# Find the partial pressure in a solution containing ethanol and 1-propanol with a total vapor pressure of "56.3 torr"? The pure vapor pressures are "100.0 torr" and "37.6 torr", respectively, and the solution has a mol fraction of 0.300 of ethanol.

Aug 15, 2017

Consider Raoult's law compared with Dalton's law of partial pressures:

P_i = overbrace(chi_(i(l))P_i^"*")^"Raoult's Law" = underbrace(chi_(i(v))P_"tot")_"Dalton's Law",

where:

• ${P}_{i}$ is the partial vapor pressure of component $i$, i.e. in a mixture.
• ${P}_{i}^{\text{*}}$ is the pure vapor pressure of component $i$, i.e. by itself.
• ${\chi}_{i \left(l\right)} = {n}_{i} / \left({\sum}_{k}^{N} {n}_{k}\right)$ is the mol fraction of component $i$ in the liquid phase.
• ${\chi}_{i \left(v\right)}$ is the mol fraction in the vapor phase.
• ${P}_{\text{tot}}$ is the total pressure.

As these are assumed to produce ideal-gas vapors, and to form ideal solutions, designate ethanol as $\boldsymbol{1}$ and 1-propanol as $\boldsymbol{2}$, and Raoult's law thus applies.

Since you have not designated whether the mol fraction is in the liquid phase or the vapor phase, I will straight up assume that it is the liquid phase... you will have to verify whether it is the vapor phase or not...

Note that mol fractions for all components in a mixture must add up to $1$ (why?). i.e.

${\chi}_{1 \left(l\right)} + {\chi}_{2 \left(l\right)} = 1$,
${\chi}_{1 \left(v\right)} + {\chi}_{2 \left(v\right)} = 1$,

${\chi}_{1 \left(l\right)} \ne {\chi}_{1 \left(v\right)}$,
${\chi}_{2 \left(l\right)} \ne {\chi}_{2 \left(v\right)}$,

for a binary mixture. Thus...

$\textcolor{b l u e}{{P}_{1}} = \overbrace{{\underbrace{0.300}}_{{\chi}_{1 \left(l\right)}} \cdot \text{100 torr")^"Ethanol" = color(blue)(ul"30.0 torr}}$

$\textcolor{b l u e}{{P}_{2}} = \overbrace{{\underbrace{\left(1 - 0.300\right)}}_{{\chi}_{2 \left(l\right)}} \cdot \text{37.6 torr")^"1-propanol" = color(blue)(ul"26.3 torr}}$

What is the total pressure when they are mixed together? Well...

${P}_{\text{tot" = P_1 + P_2 = ul"56.3 torr}}$

$\text{30.0 torr" = chi_(1(v))P_"tot}$ $\text{ "" "" "" "" "" "" "" } \boldsymbol{\left(1\right)}$

$\text{26.3 torr" = chi_(2(v))P_"tot" = (1 - chi_(1(v)))P_"tot}$ $\text{ } \boldsymbol{\left(2\right)}$

What then is the mol fraction of ethanol $\left(1\right)$ in the vapor phase? What about 1-propanol $\left(2\right)$ in the vapor phase?