To solve this problem, we need to determine the freezing point of a solution containing 0.500 moles of a nonelectrolyte solute dissolved in 500 grams of ether. We will use the freezing point depression formula and data provided:
1. Moles of Solute: [tex]\(0.500\)[/tex] moles
2. Mass of Ether: [tex]\(500 \ \text{grams}\)[/tex]
3. Molal Freezing Point Depression Constant ([tex]\(K_f\)[/tex]) for Ether: [tex]\(-1.79 \ \degree C/m\)[/tex]
4. Normal Freezing Point of Ether: [tex]\(-116.3 \ \degree C\)[/tex]
### Step-by-Step Solution:
1. Convert the mass of ether to kilograms:
[tex]\[
\text{Mass of ether in kg} = \frac{500 \ \text{grams}}{1000 \ \text{grams/kg}} = 0.500 \ \text{kg}
\][/tex]
2. Calculate the molality of the solution:
Molality ([tex]\(m\)[/tex]) is defined as moles of solute per kilogram of solvent.
[tex]\[
\text{Molality} = \frac{0.500 \ \text{moles of solute}}{0.500 \ \text{kg of solvent}} = 1.0 \ \text{m}
\][/tex]
3. Calculate the freezing point depression ([tex]\(\Delta T_f\)[/tex]):
The freezing point depression can be calculated using:
[tex]\[
\Delta T_f = K_f \times \text{molality}
\][/tex]
Given that [tex]\(K_f = -1.79 \ \degree C/m\)[/tex]:
[tex]\[
\Delta T_f = -1.79 \ \degree C/m \times 1.0 \ \text{m} = -1.79 \ \degree C
\][/tex]
4. Determine the new freezing point of the solution:
The new freezing point ([tex]\(T_f\)[/tex]) of the solution can be found by subtracting the freezing point depression from the normal freezing point of the ether:
[tex]\[
T_f = -116.3 \ \degree C + \Delta T_f
\][/tex]
[tex]\[
T_f = -116.3 \ \degree C + (-1.79 \ \degree C)
\][/tex]
[tex]\[
T_f = -116.3 \ \degree C - 1.79 \ \degree C
\][/tex]
[tex]\[
T_f = -118.09 \ \degree C
\][/tex]
### Conclusion:
The freezing point of the solution is [tex]\(-118.09 \ \degree C\)[/tex].
So, the correct answer is:
a. [tex]\( -118.09 \ \degree C \)[/tex]
