Question

Total vapour pressure of mixture of $1$ mol of volatile component $A (P^{\circ}_{A} = 100\, mm\, Hg)$ and $3$ mol of volatile component $B (P^{\circ}_{B} = 60\, mm\, Hg)$ is $75$ mm. For such cases

• A. there is positive deviation from Raoult’s law
• B. boiling point has been lowered
• C. force of attraction between $A$ and $B$ is smaller than that between $A$ and $A$ or between $B$ and $B$
• D. all the above statements are correct

Answer 1

Answer: (D)

$P =P^{\circ}_{A}X_{A} + P^{\circ}_{B}X_{B}$
$P = 100 \times \frac{1}{4} + 60 \times \frac{3}{4}$
$P = 70\,$ mm $< \, 75\,$ mm (experimental) Thus, there is positive deviation (a) is true, the mixutre is more volative due to decrease in b.pt. Thus, (b) is true, also force of attraction is decreased hence (c) is true.