Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Calculate the molal depression constant of a solvent which has freezing point $16.6^{\circ} \mathrm{C}$ and latent heat of fusion $180.75 \mathrm{Jg}^{-1}$
ChemistrySolutionsJEE Main
Options:
  • A 2.68
  • B 3.86
  • C 4.68
  • D 2.86t6
Solution:
2290 Upvotes Verified Answer
The correct answer is: 3.86
$\begin{gathered}K_f=\frac{R T_f^2}{1000 \times L_f}, R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ T_f=273+16.6=289.6 \mathrm{~K} ; L_f=180.75 \mathrm{Jg}^{-1} \\ K_f=\frac{8.314 \times 289.6 \times 289.6}{1000 \times 180.75} = 3.86\end{gathered}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.