Question

1.0g of a non electrolyte solute(molar mass 250g mol) was dissolved in 51.2 g of benzene. If the kf for benzene is 5.12k kg mol, the freezing point of benzene will be lowered by :

Answer 1

Molality of the non electrolyte is n/mass of solvent in kg = 1*1000\250*51.2  change in freezing point = kf* m = 5.12  *1*1000/250*51.2 = 0.4 this is change in freezing point

Answer 2

Answer : The freezing point depression is, 0.4 K

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m\\\\\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M\times w_1}$

where,

$\Delta T_f$ = change in freezing point  or freezing point depression

$k_f$ = freezing point constant  for benzene = $5.12K/mole.kg$

m = molality

$w_1$ = mass of solvent (benzene) = 51.2 g

$w_2$ = mass of solute = 1.0 g

M = molar mass of solute = 250 g/mole

i = van't Hoff factor = 1 (for non-electrolyte solution)

Now put all the given values in the above formula, we get the freezing point depression.

$\Delta T_f= (5.12K/mole.kg)\times \frac{1.0g\times 1000}{250g/mole\times 51.2kg}=0.4K$

Therefore, the freezing point depression is, 0.4 K