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  3. Chemistry
  4. The vapour pressures of ethanol and methanol are 42.0 mm and 88.5 mmHg respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The mole fraction of methanol in the vapour is:

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

Solutions

Solution:

$P_{m}=P^{\circ}_{MeOH} X^{\circ}_{MeOH}+P^{\circ}_{EroH}\, X_{ErOH} $
Thus $P_{M}=88.5\times\left[\frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}}\right]+42\times\left[\frac{\frac{46}{46}}{\frac{16}{32}+\frac{46}{46}}\right]=57.5$
Now $P'_{MeOH}=P_{MeoH}^{o}. X_{MeoH\left(l\right)}=P_{m} \times X_{MeOH\left(g\right)}$
$\therefore 88.5\times\left[\frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}}\right]=57.5X_{MeoH\left(g\right)}$
$\therefore X_{MeOH}=0.513$