Let the binary solution contain:

• glucose = $x_g = 0.1$ mole fraction
• water = $x_w = 0.9$ mole fraction

Assume total moles of solution = $1$ mole

Then, $n_{\text{glucose}} = 0.1$ $n_{\text{water}} = 0.9$

Calculate masses using molar masses

• Molar mass of glucose, $C_6H_{12}O_6$: $6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180\,\text{g/mol}$

• Molar mass of water, $H_2O$: $2(1) + 16 = 18\,\text{g/mol}$

Hence, $m_{\text{glucose}} = 0.1 \times 180 = 18\,\text{g}$ $m_{\text{water}} = 0.9 \times 18 = 16.2\,\text{g}$

Total mass of solution $m_{\text{solution}} = 18 + 16.2 = 34.2\,\text{g}$

Mass percentage of water $\%\,\text{water} = \frac{16.2}{34.2} \times 100$ $= 47.37\%$

Nearest integer $\boxed{47}$

Comparison with stored correct answer

Stored correct answer = $47$

Our derived answer also equals $47$, so they agree.