Mole fraction of ethanol is ethanol water mixture is 0.25. Hence, the percentage concentration of ethanol by weight of mixture is

Options

A. 0.25
B. 0.75
C. 0.46
D. 0.54

Explanation

To solve the problem of finding the percentage concentration of ethanol by weight in an ethanol-water mixture with a mole fraction of ethanol (solute) of 0.25, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Mole Fractions**:
- Given the mole fraction of ethanol (solute) \(X_2 = 0.25\).
- The mole fraction of water (solvent) \(X_1 = 1 - X_2 = 1 - 0.25 = 0.75\).

2. **Calculate Molecular Weights**:
- **Molecular weight of ethanol (C₂H₅OH)**:
- Carbon (C): 2 × 12 g/mol = 24 g/mol
- Hydrogen (H): 6 × 1 g/mol = 6 g/mol
- Oxygen (O): 1 × 16 g/mol = 16 g/mol
- Total = 24 + 6 + 16 = 46 g/mol
- **Molecular weight of water (H₂O)**:
- Hydrogen (H): 2 × 1 g/mol = 2 g/mol
- Oxygen (O): 1 × 16 g/mol = 16 g/mol
- Total = 2 + 16 = 18 g/mol

3. **Set Up Mole Ratio**:
- Let \(N_2\) be the number of moles of ethanol and \(N_1\) be the number of moles of water.
- From the mole fractions, we have:
\[ \frac{N_1}{N_2} = \frac{X_1}{X_2} = \frac{0.75}{0.25} = 3 \]
- This means \(N_1 = 3N_2\).

4. **Calculate Moles in Terms of Mass**:
- The number of moles can be expressed in terms of mass:
\[ N_1 = \frac{W_1}{M_1} \quad \text{and} \quad N_2 = \frac{W_2}{M_2} \]
- Where \(W_1\) is the mass of water, \(W_2\) is the mass of ethanol, \(M_1 = 18\) g/mol (water), and \(M_2 = 46\) g/mol (ethanol).
- Substituting into the mole ratio:
\[ \frac{W_1/18}{W_2/46} = 3 \implies \frac{W_1}{W_2} = 3 \cdot \frac{18}{46} = \frac{54}{46} = \frac{27}{23} \]
- This gives us the ratio of masses:
\[ W_1 = \frac{27}{23} W_2 \]

5. **Calculate Total Mass**:
- Total mass of the mixture \(W = W_1 + W_2\):
\[ W = W_2 + \frac{27}{23} W_2 = \frac{23 + 27}{23} W_2 = \frac{50}{23} W_2 \]

6. **Calculate Weight Percent of Ethanol**:
- The weight percent of ethanol is given by:
\[ \text{Weight percent} = \left(\frac{W_2}{W_1 + W_2}\right) \times 100 \]
- Substituting the total mass:
\[ \text{Weight percent} = \left(\frac{W_2}{\frac{50}{23} W_2}\right) \times 100 = \left(\frac{23}{50}\right) \times 100 = 46\% \]

### Final Answer:
The percentage concentration of ethanol by weight in the mixture is **46%**.