Question

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. Let x1 and x2 represent the mole fractions of 1‑propanol and 2‑propanol in a liquid mixture, respectively, and y 1 and y 2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x1 = 0.540 , calculate the composition of the vapor phase at 25 ⁰C.

Answer 1

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.