An aqueous solution of Methyl alcohol contains 48g of alcohol. The mole fraction of alcohol is 0.6. The weight of water in it is

Options

A. 27 g
B. 2.7 g
C. 18 g
D. 1.8 g

Explanation

To find the weight of water in the aqueous solution of methyl alcohol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Weight of methyl alcohol (CH₃OH) = 48 g
- Mole fraction of methyl alcohol (X₁) = 0.6

2. **Calculate the Mole Fraction of Water:**
- The mole fraction of water (X₂) can be calculated as:
\[ X_2 = 1 - X_1 = 1 - 0.6 = 0.4 \]

3. **Set Up the Mole Fraction Equation:**
- The mole fraction of methyl alcohol is given by:
\[ X_1 = \frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}} \]
- Where \( n_{CH_3OH} \) is the number of moles of methyl alcohol and \( n_{H_2O} \) is the number of moles of water.

4. **Express Moles in Terms of Each Other:**
- From the mole fraction equation:
\[ 0.6 = \frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}} \]
- Rearranging gives:
\[ n_{CH_3OH} = 0.6(n_{CH_3OH} + n_{H_2O}) \]
- This can be simplified to:
\[ 0.4 n_{CH_3OH} = 0.6 n_{H_2O} \]
- Therefore:
\[ n_{H_2O} = \frac{0.4}{0.6} n_{CH_3OH} = \frac{2}{3} n_{CH_3OH} \]

5. **Calculate Moles of Methyl Alcohol:**
- The number of moles of methyl alcohol can be calculated using its molar mass:
- Molar mass of CH₃OH = 12 (C) + 4 (H) + 16 (O) = 32 g/mol
- Thus:
\[ n_{CH_3OH} = \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ moles} \]

6. **Calculate Moles of Water:**
- Substitute \( n_{CH_3OH} \) into the equation for \( n_{H_2O} \):
\[ n_{H_2O} = \frac{2}{3} \times 1.5 = 1 \text{ mole} \]

7. **Calculate the Weight of Water:**
- The molar mass of water (H₂O) = 2 (H) + 16 (O) = 18 g/mol
- Therefore, the weight of water is:
\[ \text{Weight of } H_2O = n_{H_2O} \times \text{Molar mass of } H_2O = 1 \text{ mole} \times 18 \text{ g/mol} = 18 \text{ g} \]

### Final Answer:
The weight of water in the solution is **18 grams**.