Question

The vapour pressure of pure water at ${20}^{o}C$ is $20$ mm $Hg$. A solution of sucrose whose molar mass is $342$ g/mole is prepared by dissolving $75$ g of it in $1000$ g of water. The relative lowering of the vapor pressure of the solution was measured as $0.0039$. What is the vapour pressure of the solution when the solute was added?

• A. $14.660$ mm
• B. $10.005$ mm
• C. $25.220$ mm
• D. $19.922$ mm

Answer 1

The correct option is D $19.922$ mm

we have,

$\frac{P_B^0-P_t}{P_B^0}=x_A$

putting the values we have,

$\frac{P_B^0-P_t}{P_B^0}=0.0039$

$\frac{20-P_t}{20}=0.0039$

so we get, $P_t=19.922mm$