Question

The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute  was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute.
[Kb for benzene = 2.53 K kg mol^-1]

Answer 1

Given:

Kb for benzene = 2.53 K kg mol^–1
Mass of solute = W2 = 1.8 g = 1.8 x 10^–3 kg
Mass of solvent = W1 = 90 g = 90 x 10^–3 kg
Boiling point of solution = Tb = 354.11 K
Boiling point of pure solvent = `T_b^'` = 353.23 K

To Find: Molar mass of solute (M2)

`Delta T_b=(K_bxxW_2)/(M_2xxW_1)`

→ Elevation in boiling point (`Delta`Tb) = `T_b - T_b^'`

`Delta T_b=354.11-353.23`

`Delta T_b=0.88K`

From Formula

`M_2=(K_bxxW_2)/(DeltaT_bxxW_1)`

`M_2=(2.53xx1.8xx10^-3)/(0.88xx90xx10^-3)=4.554/79.2`

=0.0575 kg `mol^-1`

=57.5 g `mol^-1`