Question

Ethanol and methanol form a solution that is very nearly ideal. The vapour pressure of ethanol is $44.5mm$ and that of methanol is $88.7mm$ at ${20}^{o}C$. Calculate the mole fraction of methanol and ethanol in a solution obtained by mixing $60g$ of ethanol with $40g$ of methanol.

Answer 1

$P_{ethanol}^0=44.5mm$ of $Hg$ & $P_{Methanol}^0=88.7mm$ of $Hg$

$n_{ethanol}=\frac{W_{ethanol}}{M_{ethanol}}=\frac{60g}{46g}=1.30$ &

$n_{Methanol}=\frac{W_{methanol}}{M_{methanol}}=\frac{40g}{32}=1.25$

$\begin{array}{c}\therefore X_{ethanol}=\frac{1.3}{1.3+1.25}=0.51\\ X_{methanol}=\frac{1.25}{1.3+1.25}=0.49\end{array}\}$

$P_{total}=X_{eth}.P_{eth}^0+X_{met}{P}_{meth}^0$

$P_{total}=0.51\times 44.5+0.49\times 88.7=66.16mm$ of $Hg$

$\therefore P_{ethanol}=X_{eth}.P_{eth}^0=0.51\times 44.5=22.7mm$ of $Hg$

$P_{Methanol}^{\prime }=X_{meth}.P_{meth}^0=0.49\times 88.7=43.46mm$ of $Hg$

$\therefore X_{ethanol}=\frac{P_{ethanol}^{\prime }}{P_{total}}=\frac{22.7}{66.16}=0.34$