Question

The boiling point of benzene is353.3 K. When 1.80 $g$ of a non-volatile solute was dissolved in 90 $g$ of benzene, the boiling point is raised to 354.1 Κ given that $K_b$ for benzene is 2.52 K kg mol$^{-1}$, the molar mass of the solute is

• A. 15 g mol$^{-1}$
• B. 20 g mol$^{-1}$
• C. 25 g mol$^{-1}$
• D. 63 g mol$^{-1}$

Answer 1

Answer: (D)

63 g mol$^{-1}$

Answer 2

To find the molar mass of the solute, we use the boiling point elevation formula:

$\Delta T_b = K_b \cdot m$

Where:

• $\Delta T_b$ is the boiling point elevation.
• $K_b$ is the ebullioscopic constant of the solvent.
• $m$ is the molality of the solution.

First, calculate $\Delta T_b$:

$\Delta T_b = T_{boiling, solution} - T_{boiling, pure solvent} = 354.1 K - 353.3 K = 0.8 K$

Next, we know that molality ($m$) can be expressed as:

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

Where:

• $w_2$ is the mass of the solute (1.80 g).
• $M_2$ is the molar mass of the solute (what we want to find).
• $w_1$ is the mass of the solvent (90 g).

So, we can rewrite the boiling point elevation formula as:

$\Delta T_b = K_b \cdot \frac{w_2 \cdot 1000}{M_2 \cdot w_1}$

Now, rearrange to solve for $M_2$:

$M_2 = \frac{K_b \cdot w_2 \cdot 1000}{\Delta T_b \cdot w_1}$

Plug in the given values:

$M_2 = \frac{2.52 \text{ K kg mol}^{-1} \cdot 1.80 \text{ g} \cdot 1000}{0.8 \text{ K} \cdot 90 \text{ g}} = \frac{2.52 \cdot 1.80 \cdot 1000}{0.8 \cdot 90} = \frac{4536}{72} = 63 \text{ g/mol}$

Therefore, the molar mass of the solute is 63 g/mol.