99 Percentile Qs Bank for JEE MainChemistrySolutions
Calculate the molal depression constant of a solution, which freezes at 15^ C . The latent heat of fusion is 180.7 Jg ⁻¹ .
Options
- A3.81 ~K molal ⁻¹
- B0.381 k molal ⁻¹
- C1.90 ~K molal ⁻¹
- D0.19 ~K molal ⁻¹
Correct answer
A. 3.81 ~K molal ⁻¹
Step-by-step solution
K_f= R T_f^2 1000 L_f Where, aligned K_f & = molal depression constant R & = gas constant =8.314 JK ⁻¹ ~mol ⁻¹ T_f & = freezing point =273+15=288 ~K L_f & = latent heat of fusion =180.73 Jg ⁻¹ K_f & = 8.314 (288)^2 1000 180.7 = 8.314 82944 1807 & = 68959 18070 =3.81 ~K molal ⁻¹ aligned