begin{aligned} \text{mass of solution} &= 4 + 46 = 50\ \text{g} \\ w\% &= \frac{4}{50} \times 100 = 8\% \end{aligned} \]

Dividing by the 46 g of ethanol gives about 8.7%, and 4% treats the 4 g as a share of 100 g. 10% and 12% have no basis in the data.

Answer: A

The relationship is \(m = c/[\rho - cM_{2}/1000]\) mol kg^-1.

Take 1 L (1000 cm³) of solution. It contains \(c\) mol of solute, of mass \(cM_{2}\) g, and its total mass is \(1000\rho\) g. The solvent mass is the difference:

\[ \begin{aligned} \text{solvent} &= (1000\rho - cM_{2})\ \text{g} \\ m &= \frac{c \times 1000}{1000\rho - cM_{2}} = \frac{c}{\rho - cM_{2}/1000} \end{aligned} \]

The option \(\frac{1000c}{\rho - cM_{2}}\) mixes grams and kilograms, and \(\frac{1000c}{1000\rho + cM_{2}}\) adds the solute mass instead of subtracting it. \(\frac{1000c}{\rho}\) ignores the solute's own mass.

Answer: A

The molarity is 0.04 M.

Molarity is moles of solute per litre of solution.

\[ \begin{aligned} n &= \frac{0.4\ \text{g}}{40\ \text{g mol}^{-1}} = 0.01\ \text{mol} \\ M &= \frac{0.01\ \text{mol}}{0.250\ \text{L}} = 0.04\ \text{M} \end{aligned} \]

The other values come from slips with the volume: putting 0.01 mol in 0.025 L (a misplaced decimal) gives 0.40 M, and halving or doubling the 0.25 L gives 0.08 M or 0.02 M.