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7. If 0.500 mol of a nonelectrolyte solute are dissolved in 500 g of ether, what is the freezing point of the solution? The normal freezing point of ether is [tex]-116.3^{\circ} C[/tex], and the molal freezing point constant of ether is [tex]-1.79^{\circ} C / m[/tex].

a. [tex]-118.09^{\circ} C[/tex]
b. [tex]-114.51^{\circ} C[/tex]
c. [tex]114.51^{\circ} C[/tex]
d. None of the above



Answer :

GinnyAnswer
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]

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