The boiling point of benzene is 353 K when 1.8 g of a nonvolatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354 K. What is the molar mass of solute if $K_{b}$ for benzene is $2.5~K~kg~mol^{-1}$?

Question

Accepted Answer

To determine the molar mass of the solute, we can use the formula for boiling point elevation:

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

Where:
- $\Delta T_b$ is the change in boiling point.
- $i$ is the van't Hoff factor (which is 1 for a nonvolatile solute).
- $K_b$ is the ebullioscopic constant of the solvent (benzene in this case).
- $m$ is the molality of the solution.

Given:
- $\Delta T_b = 354 \, \text{K} - 353 \, \text{K} = 1 \, \text{K}$
- $K_b = 2.5 \, \text{K kg mol}^{-1}$
- Mass of solute = 1.8 g
- Mass of solvent (benzene) = 90 g = 0.09 kg

First, calculate the molality ($m$):

$$
m = \frac{\Delta T_b}{K_b} = \frac{1 \, \text{K}}{2.5 \, \text{K kg mol}^{-1}} = 0.4 \, \text{mol kg}^{-1}
$$

Molality ($m$) is also defined as:

$$
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
$$

Let $M$ be the molar mass of the solute. Then, the moles of solute is $\frac{1.8}{M}$.

Substitute the values into the molality equation:

$$
0.4 = \frac{1.8/M}{0.09}
$$

Solving for $M$:

$$
0.4 = \frac{1.8}{0.09M}
$$

$$
0.4 \times 0.09M = 1.8
$$

$$
0.036M = 1.8
$$

$$
M = \frac{1.8}{0.036} = 50 \, \text{g mol}^{-1}
$$

Therefore, the molar mass of the solute is $50 \, \text{g mol}^{-1}$, which corresponds to Option 2.

**Explanation for Other Options:**

1. **Option 1: $45 \, \text{g mol}^{-1}$** - This option is incorrect because the calculated molar mass using the given data and formula does not match 45 g/mol. The calculations clearly show that the molar mass is 50 g/mol.

2. **Option 3: $55 \, \text{g mol}^{-1}$** - This option is incorrect for the same reason as Option 1. The calculations do not support a molar mass of 55 g/mol.

3. **Option 4: $60 \, \text{g mol}^{-1}$** - Similarly, this option is incorrect because the calculated molar mass is not 60 g/mol. The correct calculation yields 50 g/mol.

In conclusion, the correct answer is Option 2, $50 \, \text{g mol}^{-1}$, as supported by the calculations based on the given data and formula for boiling point elevation.