Question:
When 0.01 mole of nonvolatile solute is dissolved in certain solvent calculate the mass of solvent in kg if $\Delta \text{T}_{\text{b}} = 0.6 \text{ K}$ and $\text{K}_{\text{b}}$ for solvent $= 2 \text{ K kg mol}^{-1}$
When 0.01 mole of nonvolatile solute is dissolved in certain solvent calculate the mass of solvent in kg if $\Delta \text{T}_{\text{b}} = 0.6 \text{ K}$ and $\text{K}_{\text{b}}$ for solvent $= 2 \text{ K kg mol}^{-1}$
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Mass of solvent (kg) $= \frac{K_b \times n_{solute}}{\Delta T_b}$.
Updated On: Apr 26, 2026
- $0.014 \text{ kg}$
- $0.028 \text{ kg}$
- $0.033 \text{ kg}$
- $0.045 \text{ kg}$
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The Correct Option is C
Solution and Explanation
Step 1: Formula
$\Delta T_b = K_b \times m$, where $m = \frac{n_{solute}}{W_{solvent(kg)}}$
Step 2: Substitution
$0.6 = 2 \times \frac{0.01}{W_{solvent}}$
Step 3: Calculation
$W_{solvent} = \frac{2 \times 0.01}{0.6} = \frac{0.02}{0.6}$
$W_{solvent} = \frac{1}{30} \approx 0.0333 \text{ kg}$.
Final Answer: (C)
$\Delta T_b = K_b \times m$, where $m = \frac{n_{solute}}{W_{solvent(kg)}}$
Step 2: Substitution
$0.6 = 2 \times \frac{0.01}{W_{solvent}}$
Step 3: Calculation
$W_{solvent} = \frac{2 \times 0.01}{0.6} = \frac{0.02}{0.6}$
$W_{solvent} = \frac{1}{30} \approx 0.0333 \text{ kg}$.
Final Answer: (C)
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