Calculate the mole fraction of water in mixture of 12 g water , 108 g acetic acid and 92 g ethanol.

Explanation

To calculate the mole fraction of water in a mixture of 12 g of water, 108 g of acetic acid, and 92 g of ethanol, we will follow these steps:

### Step 1: Calculate the number of moles of each component.

**Formula for number of moles:**
\[ \text{Number of moles} = \frac{\text{Weight (g)}}{\text{Molecular Weight (g/mol)}} \]

**1. For Water (H₂O):**
- Weight = 12 g
- Molecular Weight of water = 18 g/mol
\[ \text{Moles of water} = \frac{12 \, \text{g}}{18 \, \text{g/mol}} = 0.67 \, \text{moles} \]

**2. For Acetic Acid (CH₃COOH):**
- Weight = 108 g
- Molecular Weight of acetic acid = 60 g/mol
\[ \text{Moles of acetic acid} = \frac{108 \, \text{g}}{60 \, \text{g/mol}} = 1.8 \, \text{moles} \]

**3. For Ethanol (C₂H₅OH):**
- Weight = 92 g
- Molecular Weight of ethanol = 46 g/mol
\[ \text{Moles of ethanol} = \frac{92 \, \text{g}}{46 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 2: Calculate the total number of moles.

\[ \text{Total moles} = \text{Moles of water} + \text{Moles of acetic acid} + \text{Moles of ethanol} \]
\[ \text{Total moles} = 0.67 + 1.8 + 2 = 4.47 \, \text{moles} \]

### Step 3: Calculate the mole fraction of water.

**Mole fraction formula:**
\[ \text{Mole fraction of water} = \frac{\text{Moles of water}}{\text{Total moles}} \]
\[ \text{Mole fraction of water} = \frac{0.67}{4.47} \approx 0.14988 \]

### Step 4: Round off the answer.

Rounding off to two decimal places:
\[ \text{Mole fraction of water} \approx 0.15 \]

### Final Answer:
The mole fraction of water in the mixture is approximately **0.15**.

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