Question

A solution is made by dissolving 0.555 mol of nonelectrolyte solute in 769 g of benzene. Calculate the freezing point, Tr, and boiling point, Tb, of the solution. Constants may be found here. Number T5.50 Number T80.101

Answer 1

ANSWER:

Given,

no. mole of solute = 0.555 mol

weight of solvent = 769 g

K_f = 5.12 ^oC kg/mol

K_b = 2.53 ^oC kg/mol

Now,

molality, m = {No. of milimoles of solute x 1000} / {weight of solvent in grams}

= {0.555 mol x 1000} /{769 g}

= 0.722 m

= 0.722 mol/kg

Now,

$\Delta$ T_f = K_f x m = 5.12 ^oC kg/mol x 0.722 mol/kg = 3.70 ^oC

Hence, freezing point of solution T_f = freezing point of benzene - $\Delta$ T_f

= 5.5 ^oC - 3.70 ^oC

= 1.8 ^oC

And,

$\Delta$ T_b = K_b x m = 2.53 ^oC kg/mol x 0.722 mol/kg = 1.83 ^oC

Hence, boiling point of solution T_b = boiling point of benzene + $\Delta$ T_b

= 80.1 ^oC + 1.83 ^oC

= 81.93 ^oC

Therefore, required answer is :

T_f = 1.83 ^oC

T_b = 81.93 ^oC