Question

Calculate the molal elevation constant of water, it being given that 0.1 molal aqueous solution of a substance boils at 100.052$^\circ$C.

Hint:

The value 0.52 K kg/mol is a standard constant for water. If you get a calculation result far from this, check your math!

Answer 1

Given:
Molality of solution, $m = 0.1$ molal.
Boiling point of solution, $T_b = 100.052^\circ$C.
Boiling point of pure water, $T_b^0 = 100^\circ$C.
Calculate elevation in boiling point ($\Delta T_b$):
$\Delta T_b = T_b - T_b^0 = 100.052 - 100 = 0.052^\circ$C (or 0.052 K).
Formula for elevation in boiling point:
$\Delta T_b = K_b \times m$
Rearrange to solve for $K_b$:
$K_b = \frac{\Delta T_b}{m}$
$K_b = \frac{0.052}{0.1}$
$K_b = 0.52$ K kg mol$^{-1}$.