Mole fraction of `C_(3)H_(5)(OH)_(3)` in a solution of `36g` of water and `46g` of glycerine is:

Options

A. `0.46`
B. `0.36`
C. `0.20`
D. `0.40`

Explanation

To find the mole fraction of glycerin (C₃H₅(OH)₃) in a solution containing 36 g of water and 46 g of glycerin, follow these steps:

### Step 1: Calculate the number of moles of glycerin (C₃H₅(OH)₃)
- The molecular weight of glycerin (C₃H₅(OH)₃) is 92 g/mol.
- The formula to calculate the number of moles is:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]
- For glycerin:
\[ \text{Number of moles of glycerin} = \frac{46 \text{ g}}{92 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 2: Calculate the number of moles of water (H₂O)
- The molecular weight of water (H₂O) is 18 g/mol.
- Using the same formula:
\[ \text{Number of moles of water} = \frac{36 \text{ g}}{18 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Calculate the total number of moles in the solution
- Total number of moles = Number of moles of glycerin + Number of moles of water
\[ \text{Total moles} = 0.5 + 2 = 2.5 \text{ moles} \]

### Step 4: Calculate the mole fraction of glycerin
- The mole fraction (X) of glycerin is given by:
\[ X_{\text{glycerin}} = \frac{\text{Number of moles of glycerin}}{\text{Total number of moles}} \]
- Substituting the values:
\[ X_{\text{glycerin}} = \frac{0.5}{2.5} = 0.2 \]

### Final Answer:
The mole fraction of glycerin in the solution is **0.2**.

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