## Elevation of boiling point source : ThoughtCo

## Elevation of boiling point – Numerical

### Solution )

Kb = 2.65 , ΔTb = 0.2680C

m of dichloro benzene = ?

w = 1.5 gm.

W = 100 gm

### Solution )

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

w =  0.513 gm.

W = 50 gm

#### Kb  = 0.402 x 50 x  128 /  1000 x 0.513

Kb = 5.02  Ans.

(ii)    Kb = 5.02 , ΔTb = 0.6500C

m  = ?

w = 0.625 gm.

W = 50 gm

### 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 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

#### ΔTb = i.Kb.molality

i = 2 , because KCl gives two ions in solution

Kb = 0.52

### Solution )

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

#### R = 2 calorie

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

Kb= R.Tb2 /1000 lv

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

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