Question

I have to find the boiling point of 0.17 m NaCl solution using the mass percent and the van't Hoff factors given to me. What exactly do I do to answer this?

Answer 1

My mass percent of NaCl:
1)0.0070
2)0.500
3)1.00
4)2.00

Answer 2

van't Hoff factors:
1)1.94
2)1.87
3)1.84
4)1.83

Answer 3

This is the normal formula used
ΔT = i Kb m
where ΔT is the change in temperature
i is the van't Hoff factor
Kb is the boiling point constant
and m is the moles/kg of solute

Answer 4

So, where does my mass percent play a role in this equation?

Answer 5

ΔTb = i Kbm
Kb = 0.512
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.17m
m(Cl-) = 0.17m
m(NaCl(aq)) = 0.17 + 0.17 = 0.34m
ΔTb = i 0.34 x 0.512 = i 0.17408 C

For the cases you have given,
Mass % = Mass of Solute/Mass of Solvent
Mass of solvent= Mass % x Mass of Solution

Assume, 1 ltr H2O i.e. 1000gm,
So, Mass of NaCl =1000 x Mass %

Then,
Molality(m) = moles solute particles per kg of solution
= 1000 x Mass % / Gram Molecular Mass of Nacl
=1000 x Mass %/58

Put values & have answers............ :D