Depression in freezing point source : surfguppy.com

## Depression in freezing point –

### K = 1.86

m of glycerine (CH2 OH-CHOH-CH2OH) =92

### Depression in freezing point  ΔTf  = 0 – (-10)  = 10

Volume of solvent ‘V’ = 40 dm3 =40 x 1000 =40000 cc

weight of solvent ‘W’ = volume x density

weight of solvent ‘W’= 40000 x 1=40000 gm

#### w = ΔTf .W.m/1000 Kf

w = 10 x 40000 x 92 /1000 x 1.86

w= 19784.94 gm

w = 19.78 Kg.  Ans.

### Solution )

Kf = 1.86

m of ethylene glycol  (CH2OH-CH2OH) = 62

w =50 gm.

#### W =1000 Kf. w /ΔTf .m

W= 1000 x 1.86 x 50 / 9.3 x 62

W( weight of solvent ) = 161.29 gm.

Total solvent taken = 200 gm.

weight of ice separated =200 – 161.29 = 38.71 gm. Ans.

### Solution )

ΔTf =Kf x molality

ΔTb =Kb x molality

ΔTb / ΔTf  =Kb x molality / Kf x molality

ΔTb / ΔTf   = Kb /Kf

ΔTb / ΔTf  =( R Tb2 /1000 x lv ) / ( R Tf2 /1000 x lf )

#### ΔTb / ΔTf   = Tb2 x lf / lv x Tf2

lv = 540 calorie /gm

lf = 80 calorie /gm

Boiling point of solution  =100.1 0C =100.1 + 273 =373.1 K

Boiling point of solvent  =100 0 C

Tf = freezing point of solvent (water)=0 0 C  =0 + 273 = 273 K

Tb =Boiling point of solvent (water) = 100 0C = 100 + 273 = 373 K

#### ΔTb / ΔTf   = Tb2 x lf / lv x Tf2

(373.1 – 373 ) / ΔTf  = 373 x 373 x 80 /540 x 273 x 273

### Solution )

w = 0.48 gm ,  ΔTf =1.8 0C , Kf = 50 ,W =10.6 gm , m = ?

m =100 Kf. w /ΔTf .W

m = 100 x 50 x 0.48 / 1.8 x 10.6

m =125.79 Ans.