Question:
The mole fraction of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ in an aqueous binary solution is $0.1$. The mass percentage of water in it, to the nearest integer, is
Solution:
$\mathrm{X}_{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}=0.1$
Let total mole is 1 mol then mole of glucose will be $0.1$ and mole of water will be $0.9$
so mass $\%$ of water $=\frac{0.9 \times 18}{0.1 \times 180+0.9 \times 18} \times 100$
$=47.36$
Ans : 47