Question

Pls solve
The molal freezing pt. Constant of water is 1.86 K kg/MOL.therefore the freezing point of 0.1 molar NaCl solution in water is expected to be:
________

Answer 1

dTf = i*Kf *m

i=2;

=2*1.86*0.1

=0.372

Freezing point =0-0.372

= - 0.372°C

Answer 2

Answer:

-0.372°c

Explanation:

∆Tf=kf×molality

=1.86×0.1

=0.186

But Nacl dissociate Na+ and cl- so total no. Of ion is 2

=0.186×2

=0.372

Therfore change in freezing point of the solution = 0-0.372

= -0.372°C