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:
$T_f=5.6+273K=278.6K$
$K_f=5.1,{\Delta }_{fus}H=?$
$\Delta H_{fus}=\frac{R\times M\times T_f^2}{1000\times K_f}$
$=\frac{1.987\times 78\times {\left(278.6\right)}^2}{1000\times 5.1}$
$=2358.76cal$