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 ___________. 

m. f. of glucose $=\frac{molesofglu\cos e}{\left(molesofglu\cos e+molesof{H}_{2}O\right)}$

$0.1=\left(\frac{1}{10}\right)=\left(\frac{1}{1+9}\right)$

So glucose moles = 1, (h glucose)        ${\mathrm{hH}}_2\mathrm{O}=9$

$\begin{array}{ll}\text{ W glucose }=1\times 180\mathrm{grm}& {\mathrm{WH}}_2\mathrm{O}=9\times 18\text{ grmas }\end{array}$

$\begin{array}{l}\text{ Mass \% of }{\mathrm{H}}_2\mathrm{O}=\left(\frac{\omega {\mathrm{H}}_2\mathrm{O}\times 100}{\omega {\mathrm{H}}_2\mathrm{O}+\omega \mathrm{glucose}}\right)\\ =\left(\frac{9\times 18\times 100}{(9\times 18)+180}\right) =\left(\frac{162\times 100}{342}\right)=47.368\end{array}$