Depression in freezing point numerical part 2

Depression in freezing point

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Depression in freezing point –

Question 1)  It has been found that minimum temperature recorded in a hill station is -10 ⁰ C. Calculate the amount of glycerine to be added to 40 dm³ water used in car radiator, so that it does not freeze. Δ H _fusion = 6.01 KJ/mole

Solution )

K = 1.86

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

ΔT_f =freezing point of solvent – freezing point of solution

freezing point of solvent = 0 ⁰C

freezing point of solution =  – 10 ⁰C

Depression in freezing point  ΔT_f  = 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 = ΔT_f .W.m/1000 K_f

w = 10 x 40000 x 92 /1000 x 1.86

w= 19784.94 gm

w = 19.78 Kg.  Ans.

Question 2)  Calculate the amount of ice that will separate out on cooling a solution containing 50 gm of ethylene glycol in 200 gm of water to -9.3 ⁰C. (K_f for water =1.86 K Kg /mole )

Solution )

K_f = 1.86

m of ethylene glycol  (CH₂OH-CH₂OH) = 62

ΔT_f =freezing point of solvent – freezing point of solution

freezing point of solvent = 0 ⁰C

freezing point of solution =  – 9.3 ⁰C

ΔT_f = 0 – (-9.3)  = 9.3

w =50 gm.

W =1000 Kf. w /ΔT_f .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.

Question 3) If boiling point of an aqueous solution is 100.1 ⁰ C. What is its freezing point ? Given latent heat of fusion  and vaporisation of water are 80 calorie/gm and 540 calorie /gm respectively.

Solution )

ΔT_f =K_f x molality

ΔT_b =K_b x molality

ΔT_b / ΔT_f  =K_b x molality / K_f x molality 

ΔT_b / ΔT_f   = K_b /K_f               

ΔT_b / ΔT_f  =( R T_b² /1000 x l_v ) / ( R T_f² /1000 x l_f )

  ΔT_b / ΔT_f   = T_b² x l_f / l_v x T_f²

l_v = 540 calorie /gm

l_f = 80 calorie /gm

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

Boiling point of solvent  =100 ⁰ C

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

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

ΔT_b / ΔT_f   = T_b² x l_f / l_v x T_f²

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

depression in freezing point (ΔT_f  )= 0.36 

ΔT_f =freezing point of solvent – freezing point of solution

 freezing point of solution = freezing point of solvent –  ΔT_f

freezing point of solution= 273 – 0.36 =272.64 K or – 0.36 ⁰C Ans.

Question 4)   0.48 gm of an organic compound dissolved in 10.6 gm of benzene lowered the freezing point by 1.8 ⁰ C . find the molecular weight of compound. ( molecular depression constant for benzene = 50 )

Solution )

w = 0.48 gm ,  ΔT_f =1.8 ⁰C , K_f = 50 ,W =10.6 gm , m = ?

m =100 K_f. w /ΔT_f .W

m = 100 x 50 x 0.48 / 1.8 x 10.6

m =125.79 Ans.