Sure, let's find the boiling point of an aqueous solution containing a nonelectrolyte that freezes at -6.51°C. Here is the step-by-step approach:
1. Determine the Freezing Point Depression:
The freezing point of pure water is 0°C. Here, the solution freezes at -6.51°C. Therefore, the freezing point depression, [tex]\(\Delta T_f\)[/tex], is 6.51°C.
2. Given Constants for Water:
- The freezing point depression constant for water, [tex]\(K_f\)[/tex] is 1.86°C·kg/mol
- The boiling point elevation constant for water, [tex]\(K_b\)[/tex] is 0.512°C·kg/mol
3. Calculate Molality:
The freezing point depression equation is:
[tex]\[
\Delta T_f = K_f \times m
\][/tex]
where [tex]\( m \)[/tex] is the molality of the solution. Rearrange this equation to find the molality:
[tex]\[
m = \frac{\Delta T_f}{K_f}
\][/tex]
Substituting the given values:
[tex]\[
m = \frac{6.51}{1.86} \approx 3.5 \, \text{mol/kg}
\][/tex]
4. Calculate the Boiling Point Elevation:
The boiling point elevation equation is:
[tex]\[
\Delta T_b = K_b \times m
\][/tex]
Substitute the values into the equation:
[tex]\[
\Delta T_b = 0.512 \times 3.5 \approx 1.79 \, \text{°C}
\][/tex]
5. Determine the Boiling Point of the Solution:
The normal boiling point of pure water is 100°C. The boiling point of the solution can be found by adding the boiling point elevation to the normal boiling point:
[tex]\[
\text{Boiling point of the solution} = 100 \, \text{°C} + 1.79 \, \text{°C} \approx 101.79 \, \text{°C}
\][/tex]
Therefore, the boiling point of the solution is approximately 101.8°C.
So, the correct answer is:
b. 101.8°C
