When `1.22 g C_(6)H_(5)COOH` is added into two solvents, the following data of `DeltaT_(b)` and `K_(b)` are obtained:
i. In `100 g CH_(3)COCH_(3)`,`DeltaT_(b)=0.17`,`K_(b)=1.7 kg K mol^(-1)`.
ii. In `100 g` benzene,`DeltaT_(b)=0.13` and `K_(b)=2.6 kg K mol^(-1)`.
Find out the molecular weight of `C_(6)H_(5)COOH` in both cases and interpret the results.

Explanation

To find the molecular weight of benzoic acid (C₆H₅COOH) in two different solvents, we can use the formula for boiling point elevation, which is given by:

\[ \Delta T_b = i \cdot K_b \cdot m \]

where:
- \(\Delta T_b\) = boiling point elevation
- \(i\) = van 't Hoff factor (number of particles the solute breaks into)
- \(K_b\) = ebullioscopic constant of the solvent
- \(m\) = molality of the solution

### Step 1: Calculate the molality of the solution

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent.

Given:
- Mass of benzoic acid = 1.22 g
- Mass of solvent = 100 g = 0.1 kg

The number of moles of benzoic acid can be expressed as:

\[ \text{Moles of benzoic acid} = \frac{1.22 \text{ g}}{M \text{ g/mol}} \]

where \(M\) is the molar mass of benzoic acid.

Thus, the molality (\(m\)) can be expressed as:

\[ m = \frac{\text{Moles of benzoic acid}}{\text{Mass of solvent in kg}} = \frac{1.22/M}{0.1} = \frac{12.2}{M} \]

### Step 2: Apply the boiling point elevation formula for acetone

For the first solvent (acetone):
- \(\Delta T_b = 0.17\) °C
- \(K_b = 1.7\) kg K mol⁻¹

Using the boiling point elevation formula:

\[ 0.17 = i \cdot 1.7 \cdot \frac{12.2}{M} \]

Assuming \(i = 1\) (benzoic acid does not dissociate in acetone):

\[ 0.17 = 1.7 \cdot \frac{12.2}{M} \]

Rearranging to solve for \(M\):

\[ M = 1.7 \cdot 12.2 / 0.17 \]

Calculating \(M\):

\[ M = \frac{20.74}{0.17} \approx 122 \text{ g/mol} \]

### Step 3: Apply the boiling point elevation formula for benzene

For the second solvent (benzene):
- \(\Delta T_b = 0.13\) °C
- \(K_b = 2.6\) kg K mol⁻¹

Using the boiling point elevation formula:

\[ 0.13 = i \cdot 2.6 \cdot \frac{12.2}{M} \]

Assuming \(i = 1\) initially:

\[ 0.13 = 2.6 \cdot \frac{12.2}{M} \]

Rearranging to solve for \(M\):

\[ M = 2.6 \cdot 12.2 / 0.13 \]

Calculating \(M\):

\[ M = \frac{31.72}{0.13} \approx 244 \text{ g/mol} \]

### Step 4: Interpret the results

1. In acetone, the calculated molecular weight of benzoic acid is approximately **122 g/mol**, indicating that benzoic acid exists as a **monomer** in this solvent.
2. In benzene, the calculated molecular weight is approximately **244 g/mol**, suggesting that benzoic acid exists as a **dimer** in this solvent.