To solve the problem, we will follow these steps:
### Step 1: Calculate the vapor pressure lowering
We know that the vapor pressure of pure water at 100°C (P₀) is 760 mmHg, and the vapor pressure of the solution (P_solution) is given as 732 mmHg. We can use the formula for relative lowering of vapor pressure:
\[
\Delta P = P_0 - P_{solution}
\]
Substituting the values:
\[
\Delta P = 760 \, \text{mmHg} - 732 \, \text{mmHg} = 28 \, \text{mmHg}
\]
### Step 2: Calculate the mole fraction of the solute
Using Raoult's Law, we can relate the lowering of vapor pressure to the mole fraction of the solute (x_solute):
\[
\frac{\Delta P}{P_0} = x_{solute}
\]
Substituting the values:
\[
x_{solute} = \frac{28 \, \text{mmHg}}{760 \, \text{mmHg}} \approx 0.0368
\]
### Step 3: Calculate the mole fraction of the solvent
Since the mole fraction of the solute and solvent must add up to 1, we can find the mole fraction of the solvent (x_solvent):
\[
x_{solvent} = 1 - x_{solute} \approx 1 - 0.0368 \approx 0.9632
\]
### Step 4: Relate mole fractions to moles
Let n_solute be the number of moles of the solute and n_solvent be the number of moles of water. The mass of water is given as 100 g, and the molar mass of water is approximately 18 g/mol:
\[
n_{solvent} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} \approx 5.56 \, \text{mol}
\]
Using the mole fraction of the solvent:
\[
x_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}
\]
Substituting the values:
\[
0.9632 = \frac{5.56}{5.56 + n_{solute}}
\]
### Step 5: Solve for n_solute
Rearranging the equation to find n_solute:
\[
0.9632(5.56 + n_{solute}) = 5.56
\]
Expanding and solving for n_solute:
\[
5.36 + 0.9632 \cdot n_{solute} = 5.56
\]
\[
0.9632 \cdot n_{solute} = 0.20
\]
\[
n_{solute} \approx \frac{0.20}{0.9632} \approx 0.2077 \, \text{mol}
\]
### Step 6: Calculate the molar mass of the solute
We know the mass of the solute is 6.5 g. The molar mass (M) can be calculated as:
\[
M = \frac{\text{mass}}{n_{solute}} = \frac{6.5 \, \text{g}}{0.2077 \, \text{mol}} \approx 31.3 \, \text{g/mol}
\]
### Step 7: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[
m = \frac{n_{solute}}{mass_{solvent} \, (kg)} = \frac{0.2077 \, \text{mol}}{0.1 \, \text{kg}} \approx 2.077 \, \text{mol/kg}
\]
### Step 8: Calculate the elevation in boiling point
Using the formula for boiling point elevation:
\[
\Delta T_b = K_b \cdot m
\]
Substituting the values (K_b = 0.52 °C kg/mol):
\[
\Delta T_b = 0.52 \cdot 2.077 \approx 1.08 \, °C
\]
### Step 9: Calculate the new boiling point
The new boiling point (T_b') can be calculated as:
\[
T_b' = T_b + \Delta T_b = 100 + 1.08 \approx 101.08 \, °C
\]
### Final Answer
The boiling point of the solution will be approximately **101.08 °C**.
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