A $$5.2$$ molal aqueous solution of methyl alcohol, $$\text{CH}_3\text{OH}$$, is supplied. What is the mole fraction of methyl alcohol in the solution?

Molality ($$m$$) is defined as the number of moles of solute present in $$1\,\text{kg}$$ of solvent.

Given molality of the solution $$m = 5.2\,\text{mol kg}^{-1}$$, so

moles of methyl alcohol (solute) $$n_{\text{CH}_3\text{OH}} = 5.2$$

mass of water (solvent) taken for definition $$= 1\,\text{kg} = 1000\,\text{g}$$

Moles of water:

$$n_{\text{H}_2\text{O}} = \frac{1000\,\text{g}}{18\,\text{g mol}^{-1}} = 55.56$$

Total moles present in the solution:

$$n_{\text{total}} = n_{\text{CH}_3\text{OH}} + n_{\text{H}_2\text{O}} = 5.2 + 55.56 = 60.76$$

Mole fraction of methyl alcohol:

$$\chi_{\text{CH}_3\text{OH}} = \frac{n_{\text{CH}_3\text{OH}}}{n_{\text{total}}} = \frac{5.2}{60.76} = 0.0856 \approx 0.086$$

Therefore, the mole fraction of methyl alcohol in the solution is $$0.086$$.

Option B which is: $$0.086$$