Elevation of boiling point -Numerical part 1

Elevation of boiling point-

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Elevation of boiling point- Numerical

Question 1)  The boiling point of pure water is 100⁰C. Calculate the boiling point of an aqueous solution containing 0.6 gm of urea in 100 gm of water.(K_b of water =0.52 K/Molal)

Solution )

K_b = o.52 , ΔT_b = ? 

m  of urea (NH₂CONH₂) = 14 + 2 +12 + 16 + 14 +2 =60

w = 0.6 gm.

 W = 100 gm 

ΔT_b =1000 K_b. w /W.m

ΔT_b = 1000 x 0.52 x  0.6 /  100 x  60

ΔT_b = 0.052⁰C

Elevation in boiling point (ΔT_b ) = boiling point of solution – boiling point of  pure solvent

boiling point of solvent = 100⁰C

boiling point of solution =  100 + 0.052

boiling point of solution =  100 .052⁰C  Ans.

Question 2)   A solution containing 12.5 gm of a non electrolyte substance in 175 gm of water gave boiling point elevation of 0.70 K. Calculate the molar mass of the substance ? .(K_b of water =0.52 K Kg/Mole)

Solution )

K_b = o.52 , ΔT_b = 0.70 K

m = ?

w = 12.5 gm.

 W = 175 gm 

m =1000 K_b. w /W.ΔTb

m  = 1000 x 0.52 x  12.5 /  175 x 0.70

m =  53.06 Ans

Question 3)  On dissolving 3.24 gm of sulphur in 40 gm of benzene , boiling point of solution was higher than that of benzene by 0.81 K .What is the molecular formula of sulphur ? (Atomic weight of sulphur = 32)

(K_b for benzene = 2.53 K Kg/Mole)

Solution )

K_b = 2.53 , ΔT_b = 0.81 K

m = ?

w = 3.24 gm.

 W = 40 gm 

m =1000 K_b. w /W.ΔTb

m  = 1000 x 2.53 x  3.24 /  40 x 0.81

m = 253 Ans.

Atomic weight of sulphur = 32

molecular formula of sulphur = Sx

molecular formula of sulphur = 32 x

Hence ,

32 x = 253

   x= 253 /32= 7.91 =8.0

molecular formula of sulphur =Sx =  S₈ Ans.

 

Question 4 ) An aqueous solution of glucose containing 12 gm dissolved in 100 gm of water is found to boil at
100.34⁰C.While boiling point of pure water is 100⁰C. Calculate molal elevation constant for water ?

Solution )

 Elevation in boiling point (ΔT_b ) = boiling point of solution – boiling point of  pure solvent

boiling point of solvent = 100⁰C

boiling point of solution =  100.34⁰C

Elevation in boiling point (ΔT_b ) = 100.34 -100=0.34⁰C

 m of glucose (C₆H₁₂O₆)= 12 x6 +12+ 16 x6 = 180

w = 12 gm.

 W = 100 gm 

K_b =ΔT_b.W.m / 1000 .w

Kb  = 0.34 x 100 x  180 /  1000 x 12

K_b = 0.51 Ans.