0.48 g of a substance is dissolved in 10.6 g of C6H6. The freezing point of benzene is lowered by 1.8°C What will be the mol. wt. of the substance (kf for benzene = 5)

Question

0.48 g of a substance is dissolved in 10.6 g of C6H6. The freezing point of benzene is lowered by 1.8°C What will be the mol. wt. of the substance (kf for benzene = 5) (A) 250.2 (B) 90.8 (C) 125.79 (D) 102.5

Tardigrade Admin · Accepted Answer

Δ Tf =(1000 Kf w/mw) m=(1000 kf w/Δ Tf w) =(1000×5×0.48/1.8×10.6) =125.79

Q. $0.48 g$ of a substance is dissolved in $10.6 g$ of $C_{6} H_{6}$ . The freezing point of benzene is lowered by $1. 8^{\circ} C$ What will be the mol. wt. of the substance $(k_{f}$ for benzene = 5)

$250.2$

$90.8$

$125.79$

$102.5$

$Δ T_{f} = \frac{1000 K _{f} w}{m w}$
$m = \frac{1000 k _{f} w}{Δ T _{f} w}$
$= \frac{1000 \times 5 \times 0.48}{1.8 \times 10.6}$
$= 125.79$

Q. 0.48g of a substance is dissolved in 10.6g of C6​H6​. The freezing point of benzene is lowered by 1.8∘C What will be the mol. wt. of the substance (kf​ for benzene = 5)

250.2

90.8

125.79

102.5