Calculate the boiling point of a solution containing `0.61 g` of benzoic acid in `50 g` of carbon disulphide assuming `84%` dimerization of the acid. The boiling point and `K_(b)` of `CS_(2)` are `46.2^(@)C` and `2.3 kg mol^(-1)`.

Explanation

To calculate the boiling point of a solution containing 0.61 g of benzoic acid in 50 g of carbon disulfide (CS₂) with 84% dimerization of the acid, we will follow these steps:

### Step 1: Calculate the number of moles of benzoic acid
First, we need to find the number of moles of benzoic acid (C₇H₆O₂) using its molecular weight.

**Molecular weight of benzoic acid (C₇H₆O₂)** = 122 g/mol.

\[ \text{Moles of benzoic acid} = \frac{\text{mass}}{\text{molecular weight}} = \frac{0.61 \text{ g}}{122 \text{ g/mol}} \approx 0.005 \text{ mol} \]

### Step 2: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

**Mass of solvent (CS₂)** = 50 g = 0.050 kg.

\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.005 \text{ mol}}{0.050 \text{ kg}} = 0.1 \text{ mol/kg} \]

### Step 3: Calculate the change in boiling point (ΔT_b)
The change in boiling point can be calculated using the formula:

\[ \Delta T_b = K_b \times m \]

Where \( K_b \) for CS₂ is given as 2.3 °C kg/mol.

\[ \Delta T_b = 2.3 \text{ °C kg/mol} \times 0.1 \text{ mol/kg} = 0.23 \text{ °C} \]

### Step 4: Adjust for dimerization
Since benzoic acid dimerizes, we need to adjust the effective molality. The degree of dimerization (α) is given as 84% (0.84).

Let \( n \) be the number of particles formed from one molecule of solute. For dimerization, \( n = 2 \).

Using the formula for effective molality considering dimerization:

\[ \frac{\Delta T_b^{\text{experimental}}}{\Delta T_b^{\text{theoretical}}} = 1 - \alpha + \frac{\alpha}{n} \]

Substituting the values:

\[ \frac{\Delta T_b^{\text{experimental}}}{0.23} = 1 - 0.84 + \frac{0.84}{2} \]

Calculating the right-hand side:

\[ 1 - 0.84 + 0.42 = 0.58 \]

Thus,

\[ \Delta T_b^{\text{experimental}} = 0.23 \times 0.58 \approx 0.1334 \text{ °C} \]

### Step 5: Calculate the boiling point of the solution
The boiling point of the pure solvent (CS₂) is given as 46.2 °C. Therefore, the boiling point of the solution (T_b) can be calculated as:

\[ T_b = T_{b0} + \Delta T_b^{\text{experimental}} = 46.2 \text{ °C} + 0.1334 \text{ °C} \approx 46.33 \text{ °C} \]

### Final Answer
The boiling point of the solution is approximately **46.33 °C**.

---