Question

The mole fraction of glucose $\left(C_{6} H_{12} O_{6}\right)$ in an aqueous binary solution is $0.1$ . The mass percentage of water in it, to the nearest integer, is

Answer 1

$X_{\text{Glucose} }=0.1$
mass $\%$ of glucose $=\frac{0 . 1 \times 180}{0 . 1 \times 180 + 0 . 9 \times 18}\times 100$
$=\frac{1800}{18 + 16 . 2}$
$=\frac{1800}{34 . 2}\%$
$=52.63\%$
$=53\%$
$\therefore $ mass $\%$ of $H_{2}O=47\%$