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Depression in freezing point numerical part 3

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 0C . Determine molecular mass of solute?  (Kf = 1.86 )

Solution )

w = 3.0 gm ,  ΔTf =1.45 0 C , Kf = 1.86 ,W =22 gm , m = ?

m =1000 K. w /ΔTf .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 Kf for nitrobenzene ?

Solution )

ΔTf = 2.0 , molality = 0.25 

KfΔTf / molality

Kf = 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’ ,

w1= 4 gm,W1 = 100 gm ,

ΔTf1 = 0.1 

For ‘B’

w2= 4 gm,W2 = 100 gm ,

ΔTf2 = 0.2 

m = 1000 Kf. w/ ΔTf.W

Because values of Kf , W and w  are same . Therefore ,

mA /mB =   ΔTf2 / ΔTf1

= 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 [ CH3COCH3] = 12+ 3+12+ 16 + 12 + 3 = 58

Depression in freezing point (ΔTf ) = freezing point of solvent – freezing point of solution

freezing point of solvent = 278.4 K

freezing point of solution =  277.12 K

ΔTf =  278.4 – 277.12  = 1.28 

m =1000 K. w /ΔTf .W

Kf = ΔTf .W.m / 1000 w

 Kf = 1.28 x 100 x 58 /1000 x 1.4

Kf = 5.30

For ‘A’,

w = 2.8 gm ,

W = 100 gm , m of ‘A’ =?

Depression in freezing point (ΔTf  ) = freezing point of solvent – freezing point of solution

freezing point of solvent = 278.4 K

freezing point of solution =  277.76 K

ΔTf =  278.4 – 277.76  = o.64

m =1000 K. w /ΔTf .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 1.0150C . Kf for C10H8 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 (ΔTf  ) =  1.015 

m =1000 K. w /ΔTf .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.

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