The vapour pressure lowering caused by addition of 100 g of sucrose (molecular mass = 342) to 1000 g of wate,if the vapour pressure of pure water at `25^(@)C` is 23.8mm Hg,is

Options

A. 0.12 mmHg
B. 0.125 mmHg
C. 1.15 mmHg
D. 1.25 mmHg

Answer

To solve the problem of vapor pressure lowering caused by the addition of 100 g of sucrose to 1000 g of water, we will follow these steps:

### Step 1: Identify the given data
- Mass of sucrose (solute) = 100 g
- Molecular mass of sucrose = 342 g/mol
- Mass of water (solvent) = 1000 g
- Vapor pressure of pure water at 25°C (P₀) = 23.8 mm Hg

### Step 2: Calculate the number of moles of sucrose
To find the number of moles of sucrose, we use the formula:
\[ \text{Number of moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} \]
Substituting the values:
\[ \text{Number of moles of sucrose} = \frac{100 \, \text{g}}{342 \, \text{g/mol}} \approx 0.292 \, \text{mol} \]

### Step 3: Calculate the number of moles of water
To find the number of moles of water, we use the formula:
\[ \text{Number of moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} \]
The molar mass of water is 18 g/mol. Thus:
\[ \text{Number of moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 4: Calculate the total number of moles in the solution
Total moles = moles of sucrose + moles of water
\[ \text{Total moles} = 0.292 + 55.56 \approx 55.852 \, \text{mol} \]

### Step 5: Calculate the mole fraction of sucrose
The mole fraction of sucrose (solute) is given by:
\[ \text{Mole fraction of sucrose} = \frac{\text{moles of sucrose}}{\text{total moles}} \]
Substituting the values:
\[ \text{Mole fraction of sucrose} = \frac{0.292}{55.852} \approx 0.00522 \]

### Step 6: Calculate the lowering of vapor pressure (ΔP)
According to Raoult's law, the lowering of vapor pressure is given by:
\[ \Delta P = P₀ \times \text{mole fraction of sucrose} \]
Substituting the values:
\[ \Delta P = 23.8 \, \text{mm Hg} \times 0.00522 \approx 0.124 \, \text{mm Hg} \]

### Conclusion
The vapor pressure lowering caused by the addition of 100 g of sucrose to 1000 g of water is approximately **0.125 mm Hg**.

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