"Solutions" — Practice

Question 1

An aqueous solution of a salt is 15.0 per cent by mass and has density $1.12\ \mathrm{g\ mL^{-1}}$. The salt's molar mass is $75.0\ \mathrm{g\ mol^{-1}}$. Find the molality and the molarity.

Accepted Answer

Take 100 g of solution: 15.0 g of salt and 85.0 g of water. The amount of salt is $15.0/75.0 = 0.200$ mol, so the molality is $0.200/0.0850 = 2.35\ \mathrm{mol\ kg^{-1}}$. The 100 g of solution occupies $100/1.12 = 89.3$ mL, so the molarity is $0.200/0.0893 = 2.24\ \mathrm{mol\ L^{-1}}$. Molarity is the smaller of the two because a litre of this solution contains less than a kilogram of water once the salt has taken up room.

Question 2

The Henry constant for $\mathrm{CO_2}$ in water at 298 K is $1.67 	imes 10^{3}$ bar per unit mole fraction. A soft drink is bottled under a $\mathrm{CO_2}$ pressure of 2.5 bar. Find the mole fraction of dissolved $\mathrm{CO_2}$, and explain what happens when the cap is opened.

Accepted Answer

By Henry's law, $x = p/K_H = 2.5/1670 = 1.5 	imes 10^{-3}$. Opening the cap drops the $\mathrm{CO_2}$ partial pressure from 2.5 bar to about $3 	imes 10^{-4}$ bar, its value in ordinary air, so the equilibrium solubility falls by roughly four orders of magnitude. The solution is now enormously supersaturated and the excess gas leaves as bubbles. Warming the drink makes it worse, because $K_H$ increases with temperature and a larger $K_H$ means a less soluble gas.

Question 3

At 300 K, pure benzene has a vapour pressure of 75 mmHg and pure toluene 22 mmHg. Find the total vapour pressure of an equimolar mixture and the mole fraction of benzene in the vapour.

Accepted Answer

By Raoult's law the partial pressures are $0.50 	imes 75 = 37.5$ mmHg and $0.50 	imes 22 = 11$ mmHg, giving $p_{total} = 48.5$ mmHg. The vapour mole fraction of benzene is $y = 37.5/48.5 = 0.773$. The liquid is half benzene while the vapour is 77 per cent benzene, and benzene is the more volatile component, so the enrichment runs in the expected direction. Repeating that step over many plates of a column is fractional distillation.

Question 4

The vapour pressure of pure water at 298 K is 23.8 mmHg. Find the vapour pressure of a solution of 50.0 g of urea, $M = 60\ \mathrm{g\ mol^{-1}}$, in 850 g of water.

Accepted Answer

The amounts are $n_{	ext{urea}} = 50.0/60 = 0.833$ mol and $n_{	ext{water}} = 850/18 = 47.2$ mol, so $x_B = 0.833/48.06 = 0.0173$. The relative lowering equals that mole fraction, so $\Delta p = 0.0173 	imes 23.8 = 0.413$ mmHg and the solution's vapour pressure is $23.8 - 0.413 = 23.4$ mmHg. Note that the solvent's mole fraction is never needed, which makes this the simplest of the four properties to apply.

Question 5

A mixture of chloroform and acetone becomes warm on mixing and its total vapour pressure is lower than Raoult's law predicts. Explain both observations and name the kind of azeotrope this pair can form.

Accepted Answer

Chloroform's hydrogen forms a hydrogen bond with acetone's carbonyl oxygen, an interaction present in neither pure liquid. The new solute-solvent attraction is therefore stronger than the ones it replaced, so energy is released and mixing is exothermic, $\Delta_{mix}H < 0$. Molecules held more tightly escape less readily, so the vapour pressure falls below the Raoult line, which is negative deviation, and the volume contracts, $\Delta_{mix}V < 0$. A negative deviation this large produces a maximum boiling azeotrope.

Question 6

Explain why absolute alcohol cannot be obtained by fractional distillation of an ethanol and water mixture.

Accepted Answer

Because ethanol and water show positive deviation from Raoult's law, large enough to produce a minimum boiling azeotrope at 95.6 per cent ethanol by mass. At that composition the vapour and the liquid have identical composition, so there is nothing left to enrich and no number of plates in the column pushes it further. The cause is that ethanol hydrogen bonds strongly to itself and adding water disrupts that network, raising the escaping tendency of both components above what Raoult predicts.

Question 7

Dissolving 1.80 g of a non-volatile solute in 50.0 g of water raises the boiling point by 0.104 K. Find the molar mass, given $K_b = 0.52\ \mathrm{K\ kg\ mol^{-1}}$.

Accepted Answer

Using $M_B = K_b w_B 	imes 1000 / (\Delta T_b\, w_A)$, the calculation is $0.52 	imes 1.80 	imes 1000 / (0.104 	imes 50.0)$. The numerator is 936 and the denominator is 5.2, giving $M_B = 180\ \mathrm{g\ mol^{-1}}$. The solute is glucose, though the method never needed to know that, which is precisely the point of a colligative measurement.

