Question

A non-volatile solute A weighing \(60 \, g\) when dissolved in \(212 \, g\) of the solvent xylene reduces its vapour pressure to \(60\%\). What is the molar mass of A in \(g \, mol^{-1}\)? Given molar mass of xylene \(= 106 \, g \, mol^{-1}\).

• A. \(56\)
• B. \(126\)
• C. \(60\)
• D. \(45\)

Hint:

For non-volatile solutes, relative lowering of vapour pressure equals mole fraction of solute or solvent depending on form used. Always convert masses into moles before applying Raoult’s law.

Answer 1

The Correct Option is D

Step 1: Apply Raoult’s law for vapour pressure lowering.
For a non-volatile solute:
\[ \frac{P}{P^0} = X_{\text{solvent}} \] Given vapour pressure is reduced to \(60\%\), so:
\[ \frac{P}{P^0} = 0.6 \] Thus, mole fraction of solvent is:
\[ X_{\text{solvent}} = 0.6 \]

Step 2: Write expression for mole fraction.
\[ X_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}} \] \[ 0.6 = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}} \]

Step 3: Calculate moles of solvent (xylene).
Mass of xylene is:
\[ 212 \, g \] Molar mass of xylene is:
\[ 106 \, g \, mol^{-1} \] \[ n_{\text{solvent}} = \frac{212}{106} = 2 \, mol \]

Step 4: Substitute and find moles of solute.
\[ 0.6 = \frac{2}{2 + n_{\text{solute}}} \] \[ 0.6(2 + n_{\text{solute}}) = 2 \] \[ 1.2 + 0.6n_{\text{solute}} = 2 \] \[ 0.6n_{\text{solute}} = 0.8 \] \[ n_{\text{solute}} = \frac{0.8}{0.6} = 1.333 \, mol \]

Step 5: Calculate molar mass of solute.
Given mass of solute is:
\[ 60 \, g \] \[ \text{Molar mass} = \frac{\text{Mass}}{\text{Moles}} \] \[ \text{Molar mass} = \frac{60}{1.333} \] \[ \text{Molar mass} \approx 45 \, g \, mol^{-1} \]

Step 6: Select the correct option.
The calculated molar mass matches option (D).
Final Answer:
The molar mass of solute A is:
\[ \boxed{45 \, g \, mol^{-1}} \]