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 \((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)
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 \((K_f(H_2O)=1.86\,K\,kg\,mol^{-1})\)
Show 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.
Updated On: Jun 17, 2026
- 96.8
- 43.4
- 86.8
- 48.4
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
This question is based on the colligative property known as depression in freezing point.
For a non-electrolyte:
\[
\Delta T_f=iK_fm
\]
Since the solute is non-electrolytic,
\[
i=1
\]
Therefore:
\[
\Delta T_f=K_fm
\]
where
• \(\Delta T_f\) = depression in freezing point
• \(K_f\) = molal depression constant
• \(m\) = molality
Such questions are among the most important numerical problems from the Solutions chapter and are frequently asked in CUET.
Step 1: Calculate depression in freezing point. The normal freezing point of pure water is: \[ 0^\circ C \] The freezing point of solution is: \[ -1.05^\circ C \] Therefore, \[ \Delta T_f = 0-(-1.05) = 1.05^\circ C \] Hence, \[ \boxed{\Delta T_f=1.05^\circ C} \]
Step 2: Calculate molality of the solution. Using \[ \Delta T_f=K_fm \] \[ m=\frac{\Delta T_f}{K_f} \] Substituting the values: \[ m=\frac{1.05}{1.86} \] \[ m=0.5645 \] Thus, \[ \boxed{m=0.5645\,mol\,kg^{-1}} \]
Step 3: Use the definition of molality. Molality is given by: \[ m= \frac{\text{moles of solute}} {\text{mass of solvent in kg}} \] Mass of water: \[ 20.4\,g = 0.0204\,kg \] Therefore, \[ 0.5645 = \frac{\text{moles of solute}} {0.0204} \] Hence, \[ \text{moles of solute} = 0.5645\times0.0204 \] \[ =0.011516 \] Thus, \[ \boxed{\text{Moles of solute}=0.011516} \]
Step 4: Calculate molar mass. Given: \[ \text{Mass of solute}=1.0\,g \] Molar mass: \[ M= \frac{\text{Mass}} {\text{Moles}} \] \[ M= \frac{1.0}{0.011516} \] \[ M=86.83 \] Therefore, \[ \boxed{M\approx86.8\,g\,mol^{-1}} \]
Step 5: Verify with options. The obtained value is: \[ 86.8 \] which exactly matches: \[ \boxed{\text{Option (C)}} \]
Step 6: Final conclusion. The molar mass of the non-electrolytic solute is: \[ \boxed{86.8\,g\,mol^{-1}} \] Hence, \[ \boxed{\text{Option (C)}} \]
• \(\Delta T_f\) = depression in freezing point
• \(K_f\) = molal depression constant
• \(m\) = molality
Such questions are among the most important numerical problems from the Solutions chapter and are frequently asked in CUET.
Step 1: Calculate depression in freezing point. The normal freezing point of pure water is: \[ 0^\circ C \] The freezing point of solution is: \[ -1.05^\circ C \] Therefore, \[ \Delta T_f = 0-(-1.05) = 1.05^\circ C \] Hence, \[ \boxed{\Delta T_f=1.05^\circ C} \]
Step 2: Calculate molality of the solution. Using \[ \Delta T_f=K_fm \] \[ m=\frac{\Delta T_f}{K_f} \] Substituting the values: \[ m=\frac{1.05}{1.86} \] \[ m=0.5645 \] Thus, \[ \boxed{m=0.5645\,mol\,kg^{-1}} \]
Step 3: Use the definition of molality. Molality is given by: \[ m= \frac{\text{moles of solute}} {\text{mass of solvent in kg}} \] Mass of water: \[ 20.4\,g = 0.0204\,kg \] Therefore, \[ 0.5645 = \frac{\text{moles of solute}} {0.0204} \] Hence, \[ \text{moles of solute} = 0.5645\times0.0204 \] \[ =0.011516 \] Thus, \[ \boxed{\text{Moles of solute}=0.011516} \]
Step 4: Calculate molar mass. Given: \[ \text{Mass of solute}=1.0\,g \] Molar mass: \[ M= \frac{\text{Mass}} {\text{Moles}} \] \[ M= \frac{1.0}{0.011516} \] \[ M=86.83 \] Therefore, \[ \boxed{M\approx86.8\,g\,mol^{-1}} \]
Step 5: Verify with options. The obtained value is: \[ 86.8 \] which exactly matches: \[ \boxed{\text{Option (C)}} \]
Step 6: Final conclusion. The molar mass of the non-electrolytic solute is: \[ \boxed{86.8\,g\,mol^{-1}} \] Hence, \[ \boxed{\text{Option (C)}} \]
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