Question

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 12.0% NaOH by mass?

Answer 1

Sol :-

(a).

Let mass of the solution = 100 g

So, mass of solute i.e. NaOH = 12 g and

Mass of solvent i.e. H₂O = 100 g - 12 g = 88 g

Now, Number of moles of NaOH = n_NaOH = 12 g / 40 g/mol = 0.30 mol

Number of moles of H₂O = 88 g / 18 g/mol = n_H2O = 4.89 mol

Total moles = n_T = 0.30 mol + 4.89 mol = 5.19 mol

So,

Mole fraction of solute i.e. NaOH (X_NaOH) = n_NaOH / n_T = 0.30 mol / 5.19 mol = 0.058

Hence, Mole fraction of solute = 0.058

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(b).

We know,

Molality = Number of moles of solute / Mass of solvent in Kg

Molality = 0.30 mol / 0.088 Kg = 3.41 mol/kg or 3.41 m

Hence, Molality of NaOH solution = 3.41 m