Question

The vapour pressures of ethanol and methanol are $42.0\text{mm}$ and $88.5\text{mm}$ Hg, respectively. An ideal solution is formed at the same temperature by mixing $46.0\text{g}$ of ethanol with $16.0\text{g}$ of methanol. The mole fraction of ethanol in the vapour is:

• A. $0.487$
• B. $0.352$
• C. $0.892$
• D. $0.665$

Answer 1

The correct option is A $0.487$

Given, $\text{Mass of ethanol}=46\text{g}$
Moles of Ethanol$=\frac{46}{46}=1$
Given, $\text{Mass of methanol}=16\text{g}$
Moles of Methanol $=\frac{16}{32}=\frac{1}{2}=0.5$
Mole fraction$=\frac{n_{\text{solute}}}{n_{\text{solution}}}$
Mole fraction of Ethanol $\left({\chi }_E\right)=\frac{1}{1+0.5}=\frac{1}{1.5}=0.666$
Mole fraction of methanol $\left({\chi }_M\right)=\frac{0.5}{1+0.5}=\frac{0.5}{1.5}=0.333$
$\text{Partial pressure of Ethanol = vapour pressure of ethanol}\times {\chi }_E$
$=42\times 0.666=27.972$
$\text{Partial pressure of methanol = vapour pressure of methanol}\times {\chi }_E$
$=88.5\times 0.333=29.471$
$\text{Total partial pressure}=27.972+29.471=57.443$
Mole fraction of ethanol in vapour $=\frac{\text{partial pressure ethanol}}{\text{total partial pressure}}=\frac{27.942}{57.443}=0.486$