Question:
\(0.05\) mole of a non-volatile solute is dissolved in \(500\,g\) of water. What is the depression in freezing point of resultant solution?
\((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)
\(0.05\) mole of a non-volatile solute is dissolved in \(500\,g\) of water. What is the depression in freezing point of resultant solution?
\((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)
\((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)
Show Hint
For freezing point depression:
\[
\Delta T_f=K_fm
\]
where molality is
\[
m=\frac{\text{moles of solute}}{\text{kg of solvent}}
\]
Always convert solvent mass into kilograms before calculation.
Updated On: Jun 22, 2026
- \(0.047\,K\)
- \(0.372\,K\)
- \(0.093\,K\)
- \(0.186\,K\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the formula for depression in freezing point.
Depression in freezing point is given by \[ \Delta T_f=K_f\times m \] where \[ K_f=\text{molal depression constant} \] and \[ m=\text{molality of solution} \]
Step 2: Calculate the molality of the solution.
Given: \[ \text{Moles of solute}=0.05 \] Mass of water: \[ 500\,g=0.5\,kg \] Molality is \[ m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] \[ m=\frac{0.05}{0.5} \] \[ m=0.1\,mol\,kg^{-1} \]
Step 3: Calculate depression in freezing point.
Given, \[ K_f=1.86\,K\,kg\,mol^{-1} \] Therefore, \[ \Delta T_f=1.86\times 0.1 \] \[ \Delta T_f=0.186\,K \]
Step 4: Final conclusion.
Hence, the depression in freezing point is \[ \boxed{0.186\,K} \]
Depression in freezing point is given by \[ \Delta T_f=K_f\times m \] where \[ K_f=\text{molal depression constant} \] and \[ m=\text{molality of solution} \]
Step 2: Calculate the molality of the solution.
Given: \[ \text{Moles of solute}=0.05 \] Mass of water: \[ 500\,g=0.5\,kg \] Molality is \[ m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] \[ m=\frac{0.05}{0.5} \] \[ m=0.1\,mol\,kg^{-1} \]
Step 3: Calculate depression in freezing point.
Given, \[ K_f=1.86\,K\,kg\,mol^{-1} \] Therefore, \[ \Delta T_f=1.86\times 0.1 \] \[ \Delta T_f=0.186\,K \]
Step 4: Final conclusion.
Hence, the depression in freezing point is \[ \boxed{0.186\,K} \]
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