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?  (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’ ,

w₁= 4 gm,W₁ = 100 gm ,

ΔT_f1 = 0.1 

For ‘B’

w₂= 4 gm,W₂ = 100 gm ,

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

Depression in freezing point (ΔT_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 

m =1000 K_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

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

freezing point of solvent = 278.4 K

freezing point of solution =  277.76 K

ΔT_f =  278.4 – 277.76  = o.64

m =1000 K_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 1.015⁰C . K_f 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 (ΔT_f  ) =  1.015 

m =1000 K_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.