Question

A solution of 1.25 g of a non-electrolyte in 20 g of water freezes at 271.94 K. If K_f is 1.86 K kg mol⁻¹, then the molecular mass of solute will be ______.

Options

• 207.8

• 179.79

• 209.6

• 109.5

Answer 1

A solution of 1.25 g of a non-electrolyte in 20 g of water freezes at 271.94 K. If K_f is 1.86 K kg mol⁻¹, then the molecular mass of solute will be 109.5.

Explanation:

Given: Mass of solute = 1.25 g

Mass of water = 20 g

Freezing point of solution = 271.94 K

K_f = 1.86 K kg mol⁻¹

ΔT_f = 273 − 271.94 = 1.06 K

ΔT_f = K_f m

1.06 = 1.86 m

`m = 1.06/1.86`

m = 0.57 mol/kg.

`"Molality" = "Moles of solute"/"Mass of solvent in kg"`

`"Mass of solvent in kg" = 20/1000 = 0.02` kg

Moles of solute = Molality × Mass of solvent in kg

Moles of solute = 0.57 × 0.02

Moles of solute = 0.0114 mol

`"Molecular mass" = "Mass of solute"/"Moles of solute"`

= `1.25/0.0114`

= 109.65 ≈ 109.5