Question

The vapour pressure of ethanol and methanol at $298K$ temperature is $44.5mm$ and $88.7mm$ respectively. If $60gram$ ethanol is mixed with $40gram$ methanol at $298K$ temperature and prepared an ideal solution then find mole-fraction of methanol in vapour state.

Answer 1

$Numberofmolesofethanol=\frac{60}{46}=1.3$

$Numberofmolesofmethanol=\frac{40}{32}=1.25$

$Totalnumberofmoles=1.3+1.25=2.55$

$Molefractionofethanol\left({\chi }^a\right)=\frac{1.3}{2.55}=0.51$

$Molefractionofmethanol\left({\chi }^b\right)=\frac{1.25}{2.55}=0.49$

According to Raoult's law, $p=p_0^a{\chi }^a+p_0^b{\chi }^b$

$p=totalpressure$

$p_0^a=partialpressureofethanol=p_0^a\times {\chi }^a=44.5\times 0.51=22.7mmHg$

$p_0^b=partialpressureofmethanol=p_0^b\times {\chi }^b=88.7\times 0.49=43.5mmHg$

$p=(44.5\times 0.51+88.7\times 0.49)=66.158mmHg$

$\therefore molefractionofmethanolinvapourphase=\frac{43.5}{66.158}=0.657$