Question

What is elevation of boiling point? A liquid has boiling point 353.23 K. The boiling point of solution becomes 354.11 K after dissolving 1.8 g non-volatile solute of molar mass 58 g mol$^{-1$ to 90 g liquid. Calculate boiling point elevation constant for the liquid.}

Hint:

Always convert the solvent mass to kilograms when calculating molality. Remember that elevation of boiling point depends only on the number of solute particles, not their nature.

Answer 1

The elevation in boiling point is given by the formula:
\[ \Delta T_b = K_b \times m \]
Where:
- \( \Delta T_b \) = elevation in boiling point
- \( K_b \) = molal elevation constant
- \( m \) = molality of the solution

Step 1: Calculate elevation in boiling point.
\[ \Delta T_b = 354.11 - 353.23 = 0.88 \, K \]

Step 2: Calculate number of moles of solute.
Mass of solute \( = 1.8 \, g \)
Molar mass of solute \( = 58 \, g \, mol^{-1} \)
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.8}{58} = 0.0310 \, mol \]

Step 3: Calculate mass of solvent in kilograms.
Mass of solvent \( = 90 \, g = 0.090 \, kg \)

Step 4: Calculate molality of solution.
\[ m = \frac{0.0310}{0.090} = 0.344 \, mol \, kg^{-1} \]

Step 5: Calculate boiling point elevation constant.
\[ K_b = \frac{\Delta T_b}{m} = \frac{0.88}{0.344} = 2.56 \, K \, kg \, mol^{-1} \]
\[ \boxed{K_b = 2.56 \, K \, kg \, mol^{-1}} \]
Thus, the boiling point elevation constant for the liquid is \( 2.56 \, K \, kg \, mol^{-1} \).