Question

The vapour pressure of pure benzene at a certain temperature is 0.85 bar. When 0.5 g of a non-volatile solute is added to 39 g of benzene, the vapour pressure becomes 0.845 bar. What is the molar mass of the substance? ________.

• A. $85g~mol^{-1}$
• B. $127.5~g~mol^{-1}$
• C. $170~g~mol^{-1}$
• D. $210g~mol^{-1}$
• E. $145~g~mol^{-1}$

Hint:

For dilute solutions, $\frac{\Delta P}{P^{\circ}} \approx \frac{w_2 M_1}{M_2 w_1}$.

Answer 1

The Correct Option is C

Step 1: Concept
Use Raoult's Law for relative lowering of vapour pressure: $\frac{P^{\circ} - P}{P^{\circ}} = \frac{n_2}{n_1}$.

Step 2: Meaning
$P^{\circ} = 0.85$, $P = 0.845$, $w_2 = 0.5g$, $w_1 = 39g$, $M_1 = 78~g/mol$.

Step 3: Analysis
$\frac{0.85 - 0.845}{0.85} = \frac{0.5 / M_2}{39 / 78} \implies \frac{0.005}{0.85} = \frac{0.5 / M_2}{0.5}$. $\frac{1}{170} = \frac{1}{M_2}$.

Step 4: Conclusion
$M_2 = 170~g~mol^{-1}$. Final Answer: (C)