Mole fraction of ethyl alcohol in aqueous ethyl alcohol (`C_2H_5OH`) solution is 0.28. Hence percentage of ethyl alcohol by weight is:

Options

A. 49.84%
B. 58.4%
C. 69.48%
D. 74.58%

Answer

To find the percentage of ethyl alcohol by weight in an aqueous solution where the mole fraction of ethyl alcohol is 0.28, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mole Fractions**:
- Given the mole fraction of ethyl alcohol (C₂H₅OH) is 0.28.
- Therefore, the mole fraction of water (H₂O) can be calculated as:
\[ \text{Mole fraction of water} = 1 - \text{Mole fraction of ethyl alcohol} = 1 - 0.28 = 0.72 \]

2. **Calculate the Molar Masses**:
- Molar mass of ethyl alcohol (C₂H₅OH) = 46 g/mol.
- Molar mass of water (H₂O) = 18 g/mol.

3. **Calculate the Weight of Ethyl Alcohol**:
- Using the mole fraction and molar mass, we can find the weight of ethyl alcohol in a hypothetical 1 mole of solution:
\[ \text{Weight of ethyl alcohol} = \text{Molar mass of ethyl alcohol} \times \text{Mole fraction of ethyl alcohol} = 46 \, \text{g/mol} \times 0.28 = 12.88 \, \text{g} \]

4. **Calculate the Weight of Water**:
- Similarly, for water:
\[ \text{Weight of water} = \text{Molar mass of water} \times \text{Mole fraction of water} = 18 \, \text{g/mol} \times 0.72 = 12.96 \, \text{g} \]

5. **Calculate the Total Weight of the Solution**:
- Total weight of the solution is the sum of the weights of ethyl alcohol and water:
\[ \text{Total weight of solution} = \text{Weight of ethyl alcohol} + \text{Weight of water} = 12.88 \, \text{g} + 12.96 \, \text{g} = 25.84 \, \text{g} \]

6. **Calculate the Percentage of Ethyl Alcohol by Weight**:
- The weight percentage of ethyl alcohol in the solution can be calculated as:
\[ \text{Percentage of ethyl alcohol} = \left( \frac{\text{Weight of ethyl alcohol}}{\text{Total weight of solution}} \right) \times 100 = \left( \frac{12.88 \, \text{g}}{25.84 \, \text{g}} \right) \times 100 \approx 49.84\% \]

### Final Answer:
The percentage of ethyl alcohol by weight in the solution is approximately **49.84%**.

---