Total vapour pressure of mixture of $1 \mathrm{~mol}$ of volatile component $\mathrm{A}\left(P_{A}^{o}=100…

Total vapour pressure of mixture of $1 \mathrm{~mol}$ of volatile component $\mathrm{A}\left(P_{A}^{o}=100 \mathrm{~mmHg}ight)$ and $3 \mathrm{~mol}$ of volatile component $\mathrm{B}(P_{B}^{o}=60$ $\mathrm{~mmHg}$ ) is $75 \mathrm{~mm}$. For such case:
  1. there is positive deviation from Raoult's law
  2. boiling point has been lowered
  3. force of attraction between $\mathrm{A}$ and $\mathrm{B}$ is smaller than that between $\mathrm{A}$ and $\mathrm{A}$ or between $\mathrm{B}$ and $\mathrm{B}$
  4. all the above statements are correct

Solution

$P=P_{A}^{o} X_{A}+P_{B}^{o} X_{B}$
$=\frac{100}{4}+\frac{60 \times 3}{4}$
$=70 \mathrm{~mm} < 75 \mathrm{~mm}$ (experimental)
Thus, there is positive deviation (a) true mixture is more volatile due to decrease in b.p. Thus, (b) is true also force of attraction is decreased thus (c) is true. ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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