Question

The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm⁻³. Find the ratio of the molecular weights of the solute and solvent, `((MW_"solute")/(MW_"solvent"))`.

Answer 1

Given: Mole fraction of solute, `chi_"solute"` = 0.1

So, `chi_"solvent"` = 0.9

Molarity = Molality

Density of solution, ρ = 2.0 g/cm³ = 2000 g/L

Temperature, T = 298 K

Let

M₂ = molecular weight of solute

M₁​ = molecular weight of solvent

Assume total moles = 1 (for simplicity of mole fraction), then

Solute moles = 0.1

Solvent moles = 0.9

Mass of solute = 0.1 × M₂

Mass of solvent = 0.9 × M₁

Total mass of solution = 0.1 × M₂ + 0.9 × M₁

Volume = `"mass"/"density"`

= `(0.1 M_2 + 0.9 M_1)/2000 L`

Molarity = `"moles of solute"/"volume of solution in L"`

`M = 0.1/((0.1 M_2 + 0.9 M_1)/2000)`

= `200/(0.1 M_2 + 0.9 M_1)`

Molality = `"moles of solute"/"mass of solvent in kg"`

`m = 0.1/(0.0009 M_1)`

∴ `200/(0.1 M_2 + 0.9 M_1) = 0.1/(0.0009 M_1)`

200 × 0.0009 M₁ = 0.1 × (0.1 M₂ + 0.9 M₁)

⇒ 0.18 M₁ = 0.01 M₂ + 0.09 M₁

⇒ 0.09 M₁ = 0.01 M₂

⇒ `M_2/M_1 = 0.09/0.01`

⇒ `M_2/M_1 = 9`

∴ `(MW_"solute")/(MW_"solvent") = 9`