Assuming 100% ionisation, the increasing order of the freezing point of the solution will be

Options

A. `0.10 mol kg^(-1) Ba_3(PO_4)_2 lt 0.10 mol kg^(-1) Na_2SO_4 lt 0.10 mol kg^(-1) KCI`
B. `0.10 mol kg^(-1) KCI lt 0.10 mol kg^(-1) Na_2SO_4 0.10 mol kg^(-1) Ba_3(PO_4)_2`
C. `0.10 mol kg^(-1) Na_2 SO_4 lt 0.10 mol kg^(-1) Ba_3(PO_4)_2 lt 0.10 mol kg^(-1) KCI`
D. `0.10 mol kg^(-1) KCI lt 0.10 mol kg^(-1) Ba_3(PO_4)_2 lt 0.10 mol kg^(-1) Na_2SO_4`

Answer

A

Explanation

`Ba_3(PO_4)_2 i = 5`
`Na_2SO_4, i implies 3 KCI, i = 2`
For the same concentration, `DeltaT_f prop i`
`:.` Freezing point will have the order :
`Ba_3(PO_4)_2 lt Na_2 SO_4, lt KCI`.