Looking for classes? Ksquare Career Institute, Bengaluru →
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 $K_f$ of benzene is 5.12 K kg mol$^{-1}$, the freezing point of benzene will be lowered by :-
A
0.4 K
B
0.3 K
C
0.5 K
D
0.2 K
Detailed Solution
$\Delta T_f = K_f \times$ molality
Molality $= \dfrac{w_{solute} \times 1000}{M_{solute} \times w_{solvent}}$
$= \dfrac{1 \times 1000}{250 \times 51.2}$ mol kg$^{-1}$
$\Delta T_f = \left(\dfrac{1 \times 1000}{250 \times 51.2}\right) \times 5.12$
$\Delta T_f = \dfrac{5120}{12800}$
$\Delta T_f = 0.4$ K
Molality $= \dfrac{w_{solute} \times 1000}{M_{solute} \times w_{solvent}}$
$= \dfrac{1 \times 1000}{250 \times 51.2}$ mol kg$^{-1}$
$\Delta T_f = \left(\dfrac{1 \times 1000}{250 \times 51.2}\right) \times 5.12$
$\Delta T_f = \dfrac{5120}{12800}$
$\Delta T_f = 0.4$ K