The mole fraction of glucose `(C_(6)H_(12)O_(6))` in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is _______.

Explanation

To find the mass percentage of water in an aqueous binary solution where the mole fraction of glucose (C₆H₁₂O₆) is 0.1, follow these steps:

1. **Determine the mole fractions of each component:**
- Mole fraction of glucose, \( x_{\text{glucose}} = 0.1 \)
- Mole fraction of water, \( x_{\text{water}} = 1 - x_{\text{glucose}} = 1 - 0.1 = 0.9 \)

2. **Calculate the masses of glucose and water:**
- Molar mass of glucose, \( M_{\text{glucose}} = 180 \, \text{g/mol} \)
- Molar mass of water, \( M_{\text{water}} = 18 \, \text{g/mol} \)

Using the mole fractions and molar masses:
- Mass of glucose, \( m_{\text{glucose}} = x_{\text{glucose}} \times M_{\text{glucose}} = 0.1 \times 180 = 18 \, \text{g} \)
- Mass of water, \( m_{\text{water}} = x_{\text{water}} \times M_{\text{water}} = 0.9 \times 18 = 16.2 \, \text{g} \)

3. **Calculate the total mass of the solution:**
- Total mass, \( m_{\text{total}} = m_{\text{glucose}} + m_{\text{water}} = 18 \, \text{g} + 16.2 \, \text{g} = 34.2 \, \text{g} \)

4. **Calculate the mass percentage of water:**
- Mass percentage of water, \( \text{Mass \% of water} = \left( \frac{m_{\text{water}}}{m_{\text{total}}} \right) \times 100 = \left( \frac{16.2}{34.2} \right) \times 100 \approx 47.37 \% \)

Rounding to the nearest integer, the mass percentage of water is approximately \( 47 \% \).