S34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be

Options

A. `0.0099`
B. `1.1597`
C. `0.840`
D. `0.9901`

Explanation

To solve the problem of finding the relative lowering of vapor pressure when 34.2 g of cane sugar is dissolved in 180 g of water, we can follow these steps:

### Step 1: Calculate the number of moles of cane sugar (solute)
The molecular formula of cane sugar (sucrose) is C12H22O11. The molar mass can be calculated as follows:
- Carbon (C): 12 × 12 = 144 g/mol
- Hydrogen (H): 22 × 1 = 22 g/mol
- Oxygen (O): 11 × 16 = 176 g/mol

Total molar mass of cane sugar = 144 + 22 + 176 = 342 g/mol

Now, we can calculate the number of moles of cane sugar:
\[ \text{Moles of cane sugar} = \frac{\text{mass of cane sugar}}{\text{molar mass of cane sugar}} = \frac{34.2 \, \text{g}}{342 \, \text{g/mol}} \approx 0.1 \, \text{mol} \]

### Step 2: Calculate the number of moles of water (solvent)
The molar mass of water (H2O) is:
- Hydrogen (H): 2 × 1 = 2 g/mol
- Oxygen (O): 1 × 16 = 16 g/mol

Total molar mass of water = 2 + 16 = 18 g/mol

Now, we can calculate the number of moles of water:
\[ \text{Moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{180 \, \text{g}}{18 \, \text{g/mol}} = 10 \, \text{mol} \]

### Step 3: Calculate the total number of moles in the solution
Total moles in the solution = moles of cane sugar + moles of water
\[ \text{Total moles} = 0.1 \, \text{mol} + 10 \, \text{mol} = 10.1 \, \text{mol} \]

### Step 4: Calculate the relative lowering of vapor pressure
The formula for relative lowering of vapor pressure is given by:
\[ \text{Relative lowering of vapor pressure} = \frac{\text{moles of solute}}{\text{total moles in solution}} = \frac{0.1 \, \text{mol}}{10.1 \, \text{mol}} \]

Calculating this gives:
\[ \text{Relative lowering of vapor pressure} \approx \frac{0.1}{10.1} \approx 0.0099 \]

### Conclusion
The relative lowering of vapor pressure is approximately 0.0099.

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