Question

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non- volatile, non-electrolyte solid weighing 1.0 g when added to 39.0 g of benzene (molar mass 78 $g mol^{-1}$), vapour pressure of the solution is reduced to 0.845 bar. What is the molar mass of the solid substance?

• A. 270 $g mol^{-1}$
• B. 170 $g mol^{-1}$
• C. 240 $g mol^{-1}$
• D. 340 $g mol^{-1}$
• E. 370 $g mol^{-1}$

Answer 1

The Correct Option is D

Determining Molar Mass Using Raoult's Law  

Given:

• Vapour pressure of pure benzene, \( P^0 = 0.850 \, \text{bar} \)
• Vapour pressure of solution, \( P_{\text{solution}} = 0.845 \, \text{bar} \)
• Mass of solute = 1.0 g
• Mass of benzene = 39.0 g
• Molar mass of benzene = 78 g/mol

Raoult's Law:

\[ \frac{P^0 - P_{\text{solution}}}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

Let molar mass of solute = \( M \)

Then, moles of solute = \( \frac{1}{M} \)

Moles of benzene = \( \frac{39}{78} = 0.5 \)

Now, plug values into the formula:

\[ \frac{0.850 - 0.845}{0.850} = \frac{1/M}{1/M + 0.5} \Rightarrow \frac{0.005}{0.850} = \frac{1}{1 + 0.5M} \]

\[ 0.00588 = \frac{1}{1 + 0.5M} \Rightarrow 1 + 0.5M = \frac{1}{0.00588} \approx 170.07 \Rightarrow 0.5M = 169.07 \Rightarrow M \approx \mathbf{338.14 \, g/mol} \]

Correct Answer: 340 g mol^-1