Question

A solution containing 0.11 kg of Ba(NO3)2 in 0.1 kg of water boils at 100.46C calculate the degree of ionization of the salt (kb-0.5 kg.C/mol)?

Answer 1

Boiling point elevation, deltaT= i* kb*m (1), kb=0.5 kg/deg.c/mole, i= van;t Hoff factor and

boiling point elevation = boiling point of solution- boiling point of water= 100.46-100=0.46 deg.c

m= molality= moles of Ba(NO3)2/ kg of solvent

moles= mass/ molar mass of Ba(NO3)2, molar mass of Ba(NO3)2= 261.34 g/mole

moles of Ba(NO3)2= 0.11 kg *1000g/kg/261.34 =0.42 moles

molality= moles of solute/ kg of solvent = 0.42/0.1= 4.2 moles of Ba(NO3)2/ kg of water

with these values, Eq.1 when used

0.46= i*0.5 kg/mole.deg.c* 4.2 moles/kg of water

i= 0.22,the degree of ionization