The boiling point of water at `750mm Hg` is `99.63^(@)C`. How much sucrose is to be added to `500 g` of water such that it boils at `100^(@)C`.

Explanation

To solve the problem of how much sucrose needs to be added to 500 g of water to raise its boiling point from 99.63°C to 100°C at a pressure of 750 mm Hg, we can follow these steps:

### Step 1: Understand the boiling point elevation
The boiling point elevation can be calculated using the formula:
\[ \Delta T_b = K_b \cdot m \]
where:
- \(\Delta T_b\) = change in boiling point
- \(K_b\) = ebullioscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 2: Calculate \(\Delta T_b\)
Given that the boiling point of water at 750 mm Hg is 99.63°C, and we want it to boil at 100°C:
\[ \Delta T_b = 100°C - 99.63°C = 0.37°C \]

### Step 3: Find the ebullioscopic constant (\(K_b\)) for water
The ebullioscopic constant (\(K_b\)) for water is approximately 0.512 °C kg/mol.

### Step 4: Calculate the molality (\(m\))
Rearranging the boiling point elevation formula to find molality:
\[ m = \frac{\Delta T_b}{K_b} = \frac{0.37°C}{0.512 °C \, \text{kg/mol}} \approx 0.722 \, \text{mol/kg} \]

### Step 5: Calculate the number of moles of sucrose needed
The molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
For 500 g of water:
\[ \text{mass of solvent} = 500 \, \text{g} = 0.5 \, \text{kg} \]
Thus, the number of moles of sucrose (\(n\)) needed is:
\[ n = m \cdot \text{mass of solvent in kg} = 0.722 \, \text{mol/kg} \cdot 0.5 \, \text{kg} \approx 0.361 \, \text{mol} \]

### Step 6: Calculate the mass of sucrose required
The molar mass of sucrose (C12H22O11) is approximately 342 g/mol. Therefore, the mass of sucrose (\(W\)) needed is:
\[ W = n \cdot \text{molar mass} = 0.361 \, \text{mol} \cdot 342 \, \text{g/mol} \approx 123.5 \, \text{g} \]

### Conclusion
To raise the boiling point of 500 g of water to 100°C, approximately **123.5 g of sucrose** needs to be added.

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