Question

a) The boiling point of benzene is $353.23$K. When $1.80$g of a non-volatile non-ionisation solute was dissolved in $90$g of benzene, the boiling point raised to $354.11$K.

Calculate the molar mass of the solute. [$K_b$ for benzene $=2.53$K Kg mol${}^{-1}$].

b) Define:

i. The molality of a solution.
ii. Isotonic solutions.

Answer 1

a) The elevation in the boiling point $\Delta T_b=354.11-353.23=0.88$ K.

The expression for the molar mass of the solute is

$M_2=\frac{K_b\times w_2\times 1000}{\Delta T_b\times w_1}$

Substitute values in the above expression.

$M_2=\frac{2.53\times 1.8\times 1000}{0.88\times 90}=57.5$ g/mol

Hence, the molar mass of the solute is 57.5 g/mol

b) i. Molality is the number of moles of solute dissolved in $1$kg of solvent.

Molality $=\frac{\text{Number of moles of solute}}{\text{Volume of solvent (in kg)}}$

ii. Two solutions are isotonic if they have the same osmotic pressure at a given temperature.