What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol\(^{-1}\)) and the decrease in freezing point is 0.40 K?
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- 5.12 K kg mol\(^{-1}\)
- 4.43 K kg mol\(^{-1}\)
- 1.86 K kg mol\(^{-1}\)
- 3.72 K kg mol\(^{-1}\)
The Correct Option is C
Solution and Explanation
\[ \Delta T_f = K_f \times m \]
where \( K_f \) is the freezing point depression constant, and \( m \) is the molality.
\[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\frac{1}{256}}{\frac{50}{1000}} = \frac{1}{256} \times \frac{1000}{50} = 0.078125 \, \text{mol/kg} \]
\[ \Delta T_f = K_f \times m \]
Substitute \( \Delta T_f = 0.40 \, \text{K} \) and \( m = 0.078125 \, \text{mol/kg} \): \[ 0.40 = K_f \times 0.078125 \]
\[ K_f = \frac{0.40}{0.078125} = 1.86 \, \text{K kg mol}^{-1} \]
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