A solution of sucrose (molar mass =342) is prepared by dissolving 688.4 g in 1000 g of water. Calculate
The boiling point of solution.

Options

A. `273`
B. `373.104`
C. `400`
D. `500`

Explanation

To solve the problem of calculating the boiling point of a sucrose solution, we will follow these steps:

### Step 1: Calculate the number of moles of sucrose.
Given:
- Mass of sucrose (Wb) = 688.4 g
- Molar mass of sucrose = 342 g/mol

The formula to calculate the number of moles (n) is:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ n = \frac{688.4 \, \text{g}}{342 \, \text{g/mol}} \approx 2.01 \, \text{mol} \]

### Step 2: Calculate the molality of the solution.
Given:
- Mass of water (Wa) = 1000 g
- Molar mass of water = 18 g/mol

To find molality (m), we use the formula:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
First, convert the mass of water to kg:
\[ \text{mass of water in kg} = \frac{1000 \, \text{g}}{1000} = 1 \, \text{kg} \]
Now, substituting the values:
\[ m = \frac{2.01 \, \text{mol}}{1 \, \text{kg}} = 2.01 \, \text{mol/kg} \]

### Step 3: Calculate the boiling point elevation (ΔTb).
The formula for boiling point elevation is:
\[ \Delta T_b = K_b \times m \]
Where:
- \( K_b \) for water = 0.52 °C kg/mol

Substituting the values:
\[ \Delta T_b = 0.52 \, \text{°C kg/mol} \times 2.01 \, \text{mol/kg} \approx 1.05 \, \text{°C} \]

### Step 4: Calculate the boiling point of the solution.
The normal boiling point of water (Tb0) is 100 °C (or 373 K). The new boiling point (Tb) is given by:
\[ T_b = T_{b0} + \Delta T_b \]
Substituting the values:
\[ T_b = 373 \, \text{K} + 1.05 \, \text{°C} = 374.05 \, \text{K} \]

### Final Answer:
The boiling point of the sucrose solution is approximately **374.05 K**.

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