Question

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In other words, what is the boiling point elevation (increase)?

Answer 1

Answer: The elevation in boiling point is 1.024°C.

Explanation:

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point  = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.