Question

A solution is made by dissolving 3.8 moles of sodium chloride in 185 grams of water is the molal boiling point constant for water (kb) is 0.51cm what would be the boiling point of this solution

Answer 1

n(NaCl) = 3.8 mol
m(H2O) = 185 g = 0.185 kg
Kb=0.51 ⁰C*m⁻¹
T(water) = 100⁰C

M- molality of the solution = n ( mol of the solute)/m (solvent, kg) =
= 3.8 mol/0.185kg = 3.8/0.185 m

ΔT = Kb*M =0.51 ⁰C*m⁻¹ * 3.8/0.185 m = 10. ⁰C

T(boiling of the solution) = 100+10=110⁰C