The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg$$^{-1}$$) of the aqueous solution is:

For a binary solution containing only one solute and one solvent, the mole fraction of the solvent is given by the general definition

$$X_{\text{solvent}} \;=\; \frac{n_{\text{solvent}}}{n_{\text{solute}} \;+\; n_{\text{solvent}}}.$$

We are told that $$X_{\text{solvent}} = 0.8.$$ To make the arithmetic straightforward, we may choose any convenient total number of moles. The easiest choice is to take the total number of moles of the solution as $$1$$. Let us therefore write

$$n_{\text{solute}} + n_{\text{solvent}} = 1.$$

Substituting this into the definition of mole fraction, we have

$$0.8 \;=\; \frac{n_{\text{solvent}}}{1} \;=\; n_{\text{solvent}}.$$

Hence

$$n_{\text{solvent}} = 0.8,$$

and since the total is $$1,$$ the moles of solute must be

$$n_{\text{solute}} = 1 - 0.8 = 0.2.$$

Now, the solvent is water (the phrase “aqueous solution” tells us this). The molar mass of water is $$18 \text{ g mol}^{-1}.$$ Therefore, the mass of $$0.8$$ mol of water is

$$m_{\text{water}} = 0.8 \times 18 \text{ g} = 14.4 \text{ g}.$$

To use the definition of molality we need this mass in kilograms, so

$$m_{\text{water}} = 14.4 \text{ g} = 0.0144 \text{ kg}.$$

The formula for molality ($$m$$) is

$$m \;=\; \frac{\text{moles of solute}}{\text{mass of solvent in kg}}.$$

Substituting the values just found, we obtain

$$m \;=\; \frac{0.2}{0.0144} \text{ mol kg}^{-1}.$$

Carrying out the division,

$$m \;=\; 13.888\ldots \text{ mol kg}^{-1}.$$

Rounded appropriately,

$$m \;=\; 13.88 \text{ mol kg}^{-1}.$$

Hence, the correct answer is Option C.