Question

A solution containing \( 8.0 \text{ g} \) of a non-volatile solute in \( 100 \text{ g} \) of diethyl ether boils at \( 36.86 ^\circ\text{C} \) whereas pure diethyl ether boils at \( 35.60 ^\circ\text{C} \). Determine the molar mass of the solute. (\( K_b \) for diethyl ether = \( 2.02 \text{ K kg mol}^{-1} \))

Hint:

Always ensure the mass of the solvent is either converted to kg, or if you use grams, remember to include the \(1000\) multiplier in the numerator for the molality formula.

Answer 1

Concept: When a non-volatile solute is added to a volatile solvent, the vapor pressure of the solvent decreases, which causes an elevation in the boiling point. This colligative property is mathematically expressed as: \[ \Delta T_b = K_b \times m \] Where:

• \( \Delta T_b \) is the elevation in boiling point (\( T_{\text{solution}} - T_{\text{solvent}} \)).

• \( K_b \) is the molal boiling point elevation constant.

• \( m \) is the molality of the solution (\( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)).
Step 1: Extracting given data and calculating elevation in boiling point.

• Mass of solute (\( W_B \)) = \( 8.0 \text{ g} \)

• Mass of solvent (\( W_A \)) = \( 100 \text{ g} = 0.1 \text{ kg} \)

• Boiling point of solution (\( T_b \)) = \( 36.86 ^\circ\text{C} \)

• Boiling point of pure solvent (\( T_b^\circ \)) = \( 35.60 ^\circ\text{C} \)

• Ebullioscopic constant (\( K_b \)) = \( 2.02 \text{ K kg mol}^{-1} \)
Elevation in boiling point \( \Delta T_b = 36.86 - 35.60 = 1.26 \text{ K} \) (or \(^\circ\text{C}\), the difference is the same).

Step 2: Setting up the formula for Molar Mass.
Molality \( m = \frac{W_B \times 1000}{M_B \times W_A \text{ (in g)}} \) Substitute this into the boiling point elevation formula: \[ \Delta T_b = \frac{K_b \times W_B \times 1000}{M_B \times W_A} \] Rearranging to solve for the molar mass of the solute (\( M_B \)): \[ M_B = \frac{K_b \times W_B \times 1000}{\Delta T_b \times W_A} \]

Step 3: Substituting values and calculating.
\[ M_B = \frac{2.02 \times 8.0 \times 1000}{1.26 \times 100} \] \[ M_B = \frac{16160}{126} \] \[ M_B = 128.25 \text{ g mol}^{-1} \] Final Answer: The molar mass of the non-volatile solute is \( 128.25 \text{ g mol}^{-1} \).