Question

Pure benzene freezes at 5.45$^\circ$C at a certain place but a 0.374 m solution of tetrachloroethane in benzene freezes at 3.55$^\circ$C. The $K_f$ for benzene is

• A. 5.08 K $kg\, mol^{-1}$
• B. 508 K $kg \, mol^{-1}$
• C. 0.508 $K \,kg\, mol^{-1}$
• D. 50.8$^\circ$C $kg\, mol^{-1}$

Answer 1

Answer: (A)

$\Delta \, T_f = (5.45 - 3.55) = K_f\times m$
or $1.90 = k_f \times 0.374$
or $k_f = \frac{1.90}{0.374} $ = 5.08