Question

An aqueous solution 1L salt solution contains 100g of NaCI. Calcaulte the melting point and boiling point of the solution at 1 atm

Answer 1

No. of moles = Nacl

n= 100g / 58.44g/mol = 1.71116 moles

volume of solution = 1L

Molality of solution = n/volume = 1.71116 /1 = 1.71116m

(A) At pressure 1 atm the value of Kb (elevation constant) for water = 0.51^o C/m

therefore elevation in boiling point ΔTb = m x kb

ΔTb = 1.71116 m x 0.51^oC/m = 0.8726 ^oC

Boiling point of water = 100^oC = T

final Boiling point = T + ΔTb = 100 + 0.8726 = 100.8726^oC

(B) now at 1 atm pressure the freezing point depression constant Kf = 1.86^oC/m

ΔTf = m x kf

ΔTf = 1.71116 m x 1.86^oC/m = 3.18276^oC

Freezing point of water = 0^oC = Tf

Final freezing point of solution = Tf - ΔTf = 0- 3.18276^oC = -3.18276^oC