Question

Vapour pressure of $CCl_4$ at $25^{\circ}C$ is 143 mm Hg. 0.5 g of a non-volatile solute (mol. weight 65) is dissolved in 100 mL of $CCl_4$. Find the vapour pressure of the solution. (Density of $CCl_4 = 1.58\, g/cm^3$)

• A. 141.93 mm
• B. 94.39 mm
• C. 199.34 mm
• D. 143.99 mm

Answer 1

Answer: (A)

$W_{B} = 0.5\, g, \,M_{B} = 65\, amu$
$W_{A }=$ volume $×$ density $= 100 × 1.58 = 158\, g$
$M_{A}\left(CCl_{4}\right) = 12 + 35.5 × 4 = 154 \,g\, mol^{-1}$
$\frac{p^{\circ}-p_{s}}{p^{o}} = \frac{W_{B}}{M_{B}}\cdot\frac{M_{A}}{W_{A}}$
$p^{\circ}-p_{s} = \frac{W_{B}M_{A}}{M_{B}W_{A}} \times p^{\circ}$ or $p_{s} = p^{\circ} \frac{W_{B}M_{A}}{M_{B}W_{A}} \times p^{\circ }$
$= 143-\frac{0.5\times154}{65\times158}\times143 = 143 - 1.07 = 141.93\,mm$