1.00 g of a non-electrolyte solute when dissolved in 50 g of organic solvent, lowered the freezing point of organic solvent by 0.50 K. If the freezing point depression constant of organic solvent is $5\, K\, kg\, mol^{-1}$, the molar mass of the solute will be

$200\, g\, mol^{-1}$

The correct answer is Option (2) → $200\, g\, mol^{-1}$

We can solve this using the freezing point depression formula:

$\Delta T_f = K_f \cdot m$

Where:

• $\Delta T_f = 0.50 \, \text{K}$
• $K_f = 5 \, \text{K kg mol}^{-1}$
• m = molality of solute = $\frac{\text{moles of solute}}{\text{kg of solvent}}$

Step 1: Calculate molality

$m = \frac{\Delta T_f}{K_f} = \frac{0.50}{5} = 0.10 \, \text{mol/kg}$

Step 2: Calculate moles of solute

$\text{moles of solute} = m \times \text{kg of solvent} = 0.10 \times 0.050 = 0.005 \, \text{mol}$

Step 3: Calculate molar mass

$M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.00}{0.005} = 200 \, \text{g/mol}$