Mole fraction of ethyl alcohol in aqueous ethyl alcohol (`C_2H_5OH`) solution is 0.55. Hence percentage of ethyl alcohol by weight is:

Explanation

To find the percentage of ethyl alcohol (C₂H₅OH) by weight in an aqueous solution where the mole fraction of ethyl alcohol is 0.55, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mole Fraction**:
The mole fraction of ethyl alcohol (C₂H₅OH) is given as 0.55. Therefore, the mole fraction of water (H₂O) can be calculated as:
\[ \text{Mole fraction of water} = 1 - \text{Mole fraction of ethyl alcohol} = 1 - 0.55 = 0.45 \]

2. **Calculate Moles of Each Component**:
Let's assume we have 1 mole of the solution. Hence:
- Moles of ethyl alcohol = 0.55 moles
- Moles of water = 0.45 moles

3. **Calculate the Mass of Each Component**:
- The molar mass of ethyl alcohol (C₂H₅OH) is approximately 46 g/mol. Therefore, the mass of ethyl alcohol is:
\[ \text{Mass of ethyl alcohol} = 0.55 \, \text{moles} \times 46 \, \text{g/gmol} = 25.3 \, \text{grams} \]
- The molar mass of water (H₂O) is approximately 18 g/mol. Therefore, the mass of water is:
\[ \text{Mass of water} = 0.45 \, \text{moles} \times 18 \, \text{g/gmol} = 8.1 \, \text{grams} \]

4. **Calculate the Total Mass of the Solution**:
The total mass of the solution is the sum of the masses of ethyl alcohol and water:
\[ \text{Total mass of solution} = \text{Mass of ethyl alcohol} + \text{Mass of water} = 25.3 \, \text{grams} + 8.1 \, \text{grams} = 33.4 \, \text{grams} \]

5. **Calculate the Weight Percentage of Ethyl Alcohol**:
The weight percentage of ethyl alcohol in the solution can be calculated using the formula:
\[ \text{Weight percentage of ethyl alcohol} = \left( \frac{\text{Mass of ethyl alcohol}}{\text{Total mass of solution}} \right) \times 100 \]
Substituting the values:
\[ \text{Weight percentage of ethyl alcohol} = \left( \frac{25.3 \, \text{grams}}{33.4 \, \text{grams}} \right) \times 100 \approx 75.74\% \]

### Final Answer:
The percentage of ethyl alcohol by weight in the solution is approximately **75.74%**.

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