Depression in freezing point numerical part 1

Depression in freezing point

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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, K_f =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 ‘ΔT_f ‘=freezing point of solvent – freezing point of solution

freezing point of solvent =0 ⁰C

freezing point of solution = -10 ⁰C

ΔT_f = 0 – (-10) = 10

K_f = 1.86

molecular weight of ethylene glycol (CH₂OH-CH₂OH)= (12+2+16+1)2=62

ΔT_f =1000 K_f. w /m .W

w = ΔT_f .m.W/1000.K_f

= 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 . K_f for water =1.86 K Kg/mole

Solution )

w=?

W = 4 Kg.

W =4 x 1000 gm

W =  4000 gm

ΔT_f =freezing point of solvent – freezing point of solution

freezing point of solvent =0 ⁰C

freezing point of solution = – 6 ⁰C

ΔT_f = 0 – (- 6) = 6

K_f = 1.86

molecular weight of ethylene glycol (CH₂OH-CH₂OH)= (12+2+16+1)2=62

ΔT_f =1000 K_f. w /m .W

w = ΔT_f .m.W/1000.K_f

= 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 ?

(K_f for water =1.86 K Kg /mole)

Solution )

Volume of ethylene glycol = 50 cc

w= v x d

W = 50 gm.

ΔT_f =freezing point of solvent – freezing point of solution

freezing point of solvent =0 ⁰C

freezing point of solution = – 34 ⁰C

ΔT_f = 0 – (- 34) = 34

K_f = 1.86

molecular weight of ethylene glycol (CH₂OH-CH₂OH)= (12+2+16+1)2=62

ΔT_f =1000 K_f. w /m .W

w = ΔT_f .m.W/1000.K_f

= 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 . K_f 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

ΔT_f =freezing point of solvent – freezing point of solution

freezing point of solvent =5.51 ⁰C

freezing point of solution =  5.03 ⁰C

ΔT_f = 5.51 – 5.03 = 0.48

K_f =  5.12

m =1000 K_f . w /ΔT_f .W

m = 1000 x 5.12 x 0.643 / 0.48 x 43.95 =156.056

m = 156.06 Ans.