AP EAPCET 19-Aug-2021 Shift 2 Paper

Formula used,
Freezing point of solution, ΔT1=Ki× molality of solution and ΔTb=Kb× molatity of solution
Now, ratio of Kf and Kh is

ΔTf

ΔTb

=

Kf

Kb

...(i)
Given that, ΔTb=T,−T1=100.20−100=0.20°C
Kf of water =1.86K‌kg‌mol−1
Kb of water =0.512K‌kg‌mol−1
Put value in Eq. (i),
∴

ΔTf

0.20

=

1.86

0.512

⇒ΔTf=0.726
Now, ΔTf⇒T1−T2
0.726°C=0°C−T2
⇒T2=−0.726°C
Hence, freezing point of solution is −0.726°C