What freezing point depression would be predicted for 0.2 molal solution of benzoic acid in benzene if latent heat of fusion is 40.00 cal $g^{- 1}$ at 280 K (freezing point) for benzene? (assume no change in molecular state)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

${0.52}^{o}$

${0.0052}^{o}$

${0.78}^{o}$

${5.2}^{o}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

T_o=Freezing point of solvent

${ΔH}_{f} =$ Latent heat of fusion

$K_{f} = \frac{{RT}_{0}^{2}}{1000 {ΔH}_{f}} = \frac{2 \times {(280)}^{2}}{1000 \times 40} = \frac{156800}{40000} = 3 .92 K {mol}^{- 1} kg$

${ΔT}_{f} = molality \times K_{f} = 0 .2 \times 3 .92 = 0 {.78}^{o}$

JEE

NEET

Foundation JEE

Foundation NEET

CBSE