"Solutions" — Practice

Question 1

Two volatile liquids have vapour pressures of $120$ and $300$ mm Hg at $300$ K. For a solution containing $2$ mol of the first and $3$ mol of the second, find the total pressure and the mole fraction of the first in the vapour.

Accepted Answer

$x_1=0.4$, $x_2=0.6$

$p_1=0.4	imes120=48$ mm; $p_2=0.6	imes300=180$ mm

$P=228$ mm Hg

$y_1=\dfrac{48}{228}=0.211$

The liquid is $40\%$ of the first component but the vapour only $21\%$, since the vapour is enriched in the more volatile second component.

Question 2

A mixture of acetone and carbon disulphide shows a higher vapour pressure than Raoult's law predicts at every composition. Predict the signs of $\Delta H_{mix}$ and $\Delta V_{mix}$, and the type of azeotrope.

Accepted Answer

Higher than predicted vapour pressure means **positive deviation**: the A-B attraction is weaker than the A-A and B-B attractions.

Breaking the stronger interactions costs energy, so $\Delta H_{mix}>0$ and the mixture cools.

Weaker attraction lets the molecules sit further apart, so $\Delta V_{mix}>0$ and the volume expands.

Higher vapour pressure means a lower boiling point, so the mixture forms a **minimum**-boiling azeotrope.

Question 3

The vapour pressure of pure benzene at $300$ K is $100$ mm Hg. Adding $2.0$ g of a non-volatile solute to $39$ g of benzene lowers it to $98$ mm Hg. Find the molar mass of the solute.

Accepted Answer

$\dfrac{100-98}{100}=0.02=x_{solute}$

$n_{benzene}=\dfrac{39}{78}=0.5$ mol

$\dfrac{n_s}{n_s+0.5}=0.02\qquad\Rightarrow\qquad n_s=0.0102$ mol

$M=\dfrac{2.0}{0.0102}=196$ g mol$^{-1}$

Question 4

The Henry constant for nitrogen in water is $8.5	imes10^{4}$ atm at $293$ K. Find the mass of nitrogen dissolved in $1$ kg of water in contact with air at $5$ atm, with nitrogen at $78\%$ of air.

Accepted Answer

$p_{	ext{N}_2}=0.78	imes5=3.9$ atm

$x=\dfrac{3.9}{8.5	imes10^{4}}=4.59	imes10^{-5}$

In $1$ kg of water there are $55.5$ mol, so $n_{	ext{N}_2}=4.59	imes10^{-5}	imes55.5=2.55	imes10^{-3}$ mol

Mass $=2.55	imes10^{-3}	imes28=0.071$ g

At five atmospheres a diver's blood dissolves five times the surface amount, which is the origin of decompression sickness.

Question 5

Explain why a bottle of carbonated drink fizzes when opened and why it goes flat faster when warm.

Accepted Answer

The bottle is sealed under a high partial pressure of carbon dioxide, so the dissolved mole fraction is correspondingly high.

Opening it drops the partial pressure to atmospheric, so the equilibrium solubility falls sharply and the excess gas comes out of solution as bubbles.

Warming raises the Henry constant, which by the same relation lowers the solubility at any given pressure.

A warm drink therefore holds less gas to begin with and releases it faster.

Question 6

Compute $K_b$ for water from first principles, given $T_b=373$ K and $\Delta H_{vap}=40.7$ kJ mol$^{-1}$, and compare with $K_f$.

Accepted Answer

$K_b=\dfrac{8.314	imes\left(373ight)^{2}	imes18}{1000	imes40700}=\dfrac{8.314	imes139129	imes18}{4.07	imes10^{7}}$

$=\dfrac{2.082	imes10^{7}}{4.07	imes10^{7}}=0.512$ K kg mol$^{-1}$

$K_f$ computed the same way is $1.86$, larger by a factor of $3.6$.

The ratio traces to the enthalpies: vaporisation costs nearly seven times what fusion does, and it appears in the denominator.

Question 7

A solution of $1.25$ g of a compound in $50$ g of water freezes at $-0.186^{\circ}$C. Find its molar mass, taking $K_f=1.86$ K kg mol$^{-1}$.

Accepted Answer

$m=\dfrac{0.186}{1.86}=0.1$ mol kg$^{-1}$

Moles in $50$ g of water $=0.1	imes0.050=5	imes10^{-3}$

$M=\dfrac{1.25}{5	imes10^{-3}}=250$ g mol$^{-1}$

The compound must be a non-electrolyte, since any dissociation would have produced a larger depression and a smaller apparent molar mass.

Question 8

Compare the boiling point elevation and the osmotic pressure of a $10^{-4}$ M aqueous solution of a protein at $300$ K, and say which is measurable.

Accepted Answer

$\Delta T_b=0.512	imes10^{-4}=5.1	imes10^{-5}$ K, which no ordinary thermometer can resolve.

$\pi=10^{-4}	imes0.0821	imes300=2.46	imes10^{-3}$ atm, that is about $1.9$ mm Hg — easily measured as a $2.5$ cm column of water.

Osmotic pressure is more sensitive by roughly four orders of magnitude.

This is why every determination of a macromolecular mass by colligative means uses osmometry.

Question 9

A $0.5$ m aqueous solution of a weak acid $	ext{HA}$ freezes at $-1.0^{\circ}$C. Find the degree of dissociation, with $K_f=1.86$.

Accepted Answer

$i=\dfrac{\Delta T_f}{K_fm}=\dfrac{1.0}{1.86	imes0.5}=1.075$

For $	ext{HA}ightarrow	ext{H}^++	ext{A}^-$, $n=2$:

$1.075=1+\left(2-1ight)\alpha\qquad\Rightarrow\qquad \alpha=0.075$

Only $7.5\%$ dissociated, consistent with a weak acid. A strong acid would have given $i$ close to $2$.

Question 10

Acetic acid in benzene shows an observed molar mass of $100.5$ g mol$^{-1}$ against a formula mass of $60$. Find the degree of association.

Accepted Answer

$i=\dfrac{60}{100.5}=0.597$

For dimerisation, $n=2$: $i=1-\dfrac{\alpha}{2}$

$0.597=1-\dfrac{\alpha}{2}\qquad\Rightarrow\qquad \alpha=0.806$

About $81\%$ associated. In benzene there is nothing to compete for the carboxylic hydrogen bonds, so the cyclic dimer dominates.

Question 11

Two solutions are isotonic at $300$ K: one containing $6.0$ g of urea per litre and the other containing an unknown non-electrolyte at $10.0$ g per litre. Find the molar mass of the unknown.

Accepted Answer

Urea has $M=60$, so its concentration is $\dfrac{6.0}{60}=0.1$ mol L$^{-1}$.

Since both are non-electrolytes, isotonic means equal molarity.

$\dfrac{10.0}{M}=0.1\qquad\Rightarrow\qquad M=100$ g mol$^{-1}$

No temperature or gas constant was needed, because they appear identically on both sides and cancel.

Question 12

A red blood cell is placed successively in pure water, in $0.9\%$ saline and in $5\%$ saline. Describe what happens in each and explain.

Accepted Answer

In pure water the exterior is **hypotonic**, so water flows into the cell and it swells and eventually bursts.

In $0.9\%$ saline the exterior is **isotonic** at about $7.3$ atm, matching the cell contents, so there is no net flow and the cell is unchanged.

In $5\%$ saline the exterior is **hypertonic**, so water leaves the cell and it shrivels.

Intravenous fluids must therefore be isotonic, which is why the concentration of physiological saline is fixed at that value rather than chosen for convenience.

Practice — "Solutions"

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