Question

PART 1- 8 points each) 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? (K) for 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)

= 0.0701 mol

m(solvent)= 150 g

= 0.15 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.0701 mol)/(0.15 Kg)

= 0.467 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*0.512*0.467

= 0.717 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.717

= 100.717 oC

Answer: 100.717 oC