The mole fraction of a solute in a 100 molal aqueous solution is ________ $$\times 10^{-2}$$. (Round off to the Nearest Integer).
[Given: Atomic masses: H: 1.0 u, O: 16.0 u]

A 100 molal aqueous solution means 100 moles of solute are dissolved in 1000 g (i.e., 1 kg) of water. The molar mass of water is $$18$$ g/mol, so the number of moles of water in 1000 g is $$\frac{1000}{18} = \frac{500}{9}$$ mol.

The mole fraction of the solute is $$x_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{100}{100 + \frac{500}{9}} = \frac{100}{\frac{900 + 500}{9}} = \frac{100 \times 9}{1400} = \frac{900}{1400} = \frac{9}{14} \approx 0.6429$$.

Expressing this as $$\_\_ \times 10^{-2}$$, we get $$0.6429 = 64.29 \times 10^{-2}$$. Rounding to the nearest integer gives $$64 \times 10^{-2}$$. The answer is 64.