Question

0.72 g of camphor in 32 g of acetone produces an elevation of 0.25°C in the boiling point of acetone. Calculate the molecular mass of camphor. (K_b for acetone = 1.72 K kg mol⁻¹)

Answer 1

Given: Mass of camphor (w₂) = 0.72 g

Mass of acetone (w₁) = 32 g = 0.032 kg

Elevation of boiling point ΔT_b = 0.25°C

Molal elevation constant K_b = 1.72 K kg mol⁻¹

Molar mass of camphor = M₂ (to be calculated)

ΔT_b = K_b × m

`Delta T_b = K_b * w_2/(M_2 xx w_1)`

`M_2 = (w_2 * K_b)/(Delta T_b * w_1)`

= `(0.72 xx 1.72)/(0.25 xx 0.032)`

= `1.2384/0.008`

= 154.8 g/mol

∴ The molecular mass of camphor is 154.8 g/mol