## Depression in freezing point source : Slideplayer.com

## Depression in freezing point-Numerical

### Solution )

Latent heat of fusion ‘L’ = 80 calorie

freezing point of water ‘T’ = 0 + 273 = 273 K

K molar (KM) = 0.02 T2 / L

K molar (KM)  = 0.02 x 273 x 273 /80

## K molar (KM) = 18.63 K .(per 100 gm )/mole   Ans.

K molal (Km) = 0.002 T2 / L

K molal (Km)  = 0.02 x 273 x 273 /80

## K molal (Km) = 1.863 K Kg /mole  Ans.

### Solution )

w urea = 5 gm , w glucose = 10 gm

weight of solution = 100 gm

weight of solvent = weight of solution – weight of solute

weight of solvent = 100 – ( 5 + 10 ) = 85 gm

m of urea ( NH2CONH2)  = 14 +2 + 12 +16 + 14 +2 = 60

m of glucose ( C6H12O6 ) = 12 x 6 + 1 x 12 + 16 x 6= 180

ΔTf = ΔTf (urea) + ΔTf(glucose)

ΔT= [1000 K. w(urea)/ m (urea).W ]  + [1000 K. w(glucose)/ m(glucose).W ]

ΔTf  = [1000 x 1.86 x 5 /85 x 60  ]  + [ 1000 x 1.86 x 10 / 85 x 180]

ΔTf  = 1.824 + 1.216

ΔTf  = 3.04

## freezing point of solution = – 3.040C  Ans.

### Solution )

w =0.2 gm , W = 20 gm , ΔTf = 0.45 , Kf =5.12 K Kg/mole

m  = 1000 K. w /ΔTf .W

m = 1000 x 5.12 x 0.2 / 0.45 x 20

m(observed) = 113.78

2CH3COOH                 ⇌                     (CH3COOH)2

moles before association            1                                                             0

moles after association             ( 1 – ∝)                                                    ∝ /2

∝ = degree of association

m (normal) / m(observed) = Total  moles  after association / Total moles before association

m(normal ) of CH3COOH = 12 + 3 + 12 + 16 + 16 + 1 = 60

60 / 113.78 =[ ( 1 – ∝) + ∝ /2] / 1

## ∝ = 0.945  Ans.

% of ∝ =0.945 x 100

## % of ∝ = 94.5 % Ans.

### (i) For mercuric nitrate Hg (NO3 )2 ,

w = 3.24 gm , W = 1 Kg = 1000 gm , freezing point of solution = – 0.05580C , Kf = 1.86 K Kg / mole

freezing point of solvent (water) = 00C

#### m = 1000 x 1.86 x 3.24 / 0.0558 x 1000

m(observed)of Hg (NO3)2= 108

m (normal) of Hg (NO3)2 = 200 + (14 + 16 x 3)2 =324

Hg (NO3)2         ⇌              Hg++    + 2 NO3

moles before dissociation         1                                            0                  0

moles after dissociation         (  1 – ∝)                                   ∝                  2∝

Total moles after dissociation = 1- ∝ + ∝ + 2 ∝ =  1 + 2 ∝

Total moles before dissociation = 1 + 0 + 0 = 1

m (normal) / m(observed) = Total  moles  after dissociation / Total moles before dissociation

324 / 108 = (1 + 2 ∝)/ 1

108 + 216 ∝ = 324

216 ∝ = 324 -108

216 ∝ = 216

∝ (degree of dissociation ) = 1

% of ∝ =1 x 100 = 100 %

## Hg (NO3)2 dissociates 100 %.

### (ii)  For mercuric chloride  Hg Cl2 ,

w = 10.84 gm , W = 1 Kg = 1000 gm , freezing point of solution = – 0.07440C, Kf = 1.86 K Kg / mole

freezing point of solvent (water) = 00C

#### m = 1000 x 1.86 x 10.84 / 0.0744 x 1000

m(observed) of Hg Cl2= 271

m (normal) of Hg Cl2 = 200 + (35.5 x 2) = 271