Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is $$x \times 10^{-3}$$. Value of $$x$$ is _____.

We need to find the mole fraction of urea in an aqueous solution with molality 4.44 m.

Molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}}$$

For molality = 4.44 m, we have 4.44 moles of urea in 1 kg (1000 g) of water.

Moles of water = $$\frac{1000}{18} = 55.56$$ mol

The mole fraction of urea is:

$$x_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} = \frac{4.44}{4.44 + 55.56} = \frac{4.44}{60} = 0.074$$

Given that the mole fraction = $$x \times 10^{-3}$$:

$$0.074 = x \times 10^{-3}$$

$$x = \frac{0.074}{10^{-3}} = 74$$

The answer is $$\boxed{74}$$.