# Question 0e184

Sep 19, 2014

The vapour pressures are 58.5 Torr, 70.0 Torr, and 70.1 Torr.

Raoult's Law states that the vapour pressure of a solvent ${P}_{1}$ in a solution is equal to the vapour pressure of the pure solvent P_1^° times its mole fraction ${x}_{1}$.

P_1 = x_1P_1^°

Since the oxalic acid, H₂C₂O₄ (HOx), is non-volatile, its vapour pressure ${P}_{2}$ = 0.

So, according to Henry's Law, the vapour pressure of the solution is

$P = {P}_{1} + {P}_{2} = {P}_{1} + 0 = {P}_{1}$

It is often easier to solve these problems by using the mole fraction of the solute, ${x}_{2}$.

Since we have two components, ${x}_{1} + {x}_{2} = 1$.

Then P_1 = (1 – x_2)P_1^° = P_1^° -x_2 P_1^° and

ΔP = P_1^° -P_1 = x_2P_1^°

(a) ${x}_{2}$ = 0.186

ΔP = = x_2P_1^° = 0.186 × 71.9 Torr = 13.37 Torr

P = P_1^° - ΔP = (71.9 -13.37) Torr = 58.5 Torr

(b) Mass % = 12.2

We have 12.2 g HOx and 87.8 g H₂O, or

12.2 g HOx × $\text{1 mol HOx"/"90.03 g HOx}$ = 0.1355 mol HOx and $\left(\text{87.8 g H"_2"O")/("18.02 g H"_2"O}\right)$ = 4.872 mol H₂O.

${x}_{2} = {n}_{2} / \left({n}_{1} + {n}_{2}\right) = \text{0.1355 mol"/"0.1355 mol +4.872 mol} = \frac{0.1355}{5.008} = 0.02706$

ΔP = x_2P_1^° = 0.02706 × 71.9 Torr = 1.946 Torr

P = P_1^° - ΔP = (71.9 -1.946) Torr = 70.0 Torr

(c) $m$ = 1.44 mol/kg

We have "1000 g H"_2"O" × ("1 mol H"_2"O")/("18.02 g H"_2"O") = 55.49 mol H₂O

${x}_{2} = {n}_{2} / \left({n}_{1} + {n}_{2}\right) = \text{1.44 mol"/"1.44 mol +55.49 mol" = 1.44/56.93 = "0.025 29}$

ΔP = = x_2P_1^°# = 0.025 29 × 71.9 Torr = 1.818 Torr

$P$ = (71.9 -1.818) Torr = 70.1 Torr