To solve the problem of how much sucrose needs to be added to 1 kg of water to raise the boiling point from 372.63 K to 373 K at 750 torr, we can follow these steps:
### Step 1: Understand the boiling point elevation
The boiling point elevation (\( \Delta T_b \)) is the difference between the boiling point of the solution and the boiling point of the pure solvent (water in this case).
\[
\Delta T_b = T_{b,\text{solution}} - T_{b,\text{solvent}} = 373 \, \text{K} - 372.63 \, \text{K} = 0.37 \, \text{K}
\]
### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is given by:
\[
\Delta T_b = K_b \cdot m
\]
Where:
- \( K_b \) is the ebullioscopic constant of the solvent (water here).
- \( m \) is the molality of the solution.
For water, \( K_b = 0.55 \, \text{K kg/mol} \).
### Step 3: Rearrange the formula to find molality
We can rearrange the formula to solve for molality (\( m \)):
\[
m = \frac{\Delta T_b}{K_b} = \frac{0.37 \, \text{K}}{0.55 \, \text{K kg/mol}} \approx 0.6727 \, \text{mol/kg}
\]
### Step 4: Calculate the number of moles of sucrose needed
Molality is defined as the number of moles of solute per kilogram of solvent. Since we have 1 kg of water, the number of moles of sucrose (\( n \)) can be calculated as:
\[
n = m \cdot \text{mass of solvent (kg)} = 0.6727 \, \text{mol/kg} \cdot 1 \, \text{kg} = 0.6727 \, \text{mol}
\]
### Step 5: Calculate the mass of sucrose needed
To find the mass of sucrose required, we need the molar mass of sucrose. The molar mass of sucrose (\( C_{12}H_{22}O_{11} \)) is approximately 342 g/mol.
Now we can calculate the mass (\( x \)) of sucrose:
\[
x = n \cdot \text{molar mass} = 0.6727 \, \text{mol} \cdot 342 \, \text{g/mol} \approx 229.5 \, \text{g}
\]
### Final Answer
The mass of sucrose that needs to be added to 1 kg of water to raise the boiling point to 373 K is approximately **229.5 grams**.
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