Question

PART II (8 points cach) Please show all your work. 21. What is the boiling point in "C) of a solution prepared by dissolving 11.5 g of Ca(NO3)2 (formula weight - 164 g/mol) in 150 g of water? (Kfor water is 0.52°C/m)

Answer 1

Lets calculate molality first

Molar mass of Ca(NO3)2 = 164 g/mol

mass(Ca(NO3)2)= 11.5 g

use:

number of mol of Ca(NO3)2,

n = mass of Ca(NO3)2/molar mass of Ca(NO3)2

=(11.5 g)/(164 g/mol)

= 7.008*10^-2 mol

m(solvent)= 150 g

= 0.15 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(7.008*10^-2 mol)/(0.15 Kg)

= 0.4672 molal

i for Ca(NO3)2 is 3 as it dissociates Ito 1 Ca2+ and 2 NO3-

lets now calculate ΔTb

ΔTb = i*Kb*m

= 3.0*0.52*0.4672

= 0.729 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.729

= 100.729 oC

Answer: 100.73 oC