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

$\mathrm{The} \mathrm{quantity} \mathrm{of} \mathrm{the} \mathrm{moles} \mathrm{in} \mathrm{solute} \mathrm{molecules} = \frac{\mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{solute}}{\mathrm{molecular} \mathrm{weight}}$
$$\begin{gathered} \text{ Weight of the solute }=\text{ Number of moles }\times \text{ molecular weight } \\ \text{molecular weight of the glucose =180} \\ =0.1\times 180 \\ =18 \\ likewise \\ \text{ Weight of the solvent=Number of moles }\times \text{ molecular weight } \\ \text{ Molecular weight of the water }=18 \\ =0.9 \times 18 \\ =16.2 \\ Now, use the mass percentage formula. \\ \begin{array}{rl}\text{ Mole percentage of water }& =\frac{\text{ Mass of water }}{\text{ Total mass of the solution }}\times 100\\ & =\frac{16.2}{(16.2+18)}\times 100\\ & =\frac{16.2}{34.2}\times 100\\ & =47.36\\ & =47\mathrm{\%}\end{array} \end{gathered}$$

Hence, the correct option is: (A) 47%