Question:
$3$ g of benzoic acid dissolved in $25$ g of benzene shows a depression in freezing point equal to $2.5$ K. What is the experimental molar mass of benzoic acid in benzene? \ (Molal depression constant for benzene is $5$ K kg mol$^{-1}$)
$3$ g of benzoic acid dissolved in $25$ g of benzene shows a depression in freezing point equal to $2.5$ K. What is the experimental molar mass of benzoic acid in benzene? \ (Molal depression constant for benzene is $5$ K kg mol$^{-1}$)
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If observed molar mass is double normal value, it often indicates association (dimer formation), as benzoic acid does in benzene.
Updated On: Apr 24, 2026
- $122$ g mol$^{-1}$
- $244$ g mol$^{-1}$
- $240$ g mol$^{-1}$
- $242$ g mol$^{-1}$
- $250$ g mol$^{-1}$
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The Correct Option is C
Solution and Explanation
Concept:
Chemistry - Depression in Freezing Point.
[
\Delta T_f = K_f \times m
]
where $m$ = molality.
Step 1: Find molality. [ m=\frac{\Delta T_f}{K_f}=\frac{2.5}{5}=0.5 ]
Step 2: Use molality formula. [ m=\frac{\text{moles of solute}}{\text{kg of solvent}} ] Mass of benzene: [ 25g=0.025kg ] So moles of benzoic acid: [ \text{moles}=0.5 \times 0.025=0.0125 ]
Step 3: Calculate molar mass. [ \text{Molar mass}=\frac{\text{mass}}{\text{moles}}=\frac{3}{0.0125}=240 ]
Step 4: Final answer. [ \boxed{240\text{ g mol}^{-1}} ]
Hence, correct option is (C).
Step 1: Find molality. [ m=\frac{\Delta T_f}{K_f}=\frac{2.5}{5}=0.5 ]
Step 2: Use molality formula. [ m=\frac{\text{moles of solute}}{\text{kg of solvent}} ] Mass of benzene: [ 25g=0.025kg ] So moles of benzoic acid: [ \text{moles}=0.5 \times 0.025=0.0125 ]
Step 3: Calculate molar mass. [ \text{Molar mass}=\frac{\text{mass}}{\text{moles}}=\frac{3}{0.0125}=240 ]
Step 4: Final answer. [ \boxed{240\text{ g mol}^{-1}} ]
Hence, correct option is (C).
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