Elevation of boiling point -Numerical part 2

Elevation of boiling point

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Elevation of boiling point – Numerical

Question 1) The boiling point of a solution containing 1.5 gm of dichloro benzene in 100 gm of benzene was higher by
0.268⁰C.Calculate the molar mass of dichloro benzene ? ( K_b for benzene =2.65 K /molal)

Solution )

K_b = 2.65 , ΔT_b = 0.268⁰C

m of dichloro benzene = ?

w = 1.5 gm.

 W = 100 gm 

m =1000 K_b. w /W.ΔT_b

m  = 1000 x 2.65 x  1.5 /  100 x 0.268

m =  148.32  Ans

Question 2) A solution containing 0.513 gm of naphthalene ( molar mass = 128) in 50 gm of Carbon tetrachloride gives a boiling point elevation of 0.402⁰C ,while a solution of 0.625 gm of an unknown solute gives a boiling point elevation of 0.650⁰C. Find the molar mass of unknown solute?

Solution )

(i) Elevation in boiling point (ΔT_b ) = 0.402⁰C

 m of  naphthalene = 128

w =  0.513 gm.

 W = 50 gm 

K_b =ΔT_b.W.m / 1000 .w

Kb  = 0.402 x 50 x  128 /  1000 x 0.513

K_b = 5.02  Ans.

(ii)    K_b = 5.02 , ΔT_b = 0.650⁰C

m  = ?

w = 0.625 gm.

 W = 50 gm 

m =1000 K_b. w /W.ΔT_b

m  = 1000 x 5.02 x  0.625 /  50 x 0.650

m =  96.538 = 96.54  Ans

Question 3) Calculate the boiling point of one molar  aqueous solution ( density = 1.04 gm/cc ) of potassium chloride (K_b = 0.52 K Kg/mole).

Solution )

volume of solution = 1000 ml

mass of solution = v x d = 1000 x 1.04 =1040 gm

weight of KCl present in 1040 gm solution = molecular weight of KCl = 39 + 35.5 = 74.5

mass of solvent = 1040-74.5 = 965.5 gm

one molar aqueous solution means one mole of solute  in one litre  (1000 ml ) of solution.

one molar aq. KCl solution means 74.5 gm of solute in one litre of solution.

molality =1000 w /m.W = 1000 x 74.5/74.5 x 965.5

molality = 1.035

ΔT_b = i.K_b.molality

i = 2 , because KCl gives two ions in solution

K_b = 0.52

ΔT_b = 2 x 0.52 x 1.035 =1.0764 = 1.08

boiling point of solvent = 100⁰C = 100 + 273 K =373 K

ΔT_b = boiling point of solution – boiling point of solvent

1.08 = boiling point of solution – 373

boiling point of solution = 373 + 1.08 = 374.08 K  Ans.

Question 4) The boiling point of a solution containing 1 gm of substance dissolved in 83.4 gm of benzene is 80.175⁰C. The boiling point of pure benzene is 80⁰C. If latent heat of vaporisation  of benzene is 90 cal./gm. Calculate the molecular weight of solute ?

Solution )

l_v = 90 cal./gm , boiling point of benzene =80⁰C = 80 + 273 =352 K ,

boiling point of solution = 80.175⁰C

ΔT_b = boiling point of solution – boiling point of solvent

ΔT_b = 80.175 – 80 =  0.175⁰C

 R = 2 calorie

T_b = boiling point of solvent (benzene) =80⁰C = 80 + 273 =353 K ,

K_b= R.T_b² /1000 l_v

= 2 x 353 x 353 / 1000 x 90 = 2.769 =2.77

w = 1 gm , W =83.4 gm , molecular weight of solute ‘m’ = ?

m =1000 K_b. w /W.ΔT_b

m  = 1000 x 2.77 x  1.0 / 83.4 x 0.175

m =  189.79   Ans