Question

Benzene freezes at $ 5.6^°\, C $ . Its value for $ K_f $ is $ 5.1 $ . The value of $ \Delta H\,_{fus} $ is

• A. 30.24 cal
• B. 2358.72 cal
• C. 1179.36 cal
• D. 15.12 cal

Answer 1

Answer: (B)

Given:
$T_{f}=5.6+273\,K=278.6\,K$
$K_{f}=5.1, \Delta_{fus} H=?$
$\Delta H_{fus} =\frac{R\times M\times T_{f}^{2}}{1000\times K_{f}}$
$=\frac{1.987\times78\times\left(278.6\right)^{2}}{1000\times5.1}$
$=2358.76 \, cal$