Depression in freezing point-

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### Depression in freezing point- Numerical

### Question 1) The depression of freezing point of a solution containing 3 gm of a solute in 22 gm of water is 1.45 ^{0}C . Determine molecular mass of solute? (K_{f} = 1.86 )

### Solution )

*w = 3.0 gm , ΔT _{f} =1.45 ^{0} C , K_{f} = 1.86 ,W =22 gm , m = ?*

*m =1000 K _{f }. w /ΔT_{f} .W*

*m = 1000 x 1.86 x 3.0 / 1.45 x 22*

*m =174.9 = 175 Ans.*

### Question 2) The normal freezing point of nitrobenzene is 278.82 K . A 0.25 molal solution of a certain solute in nitrobenzene causes a freezing point depression of 2.0 . Calculate the value of K_{f} for nitrobenzene ?

### Solution )

*ΔT _{f} = 2.0 , molality = 0.25 *

K_{f} = *ΔT _{f} / molality*

K_{f} = 2.0 /0.25 = 8.0 K Kg /mole Ans.

### Question 3) 4.0.gm of a substance ‘A’ dissolved in 100 gm of water, depressed the freezing point by 0.1 ^{0} C . While 4.0 gm of another substance ‘B’ depressed the freezing point by 0.2 ^{0} C . Which of the two substances has higher molecular weight ?

### Solution )

### For ‘A’ ,

*w _{1}= 4 gm,W_{1} = 100 gm ,*

*ΔT*_{f1} = 0.1

_{f1}= 0.1

*For ‘B’*

*w _{2}= 4 gm,W_{2} = 100 gm ,*

*ΔT*_{f2} = 0.2

_{f2}= 0.2

*m = 1000 K _{f}. w/ ΔT_{f}.W*

*Because values of K _{f} , W and w are same . Therefore ,*

*m _{A} /m_{B} = ΔT_{f2} / ΔT_{f1}*

*= 0.2 /0.1*

*= 2 /1*

*It means molecular weight of ‘A’ is higher than ‘B’. The molecular weight of ‘A’ is two times the molecular weight of ‘B’. Ans.*

### Question 4) 1.4 gm of acetone dissolved in 100 gm of benzene gave a solution which freezes at 277.12 K. Pure benzene freezes at 278.4 K . 2.8 gm of solid (A) dissolved in 100 gm of benzene gave a solution which froze at 277.76 K . Calculate the molecular weight of ‘A’ ?

### Solution )

### For Acetone ,

*w = 1.4 gm , W = 100 gm ,m of acetone [ CH _{3}COCH_{3}] = 12+ 3+12+ 16 + 12 + 3 = 58*

*Depression in freezing point (ΔT*_{f} ) = freezing point of solvent – freezing point of solution

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

*freezing point of solvent = 278.4 K*

*freezing point of solution = 277.12 K*

*ΔT*_{f} = 278.4 – 277.12 = 1.28

_{f}= 278.4 – 277.12 = 1.28

*m =1000 K*_{f }. w /ΔT_{f} .W

_{f }. w /ΔT

_{f}.W

*K _{f} = ΔT_{f} .W.m / 1000 w*

* K _{f} = 1.28 x 100 x 58 /1000 x 1.4*

*K*_{f} = 5.30

_{f}= 5.30

*For ‘A’,*

*w = 2.8 gm ,*

*W = 100 gm , m of ‘A’ =?*

*Depression in freezing point (ΔT*_{f} ) = freezing point of solvent – freezing point of solution

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

*freezing point of solvent = 278.4 K*

*freezing point of solution = 277.76 K*

*ΔT*_{f} = 278.4 – 277.76 = o.64

_{f}= 278.4 – 277.76 = o.64

*m =1000 K*_{f }. w /ΔT_{f} .W

_{f }. w /ΔT

_{f}.W

*m = 1000 x 5.30 x 2.8 / 0.64 x 100*

*m =231.88 Ans.*

### Question 5 ) 2.71 gm of sulphur in 72 gm naphthalene shows a freezing point lowered by 1.015^{0}C . K_{f} for C_{10}H_{8} is 6.9 K Kg/mole. How many atoms of sulphur are present in a molecule of sulphur ?

### Solution )

*w = 2.71 gm ,*

*W = 72 gm , molecular weight of sulphur ‘m’ = ?*

*Depression in freezing point (ΔT*_{f} ) = 1.015

_{f}) = 1.015

*m =1000 K*_{f }. w /ΔT_{f} .W

_{f }. w /ΔT

_{f}.W

*m = 1000 x 6.9 x 2.71 / 1.015 x 72*

*m =255.8 *

*molecular weight of sulphur = 255.8*

*Atomic weight of sulphur = 32*

*Number of atoms = molecular weight of sulphur / Atomic weight of sulphur*

*Number of atoms = 255.8 /32 = 7.9 = 8 atoms Ans.*