Question

1.00 g of a non-electrolyte solute (molar mass 250 g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by:

• A. 0.4 K
• B. 0.8 K
• C. 0.12 K
• D. 0.24 K

Answer 1

Answer: (A)

0.4 K

Answer 2

$ΔT_f​ =\frac { k_f \times W_1 \times1000 }{W_2 \times M_1}$
Where,
W1 = Weight of the solute
W2 = Weight of solvent
M1 = Molar mass of solute
kf = Freezing point depression constant
$ΔT_f = \frac {5.12 \times 1\times 1000}{51.2 \times 250}$
$ΔT_f = 0.4\ K$

So, the correct option is (A): 0.4 K