Depression in freezing point
source : ktla.com
Depression in freezing point numerical-
Question 1) How many ml. of ethylene glycol (d=1.12 gm/ml) should be added to 5 litre of water to make up an antifreeze which will protect an automobile radiator down to -10 0C, Kf =1.86 K Kg/mole.
Solution –
w=? , d of ethylene glycol= 1.12 gm/ml
W =v x d
v =5 L =5 x 1000 ml= 5000 ml, density of water =1
W = 5000 x 1= 5000 gm
Depression in freezing point ‘ΔTf ‘=freezing point of solvent – freezing point of solution
freezing point of solvent =0 0C
freezing point of solution = -10 0C
ΔTf = 0 – (-10) = 10
Kf = 1.86
molecular weight of ethylene glycol (CH2OH-CH2OH)= (12+2+16+1)2=62
ΔTf =1000 Kf. w /m .W
w = ΔTf .m.W/1000.Kf
= 10 x 62 x5000/1000 x 1.86 =1666.67 gm.
v = w (mass)/d (density)
1666.67/1.12 = 1488.0 ml
1 litre =1000 ml.
v = 1488 /1000 = 1.488 litre Ans.
Question 2) In a cold climate water gets frozen causing damage to the radiators of car . Ethylene glycol is used as an antifreeze agent .Calculate the amount of ethylene glycol to be added to 4 kg. water to prevent it from freezing at -6 0C . Kf for water =1.86 K Kg/mole
Solution )
w=?
W = 4 Kg.
W =4 x 1000 gm
W = 4000 gm
ΔTf =freezing point of solvent – freezing point of solution
freezing point of solvent =0 0C
freezing point of solution = – 6 0C
ΔTf = 0 – (- 6) = 6
Kf = 1.86
molecular weight of ethylene glycol (CH2OH-CH2OH)= (12+2+16+1)2=62
ΔTf =1000 Kf. w /m .W
w = ΔTf .m.W/1000.Kf
= 6 x 62 x4000/1000 x 1.86 = 800 gm.
w = 800 gm. Ans.
Question 3 ) The freezing point of a solution containing 50 cc of ethylene glycol in 50 gm of water is found -34 0C.Assuming ideal behaviour calculate density of ethylene glycol ?
(Kf for water =1.86 K Kg /mole)
Solution )
Volume of ethylene glycol = 50 cc
w= v x d
W = 50 gm.
ΔTf =freezing point of solvent – freezing point of solution
freezing point of solvent =0 0C
freezing point of solution = – 34 0C
ΔTf = 0 – (- 34) = 34
Kf = 1.86
molecular weight of ethylene glycol (CH2OH-CH2OH)= (12+2+16+1)2=62
ΔTf =1000 Kf. w /m .W
w = ΔTf .m.W/1000.Kf
= 34 x 62 x 50/1000 x 1.86 = 56.67 gm.
w = 56.67 gm. Ans.
w = v x d
d = w /v= 56.67 /50
d = 1.133 gm /cc Ans.
Question 4) A solution of 0.643 gm of an organic compound in 50 ml of benzene (density =0.879 gm/ml) lowered its freezing point from 5.51 0C to 5.03 0C.Calculate molecular weight of solid . Kf for benzene is 5.12.
Solution )
w= 0.643 gm. , d of benzene= 0.879 gm/ml , volume of benzene = 50 ml
W =v x d
W = 50 x 0.879 = 43.95 gm
ΔTf =freezing point of solvent – freezing point of solution
freezing point of solvent =5.51 0C
freezing point of solution = 5.03 0C
ΔTf = 5.51 – 5.03 = 0.48
Kf = 5.12
m =1000 Kf . w /ΔTf .W
m = 1000 x 5.12 x 0.643 / 0.48 x 43.95 =156.056
m = 156.06 Ans.
