Question

The elevation in boiling point when 0.30 g of acetic acid is dissolved in 100 g of benzene is 0.0633°C. Calculate the molecular weight of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution?

(Given K_b for benzene = 2.53 K kg mol⁻¹, at. wt. of C = 12, H = 1, O = 16)

Answer 1

Given: Mass of solute (acetic acid), W_B = 0.30 g

Mass of solvent (benzene), W_A = 100 g = 0.1 kg

Elevation in boiling point, ΔT_b = 0.0633°C

K_b​ for benzene = 2.53 K kg mol⁻¹

Molar mass of acetic acid (normal, for reference) = 2 × 12 + 4 × 1 + 2 × 16 = 60 g/mol

`Delta T_b = (i * K_b * W_b * 1000)/(M_B * 100)`

Assuming no dissociation or association, i.e., i = 1

`0.0633 = (2.53 xx 0.30 xx 1000)/(M_B xx 100)`

`0.0633 = 759/(100 xx M_B)`

`0.0633 = 7.59/M_B`

`M_B = 7.59/0.0633`

= 119.90 g/mol

Van’t Hoff factor (i) is given by:

i = `"Normal molar mass"/"Abnormal molar mass"`

= `60/119.90`

= 0.5

As i < 1, therefore the solute (acetic acid) is associated.