`C_(6)H_(6)` freezes at `5.5^(@)C`. At what tempreature will a solution of 10.44 g of `C_(4)H_(10)` in 200 g of `C_(6)H_(6) "freeze" K_(f)(C_(6)H_(6))= 5.12^(@)C//m`

Options

A. `4.608^(@)C`
B. `0.892^(@)C`
C. `5.5^(@)C`
D. none of these

Answer

To solve the problem, we need to determine the freezing point of a solution made by dissolving 10.44 g of \(C_4H_{10}\) (butane) in 200 g of \(C_6H_6\) (benzene). We will use the formula for freezing point depression, which is given by:

\[ \Delta T_f = K_f \times m \]

Where:
- \(\Delta T_f\) = change in freezing point
- \(K_f\) = freezing point depression constant of the solvent (benzene)
- \(m\) = molality of the solution

### Step 1: Calculate the number of moles of \(C_4H_{10}\)

First, we need to find the number of moles of butane (\(C_4H_{10}\)). The molecular weight of butane is 58 g/mol.

\[ \text{Number of moles of } C_4H_{10} = \frac{\text{mass}}{\text{molar mass}} = \frac{10.44 \text{ g}}{58 \text{ g/mol}} \approx 0.18069 \text{ mol} \]

### Step 2: Calculate the molality of the solution

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent. The mass of the solvent (benzene) is 200 g, which is 0.2 kg.

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.18069 \text{ mol}}{0.2 \text{ kg}} \approx 0.90345 \text{ mol/kg} \]

### Step 3: Calculate the freezing point depression (\(\Delta T_f\))

Now, we can calculate the freezing point depression using the \(K_f\) value for benzene, which is 5.12 °C/m.

\[ \Delta T_f = K_f \times m = 5.12 \text{ °C/m} \times 0.90345 \text{ mol/kg} \approx 4.628 \text{ °C} \]

### Step 4: Calculate the new freezing point of the solution

The freezing point of pure benzene is 5.5 °C. The new freezing point (\(T_f\)) of the solution can be calculated as follows:

\[ T_f = T_{f0} - \Delta T_f = 5.5 \text{ °C} - 4.628 \text{ °C} \approx 0.872 \text{ °C} \]

### Final Answer

The freezing point of the solution is approximately **0.872 °C**.

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