Question

10. When 4.94 g of K3 [Fe(CN)6] (M= 329 g.mol" is dissolved in 100 g of water, the freezing

point is found to be - 1.1 'C. How many ions are present for each formula unit of K [Fe(CN).]
dissolved? The Kf for water is 1.86 C .kg . mol",

Answer 1

It is 4 ions.

Step-by-step explanation:

We know that depression in freezing point of a solution is directly proportional to its molality i.e,

∆
`T_f` ∝ m

∆
`T_f` =
`K_f` x m

∆
`T_f` = (
`K_f` x w x 1000 x i)/(M x W)

where ∆
`T_f` = depression in freezing point

`K_f` = molal depression freezing point constant or cryscopic constant

m = molality of solution,

w = amount of solute ,

M = Molar mass of solute,

W= amount of solvent ,

i = Vant Hoff factor

Given :

m = 4.44g ,

M = 329 ,

W = 100g,

∆
`T_f` = 1.1 ,

`K_f` = 1.86 C Kg mol

Substituting above values in formula,

1.1 = (i x 4.94 x 1000x 1.86)/ (329 x 100)

therefore, i = 3.9386 approx

= 4 ions are present.