Question

An aquous solution freezes at -0.186 then elevation of boiling point is

Answer 1

Answer:

Both freezing point depression and boiling point elevation are colligative properties, so they follow essentially the same equations which are of the form

ΔT = kf m i

where

kf is the freezing point depression constant (or boiling point elevation constant)

m is the molality of the solute

i is the van't Hoff factor (the number of particles per mole of the solute)

Since this is one solution, the values for m and i are the same. So the b.p. elevation will be

(-5.5°C)×(b.p. elevation constant)/(f.p. depression constant)

Add that to the normal b.p. of water, and there you go.