Question

The vapor pressure of benzene is $53.3 \, kP_{a}$ at $603 \,{}^{o}C$ , but it falls to $\text{51} \text{.5} \, \text{kP}_{\text{a}}$ when $\text{19} \, \text{g}$ of a non-volatile organic compound is dissolved in $\text{500} \, \text{g}$ benzene. The molar mass of the non-volatile compound is

• A. 82
• B. 85
• C. 88
• D. 92

Answer 1

Answer: (B)

$\text{P}^{\text{0}} \, \text{=} \, \text{53} \text{.3} \, \text{kPa}$

$\text{P}_{\text{s}} = 51.5 \, \text{kP}_{\text{a}}$

$\frac{P^{0} - P_{s}}{P_{s}}=\frac{n}{N}=\frac{19 \times 78}{M \times 500}$

$\frac{53.3 - 51.5}{51.5}=\frac{n}{N}=\frac{19 \times 78}{M \times 500}$

$M=\frac{19 \times 78 \times 51.5}{\left(\right. 53.3 - 51.5 \left.\right) \times 500}$

$M=85 \, g$