Mass percentage (w/w) of ethylene glycol `(HOCH_(2)-CH_(2)OH)` in a aqueous solution is 20, then mole fraction of solute is

Options

A. 0.5
B. 0.067
C. 0.1
D. 0.04

Explanation

To find the mole fraction of ethylene glycol in an aqueous solution where the mass percentage (w/w) of ethylene glycol is 20%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Mass Percentage**:
- The mass percentage (w/w) of ethylene glycol is given as 20%. This means that in 100 grams of the solution, there are 20 grams of ethylene glycol (solute) and 80 grams of water (solvent).

2. **Identify the Components**:
- Mass of ethylene glycol (solute) = 20 grams
- Mass of water (solvent) = 100 grams - 20 grams = 80 grams

3. **Calculate Moles of Ethylene Glycol**:
- The molecular formula of ethylene glycol is \(C_2H_6O_2\).
- Calculate the molar mass of ethylene glycol:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 6 = 6 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Total molar mass = 24 + 6 + 32 = 62 g/mol
- Now, calculate the number of moles of ethylene glycol:
\[ \text{Moles of ethylene glycol} = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \text{ g}}{62 \text{ g/mol}} \approx 0.3226 \text{ moles} \]

4. **Calculate Moles of Water**:
- The molar mass of water \(H_2O\) is:
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Oxygen (O): 16 g/mol × 1 = 16 g/mol
- Total molar mass = 2 + 16 = 18 g/mol
- Now, calculate the number of moles of water:
\[ \text{Moles of water} = \frac{80 \text{ g}}{18 \text{ g/mol}} \approx 4.444 \text{ moles} \]

5. **Calculate Mole Fraction of Ethylene Glycol**:
- The mole fraction of ethylene glycol (\(X_{EG}\)) is given by:
\[ X_{EG} = \frac{\text{moles of ethylene glycol}}{\text{moles of ethylene glycol} + \text{moles of water}} = \frac{0.3226}{0.3226 + 4.444} \]
- Calculate the total moles:
\[ X_{EG} = \frac{0.3226}{0.3226 + 4.444} \approx \frac{0.3226}{4.7666} \approx 0.0676 \]

### Final Answer:
The mole fraction of ethylene glycol in the solution is approximately **0.0676**.