Question

23. An aqueous solution of KCl freezes at - 0.186°C
(К = 1.86, К, 0.512). What is the elevation in
boiling point? (Assume complete dissociation)
(1) 0.512
(2) 0.0512
(3) 1.024
(4) 0.05 х 10-1​

Answer 1

Answer:

As we know that,

Depression in freezing point,∆Tf=Kf×m

∆Tf=0.186°C

Kf=1.86

∆Tf=Kf×m

0.186°=1.86×m

m=1.86/0.186

m=0.1

so, Molarity,m=0.1

Elevation in boiling point,∆Tb=Kb×m

∆Tb=0.512×0.1

∆Tb=0.0512

So, The elevation in boiling point,∆Tb=0.0512

Hope it helps you......