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 dm^{3} water used in car radiator, so that it does not freeze. *Δ H *_{fusion} = 6.01 KJ/mole

_{fusion}

### Solution )

*K = 1.86*

*m of glycerine (CH _{2} OH-CHOH-CH_{2}OH) =92*

*ΔT*_{f} =freezing point of solvent – freezing point of solution

_{f}=freezing point of solvent – freezing point of solution

*freezing point of solvent = 0 *^{0}C

^{0}C

*freezing point of solution = – 10 *^{0}C

^{0}C

### Depression in freezing point *ΔT*_{f } = 0 – (-10) = 10

_{f }= 0 – (-10) = 10

*Volume of solvent ‘V’ = 40 dm ^{3} =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}

_{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 ^{0}C. (K_{f} for water =1.86 K Kg /mole )

### Solution )

*K _{f} = 1.86*

*m of ethylene glycol (CH _{2}OH-CH_{2}OH) = 62*

*ΔT*_{f} =freezing point of solvent – freezing point of solution

_{f}=freezing point of solvent – freezing point of solution

*freezing point of solvent = 0 *^{0}C

^{0}C

*freezing point of solution = – 9.3 *^{0}C

^{0}C

*ΔT*_{f} = 0 – (-9.3) = 9.3

_{f}= 0 – (-9.3) = 9.3

*w =50 gm.*

*W =1000 Kf. w /ΔT*_{f} .m

_{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*

_{b}^{2}/1000 x l_{v}) / ( R T_{f}^{2}/1000 x l_{f})* ΔT*_{b} / ΔT_{f} = T_{b}^{2} x l_{f} / l_{v} x T_{f}^{2}

_{b}/ ΔT

_{f}= T

_{b}

^{2}x l

_{f}/ l

_{v}x T

_{f}

^{2}

*l _{v} = 540 calorie /gm*

*l _{f} = 80 calorie /gm*

*Boiling point of solution =100.1 ^{0}C =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 ^{0}C = 100 + 273 = 373 K*

*ΔT*_{b} / ΔT_{f} = T_{b}^{2} x l_{f} / l_{v} x T_{f}^{2}

_{b}/ ΔT

_{f}= T

_{b}

^{2}x l

_{f}/ l

_{v}x T

_{f}

^{2}

*(373.1 – 373 ) / ΔT _{f} = 373 x 373 x 80 /540 x 273 x 273*

*depression in freezing point (ΔT*_{f} )= 0.36

_{f})= 0.36

*ΔT*_{f} =freezing point of solvent – freezing point of solution

_{f}=freezing point of solvent – freezing point of solution

* freezing point of solution = freezing point of solvent – ΔT*_{f}

_{f}

*freezing point of solution= 273 – 0.36 =272.64 K or – 0.36 *^{0}C Ans.

^{0}C 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 ^{0}C , 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.*