Question

A solution of nonvolatile solute is obtained by dissolving 3.5 g in 100 g solvent has boiling point elevation 0.35 K. Calculate the molar mass of solute. (Molal elevation constant = $2.5\text{ K kg mol}^{-1}$)

• A. $270\text{ g mol}^{-1}$
• B. $260\text{ g mol}^{-1}$
• C. $250\text{ g mol}^{-1}$
• D. $240\text{ g mol}^{-1}$

Hint:

Logic Tip: To minimize math errors, always group the powers of 10! The ratio $3.5 / 0.35$ is exactly 10. Then $1000 / 100$ is exactly 10. Multiply those to get 100. Then simply multiply by the $K_b$ constant: $100 \times 2.5 = 250$.

Answer 1

The Correct Option is C

Concept:
The elevation in boiling point ($\Delta T_b$) of a solution containing a non-volatile solute is directly proportional to its molality ($m$). The relationship is given by the formula: $$\Delta T_b = K_b \cdot m$$ where $K_b$ is the molal elevation constant (ebullioscopic constant). Molality ($m$) is defined as the moles of solute per kilogram of solvent. Expressed in terms of mass, the expanded formula is: $$\Delta T_b = K_b \cdot \frac{W_2 \times 1000}{M_2 \times W_1}$$ where $W_2$ is the mass of the solute, $M_2$ is the molar mass of the solute, and $W_1$ is the mass of the solvent in grams.
Step 1: Identify the given parameters.
Mass of solute, $W_2 = 3.5\text{ g}$ Mass of solvent, $W_1 = 100\text{ g}$ Boiling point elevation, $\Delta T_b = 0.35\text{ K}$ Molal elevation constant, $K_b = 2.5\text{ K kg mol}^{-1}$
Step 2: Rearrange the formula and solve for the molar mass $M_2$.
Rearrange the formula to solve for $M_2$: $$M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$$ Substitute the known values into the rearranged formula: $$M_2 = \frac{2.5 \times 3.5 \times 1000}{0.35 \times 100}$$ Simplify the denominator: $$M_2 = \frac{2.5 \times 3.5 \times 1000}{35}$$ Cancel out $3.5$ and $35$: $$M_2 = \frac{2.5 \times 1000}{10} = 2.5 \times 100$$ $$M_2 = 250\text{ g mol}^{-1}$$