A solution of sucrose (molar mass =342) is prepared by dissolving 68.4 g in 1000 g of water. Calculate
The boiling point of solution.

Options

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

Explanation

To calculate the boiling point of the sucrose solution, we will follow these steps:

### Step 1: Calculate the number of moles of sucrose.
To find the number of moles, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

Given:
- Mass of sucrose = 68.4 g
- Molar mass of sucrose = 342 g/mol

\[ \text{Number of moles} = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} = 0.2 \, \text{moles} \]

### Step 2: Calculate the molality of the solution.
Molality (m) is calculated using the formula:

\[ \text{Molality} = \frac{\text{number of moles of solute}}{\text{mass of solvent (kg)}} \]

Given:
- Mass of water (solvent) = 1000 g = 1 kg

\[ \text{Molality} = \frac{0.2 \, \text{moles}}{1 \, \text{kg}} = 0.2 \, \text{mol/kg} \]

### Step 3: Calculate the boiling point elevation (\(\Delta T_b\)).
The boiling point elevation can be calculated using the formula:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \(i\) = van 't Hoff factor (for sucrose, a non-electrolyte, \(i = 1\))
- \(K_b\) = ebullioscopic constant of water = 0.52 °C kg/mol
- \(m\) = molality of the solution = 0.2 mol/kg

\[ \Delta T_b = 1 \cdot 0.52 \, \text{°C kg/mol} \cdot 0.2 \, \text{mol/kg} = 0.104 \, \text{°C} \]

### Step 4: Calculate the boiling point of the solution.
The boiling point of the solution (\(T_b\)) can be calculated using the formula:

\[ T_b = T_b^0 + \Delta T_b \]

Where:
- \(T_b^0\) = boiling point of pure water = 100 °C

\[ T_b = 100 \, \text{°C} + 0.104 \, \text{°C} = 100.104 \, \text{°C} \]

### Final Answer:
The boiling point of the sucrose solution is **100.104 °C**.

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