Question

Calculate the mass of ascorbic acid (molecular mass = 176 g mol⁻¹) to be dissolved in 75 g of acetic acid to lower its freezing point by 1.5°C. Assume that the solute neither associates nor dissociates in solution.

(K_f for acetic acid = 3.9 K kg mol⁻¹)

Answer 1

Given: Depression in freezing point (ΔT_f) = 1.5 K

Mass of solvent (acetic acid, w₁) = 75 g

Molar mass of solute (ascorbic acid, M₂) = 176 g mol⁻¹

Cryoscopic constant K_f = 3.9 K kg mol⁻¹

Formula: ΔT_f = `(K_f.  w_2.  1000)/(M_2.  w_1)`

`w_2 = (ΔT_f * M_2*w_1)/(K_f *1000)`

= `(1.5 * 176 * 75)/(3.9 * 1000)`

= `19800/3900`

= 5.076 g