Question

1.22 g of benzoic acid (C$_{6}$H$_{5}$COOH) is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point of the solution in acetone increases by 0.17°C while in benzene increases by 0.13°C. K$_{b}$ for acetone and benzene are 1.7 K kg mol$^{-1}$ and 2.6 K kg mol$^{-1}$ respectively. If x and y are the molecular weight (in gram/mole) of benzoic acid in acetone and benzene respectively, then the value of (x + y) is:

Answer 1

366

Answer 2

The boiling point elevation ($\Delta T_b$) is a colligative property given by the formula:

$$\Delta T_b = K_b \cdot m$$

where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.

Molality ($m$) is defined as:

$$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{w_2 / M_2}{w_1 / 1000}$$

Here, $w_2$ is the mass of the solute, $M_2$ is the molecular weight of the solute, and $w_1$ is the mass of the solvent in grams.

Substituting the expression for molality into the boiling point elevation formula:

$$\Delta T_b = K_b \cdot \frac{w_2}{M_2 \cdot w_1 \text{(kg)}}$$

We can rearrange this formula to solve for the observed molecular weight ($M_2$):

$$M_2 = \frac{K_b \cdot w_2}{\Delta T_b \cdot w_1 \text{(kg)}}$$

Given:

Mass of benzoic acid ($w_2$) = 1.22 g

Mass of solvent ($w_1$) = 100 g = 0.1 kg

1. Benzoic acid in Acetone:

• Boiling point elevation ($\Delta T_b$) = 0.17 °C
• $K_b$ for acetone = 1.7 K kg mol$^{-1}$
• Let 'x' be the molecular weight of benzoic acid in acetone.

Using the formula for $M_2$:

$$x = \frac{1.7 \text{ K kg mol}^{-1} \cdot 1.22 \text{ g}}{0.17 \text{ K} \cdot 0.1 \text{ kg}}$$ $$x = \frac{2.074}{0.017}$$ $$x = 122 \text{ g/mol}$$

The theoretical molecular weight of benzoic acid (C$_6$H$_5$COOH) is $7 \times 12.01 + 6 \times 1.008 + 2 \times 16.00 \approx 122.12$ g/mol. Since the observed molecular weight is close to the theoretical value, benzoic acid exists as a monomer in acetone.

2. Benzoic acid in Benzene:

• Boiling point elevation ($\Delta T_b$) = 0.13 °C
• $K_b$ for benzene = 2.6 K kg mol$^{-1}$
• Let 'y' be the molecular weight of benzoic acid in benzene.

Using the formula for $M_2$:

$$y = \frac{2.6 \text{ K kg mol}^{-1} \cdot 1.22 \text{ g}}{0.13 \text{ K} \cdot 0.1 \text{ kg}}$$ $$y = \frac{3.172}{0.013}$$ $$y = 244 \text{ g/mol}$$

The observed molecular weight 'y' (244 g/mol) is approximately twice the theoretical molecular weight of benzoic acid (122 g/mol). This indicates that benzoic acid undergoes dimerization in benzene due to hydrogen bonding, forming a cyclic dimer.

3. Value of (x + y):

$$x + y = 122 + 244 = 366$$