Question

The boiling point of water is 100C .calculate the boiling point of an aqueous solution containing 0.6g of urea in 100g of water
Kb of water is 0
52kmol

Answer 1

from elevation in boiling,

$\boxed{\bf{\Delta T_b=K_b\times m}}$

where $\Delta T_b$ is elevation in boiling point, $K_b$ is ebullioscopic constant and m is molality of solute.

mass of urea = 0.6 g

molar mass of urea (solute) = 60g/mol

so, number of mole of urea = 0.6/60 = 0.01

mass of solvent (water) = 100g

so, molality of urea = number of mole of urea/mass of solvent in kg

= 0.01 × 1000/100 = 0.1 molal

given, ebullioscopic constant, $K_b$ = 0.52K/molal or 0.52°C/molal

then, ∆Tb = 0.52 × 0.1 = 0.052K

so, boiling point of solution, Tb = T + 0.052°C

= 100°C + 0.052 °C = 100.052°C