Question

Calculate freezing point of solution obtained by dissolving 0.1gm potassium ferrocyanide

Answer 1

your question is incomplete. A complete question is ------> calculate the freezing point of a solution containing 0.1 g of K3 [Fe(CN)6] ( M. Wt= 329) in 100 g of water if its 50% ionised.(kf =1.86).

solution : given, Mass of K3[Fe(CN)6] = 0.1g

Molecular mass= 329g

Mass of solvent ( Water) = 100g

50% ionised means, degree of

dissociation (α) = 0.5

dissociation of K3[Fe(CN)6] is ....

$K_3[Fe(CN)_6]\Leftrightarrow 3K^++[Fe(CN)_6]^{3-}$

so, Vant Hoff Factor, i = i + nα - α

here, n = 3 + 1 = 4 and α = 0.5

so, i= 1 + 4 × 0.5 - 0.5 = 1 + 1.5 = 2.5

now, ∆Tf = i Kfm

where m is molality .

m = mole of solute × 1000/mass of solvent

= (0.1/329) × 1000/100

= 1/329

now, ∆Tf = 2.5 × 1.86 × 1/329

= 0.0141°C

Answer 2

T°f = freezing point of pure solvent

Tf = freezing point of solution

i = Van't Hoff Factor

Kf = Cryoscopic constant

Wb = mass of solute

Mb = molar mass of solute

Wa = mass of solvent

If all the masses are in grams

T°f - Tf = (i × Kf × Wb × 1000) / (Mb × Wa)

T°f = 0°C

Tf =?

i = 4 (potassium ferrocyanate dissociates into 3K+ and (FeCN6)- )

Kf = 1.86 KKg/mol

Wb = 0.1 g

Mb = 329 g/mol

Wa = 100 g

Tf = - 2.26 × 10 ^ - 2 ° C.