Question

1.0 g of a non-electrolytic and non-volatile solute (X) was dissolved in 20.4 g of water. At 760 mm Hg the freezing point of solution was found to be \(-1.05^\circ C\). The molar mass (in g mol\(^{-1}\)) of the solute is \((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)

• A. 96.8
• B. 43.4
• C. 86.8
• D. 48.4

Hint:

For freezing point depression problems: \[ \Delta T_f=iK_fm \] For non-electrolytes: \[ i=1 \] Therefore: \[ \Delta T_f=K_fm \] Always calculate: \[ m=\frac{\Delta T_f}{K_f} \] first, then use molality to obtain the number of moles and finally the molar mass.

Answer 1

The Correct Option is C

Step 1: Identify the property used.
A solute lowers the freezing point of water. This is a colligative property called depression in freezing point: \[ \Delta T_f = i\,K_f\,m. \] For a non-electrolyte, $i = 1$, so $\Delta T_f = K_f\,m$.
Step 2: Find the freezing point depression.
Pure water freezes at $0^\circ C$ and the solution freezes at $-1.05^\circ C$. So \[ \Delta T_f = 0 - (-1.05) = 1.05^\circ C. \]
Step 3: Find the molality.
Rearranging, \[ m = \frac{\Delta T_f}{K_f} = \frac{1.05}{1.86} = 0.5645 \ \text{mol kg}^{-1}. \]
Step 4: Find the moles of solute.
Molality is moles of solute per kg of solvent. The water is $20.4$ g $= 0.0204$ kg. So \[ \text{moles} = 0.5645 \times 0.0204 = 0.01152. \]
Step 5: Find the molar mass.
Molar mass is mass divided by moles. The solute mass is $1.0$ g, so \[ M = \frac{1.0}{0.01152} \approx 86.8 \ \text{g mol}^{-1}. \]
Step 6: State the answer.
The molar mass of the solute is \[ \boxed{86.8 \ \text{g mol}^{-1}} \]