Question #0e184

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 × `"1 mol HOx"/"90.03 g HOx"` = 0.1355 mol HOx and `("87.8 g H"_2"O")/("18.02 g H"_2"O")` = 4.872 mol H₂O.

`x_2 = n_2/(n_1 + n_2) = "0.1355 mol"/"0.1355 mol +4.872 mol" = 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/(n_1 + n_2) = "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