Question:
The mole fraction of a solute in a 100 molal aqueous solution__________________ $\times 10^{-2}$ (Round off to the Nearest Integer).
Solution:
(64)
100 molal aqueous solution means there is 100 mole solute in $1 \mathrm{~kg}=1000 \mathrm{gm}$ water. Now,
mole-fraction of $=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solute }}+\mathrm{n}_{\text {solvent }}}$
$=\frac{100}{100+\frac{1000}{18}}=\frac{1800}{2800}=0.6428$
$=64.28 \times 10^{-2}$