Calculate the van’t hoff factor (i) for mx2 at this concentration.

A 4.6 m aqueous solution of an ionic compound with the
formula MX2 has a boiling point of 101.8 ?C.

-Calculate the van't Hoff factor (i) for MX2 at this


Elivation in boiling point , Delta T_{b}= iK_{b}m


Delta T_{b} =
elivation is boiling point of water

        = Boiling point of
solution - Boiling point of water

        = 101.8 - 100.0

        = 1.8

i = van't Hoff factor = ?

Kb = elivation in boiling point of water = 0.512

m = molality of solution = 4.6 m

Plug the values we get i= frac{Delta T_{b}}{mK_{b}}=frac{1.8}{0.512times 4.6}=0.76

Therefore van't Hoff factor for MX2 is 0.76

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts