KEAM 2012 Physics and Chemistry Paper

∵ Freezing point of the aqueous solution
=−0.186∘C
∴ Depression in freezing point,
∆Tf=0∘C−(−0.186∘C)
=+0.186∘C=+0.186~K
We know that,
∆Tf=Kf⋅m.....(i)
and elevation in boiling point,
∆Tb=Kb⋅m.....(ii)
Eq. (i) Eq. (ii) gives
‌

∆Tf

∆Tb

=‌

Kf

Kb

⇒‌‌‌

0.186K

∆Tb

=‌

1.86Kkg‌mol−1

0.52Kkg‌mol−1

∆Tb=‌

0.186×0.52

1.86

=0.052K