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Elevation of boiling point -Numerical part 1

The addition of a solute to a liquid solvent will usually raise the boiling point of the solvent. Adding salt to boil water when cooking.

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.(Kb of water =0.52 K/Molal)

Solution )

Kb = o.52 , ΔTb = ? 

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

w = 0.6 gm.

 W = 100 gm 

ΔTb =1000 Kb. w /W.m

ΔTb = 1000 x 0.52 x  0.6 /  100 x  60

ΔTb = 0.0520C

Elevation in boiling point (ΔTb ) = 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 ? .(Kb of water =0.52 K Kg/Mole)

Solution )

Kb = o.52 , ΔTb = 0.70 K

m = ?

w = 12.5 gm.

 W = 175 gm 

=1000 Kb. w /W.ΔTb

 = 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)

(Kb for benzene = 2.53 K Kg/Mole)

Solution )

Kb = 2.53 , ΔTb = 0.81 K

m = ?

w = 3.24 gm.

 W = 40 gm 

=1000 Kb. w /W.ΔTb

 = 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 (ΔTb ) = boiling point of solution – boiling point of  pure solvent

boiling point of solvent = 1000C

boiling point of solution =  100.340C

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

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

w = 12 gm.

 W = 100 gm 

Kb =ΔTb.W.m / 1000 .w

Kb  = 0.34 x 100 x  180 /  1000 x 12

Kb = 0.51 Ans.

 

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