Question:
The mole fraction of a solute in a 100 molal aqueous solution $\times 10^{-2}$. (Round off to the Nearest Integer).
[Given : Atomic masses : $\mathrm{H}: 1.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}$ ]
Solution:
100 molal aqueous solution means there is 100 mole solute in $1 \mathrm{~kg}=1000 \mathrm{gm}$ water.
Now,
$=\frac{100}{100+\frac{1000}{18}}=\frac{1800}{2800}=0.6428$
$=64.28 \times 10^{-2}$
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