The vapour pressure of an aqueous solution of cane sugar (mol.wt. 342) is 756mm at `100^(@)C`. How many grams of sugar are present per 1000g of water ?

Answer

To solve the problem, we need to find out how many grams of cane sugar are present in 1000 grams of water, given the vapor pressure of the solution. We will use the concept of vapor pressure lowering and colligative properties.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Vapor pressure of pure water at 100°C, \( P_0 = 760 \, \text{mmHg} \)
- Vapor pressure of the sugar solution, \( P_s = 756 \, \text{mmHg} \)
- Molar mass of cane sugar (sucrose), \( M = 342 \, \text{g/mol} \)
- Mass of water, \( m_{\text{water}} = 1000 \, \text{g} \)

2. **Calculate the Decrease in Vapor Pressure:**
\[ \Delta P = P_0 - P_s = 760 \, \text{mmHg} - 756 \, \text{mmHg} = 4 \, \text{mmHg} \]

3. **Calculate the Mole Fraction of the Solute (Cane Sugar):**
The decrease in vapor pressure is related to the mole fraction of the solute:
\[ \Delta P = P_0 \cdot X_{\text{solute}} \]
Rearranging gives:
\[ X_{\text{solute}} = \frac{\Delta P}{P_0} = \frac{4 \, \text{mmHg}}{760 \, \text{mmHg}} = \frac{1}{190} \]

4. **Calculate the Mole Fraction of the Solvent (Water):**
Since \( X_{\text{solute}} + X_{\text{solvent}} = 1 \):
\[ X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - \frac{1}{190} = \frac{189}{190} \]

5. **Calculate Moles of Water:**
The molar mass of water is \( 18 \, \text{g/mol} \):
\[ n_{\text{water}} = \frac{m_{\text{water}}}{M_{\text{water}}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

6. **Relate Moles of Solute to Mole Fraction:**
Using the mole fraction of the solvent:
\[ X_{\text{solvent}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{solute}}} \]
Substituting the known values:
\[ \frac{189}{190} = \frac{55.56}{55.56 + n_{\text{solute}}} \]

7. **Solve for Moles of Solute:**
Cross-multiplying gives:
\[ 189(55.56 + n_{\text{solute}}) = 190 \times 55.56 \]
Expanding and solving for \( n_{\text{solute}} \):
\[ 189 \cdot 55.56 + 189 n_{\text{solute}} = 190 \cdot 55.56 \]
\[ 189 n_{\text{solute}} = 190 \cdot 55.56 - 189 \cdot 55.56 \]
\[ 189 n_{\text{solute}} = 0.94 \]
\[ n_{\text{solute}} \approx \frac{0.94}{189} \approx 0.00497 \, \text{mol} \]

8. **Convert Moles of Solute to Grams:**
\[ m_{\text{solute}} = n_{\text{solute}} \times M_{\text{solute}} = 0.00497 \, \text{mol} \times 342 \, \text{g/mol} \approx 1.70 \, \text{g} \]

### Final Answer:
Approximately **1.70 grams** of cane sugar are present in 1000 grams of water.