Question 8

Explain why freezing point depression is used more often than boiling point elevation for determining molar mass, and use the expressions for $K_b$ and $K_f$ to justify the difference in their sizes.

Accepted Answer

Two practical reasons. The depression constant is larger, 1.86 against 0.52 for water, so the same solution gives an effect more than three times bigger and the same thermometer reads it more precisely. And no heating is applied, so a thermally fragile solute is not decomposed. The size difference follows from $K_b = RM_AT_b^2/1000\Delta_{vap}H$ against $K_f = RM_AT_f^2/1000\Delta_{fus}H$: the enthalpy of fusion is far smaller than the enthalpy of vaporisation, 6008 against 40650 J mol$^{-1}$ for water, and that ratio outweighs the higher $T_b^2$ in the numerator.

Question 9

1.00 g of a protein in 100 mL of solution gives an osmotic pressure of $2.57 	imes 10^{-3}$ bar at 300 K. Find its molar mass, taking $R = 0.083\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}}$.

Accepted Answer

The molar concentration is $C = \pi/RT = 2.57 	imes 10^{-3}/24.9 = 1.03 	imes 10^{-4}\ \mathrm{mol\ L^{-1}}$. The volume is 0.100 L, so the amount of protein is $1.03 	imes 10^{-5}$ mol and $M_B = 1.00/(1.03 	imes 10^{-5}) = 9.7 	imes 10^{4}\ \mathrm{g\ mol^{-1}}$. The same solution would depress the freezing point by about 0.0002 K, which no ordinary apparatus detects, which is why osmometry is the standard method for macromolecules.

Question 10

Explain why all four colligative properties can be traced to a single cause.

Accepted Answer

All four follow from the lowering of the solvent's vapour pressure, which follows in turn from the solution being more disordered than the pure solvent, so solvent molecules are less inclined to leave it. On a phase diagram the solution's liquid curve lies below the pure solvent's, so it meets the atmospheric pressure line at a higher temperature and the steeper solid curve at a lower one, raising the boiling point and lowering the freezing point from one displacement. Osmosis is the same reluctance expressed across a membrane instead of across a surface.

Question 11

0.585 g of sodium chloride in 100 g of water depresses the freezing point by 0.348 K. Find the van't Hoff factor and the degree of dissociation, given $K_f = 1.86\ \mathrm{K\ kg\ mol^{-1}}$.

Accepted Answer

The amount of salt is $0.585/58.5 = 0.0100$ mol, so the molality is $0.100\ \mathrm{mol\ kg^{-1}}$. The expected depression without dissociation is $1.86 	imes 0.100 = 0.186$ K, so $i = 0.348/0.186 = 1.87$. With $n = 2$, the degree of dissociation is $\alpha = (i-1)/(n-1) = 0.87$, so the salt is 87 per cent dissociated as far as a particle count is concerned.

Question 12

Acetic acid in benzene gives a van't Hoff factor of 0.52. Find the degree of association, assuming dimers form, and explain why the value differs in water.

Accepted Answer

For association of two molecules into one, $\alpha = (1-i)/(1 - 	frac{1}{2}) = 0.48/0.5 = 0.96$, so 96 per cent of the acid exists as dimers, paired through a double hydrogen bond between the carboxyl groups. Benzene is non-polar and cannot compete with that interaction. Water can: hydrogen bonding to the solvent is favourable, so the acid stays monomeric and partly ionises instead, giving $i$ slightly above 1. Association and dissociation are properties of the solution, not of the solute alone.

Question 13

Arrange 0.1 molal solutions of glucose, sodium chloride and calcium chloride in increasing order of freezing point.

Accepted Answer

Depression scales with the van't Hoff factor, so the largest factor gives the lowest freezing point. Glucose does not dissociate and has $i = 1$, $\mathrm{NaCl}$ gives two ions and approaches 2, and $\mathrm{CaCl_2}$ gives three and approaches 3. The depressions therefore increase in that order, so the freezing points increase in the reverse order: $\mathrm{CaCl_2}$, then $\mathrm{NaCl}$, then glucose. The question asks for freezing point, not depression, and the two orderings are exact reverses.

Question 14

Why does the measured van't Hoff factor of sodium chloride fall short of 2 even in dilute solution?

Accepted Answer

Because dissociation being complete does not mean the ions behave independently. Oppositely charged ions attract one another and spend part of their time close enough to move as an effective pair, so the solution contains slightly fewer independent particles than a count of formula units suggests. A colligative property counts independent particles, so the observed effect sits a little below the ideal prediction. The shortfall grows with concentration, since ions are closer together, and with ionic charge, which is why $\mathrm{CaCl_2}$ departs further from 3 than $\mathrm{NaCl}$ does from 2.

Practice — "Solutions"

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