Pure benzene freezes at 5.45°C at a certain place but a 0.374 m solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf for benzene is

Question

Pure benzene freezes at 5.45°C at a certain place but a 0.374 m solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf 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°C kg mol-1

Tardigrade Admin · Accepted Answer

Δ Tf = (5.45 - 3.55) = Kf× m or 1.90 = kf × 0.374 or kf = (1.90/0.374) = 5.08

Q. 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

5.08 K $k g m o l^{- 1}$

508 K $k g m o l^{- 1}$

0.508 $K k g m o l^{- 1}$

50.8 $^{\circ}$ C $k g m o l^{- 1}$

$Δ T_{f} = (5.45 - 3.55) = K_{f} \times m$
or $1.90 = k_{f} \times 0.374$
or $k_{f} = \frac{1.90}{0.374}$ = 5.08

Q. Pure benzene freezes at 5.45∘C at a certain place but a 0.374 m solution of tetrachloroethane in benzene freezes at 3.55∘C. The Kf​ for benzene is

5.08 K kgmol−1

508 K kgmol−1

0.508 Kkgmol−1

50.8∘C kgmol−1