‌‌...‌ (i) ‌ ‌∆Tb=‌ elevation in boiling point ‌ ‌=373.202−373.15=0.052K Kf and Kb are given as, ‌Kb=0.052Kkg∕mol ‌Kf=1.86Kkg∕mol Substitute all value in Eq. (i) ‌1.86=0.052×‌
∆Tf
0.52
‌∆Tf=‌
1.86
10
‌ or ‌0.186‌‌...‌ (ii) ‌ ‌∆Tf=‌ depression in freezing point ‌ ‌∆Tf=0.186=273.15K−x ‌x=272.964K