## Elevation of boiling point

source : HyperPhysics Concepts

## Elevation of boiling point-Numerical

### Solution )

Kb =  5.2 K Kg / mole ,

boiling point of  solution = ?

m  of  sulphur  ( octaatomic ) = 32 x 8 = 256

w =  174.5 mg. = 174.5 /1000 = 0.1745 gm.

W = 78 gm

#### ΔTb = 0.045

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

#### boiling point of solution =   ?

0.045 = boiling point  of solution – 332.15

## boiling point  of solution =332.195 K Ans.

### Solution  )

Kb =  0.510  K Kg / mole , boiling point  of  solution = ?

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

w =  6.0 gm.

W = 200 gm

#### ΔTb = 0.255

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

#### boiling point of solution =   ?

0.255 = boiling point  of solution – 100

## boiling point  of solution = 100.255 0C Ans.

### Solution )

freezing point of pure solvent (water ) = 00C = 0 + 273 = 273 K

freezing point of solution = 272.33 K

ΔTf  = freezing point of pure solvent – freezing point of  solution

#### ΔTf = 0.67

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

#### boiling point of solution = 373.187 K

ΔTb = 373.187 – 373

ΔTb = 0.187

ΔTb /ΔTf  = Kb/Kf

Kf = 1.86 K Kg /mole

0.187 / 0.67 = Kb / 1.86

Kb = 0.187 x 1.86 /0.67

## Kb = 0.519 = 0.52 K Kg /mole Ans.

### Solution )

freezing point of pure solvent (water ) = 00C

freezing point of solution = ?

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

#### boiling point of solution = 100.18 0C

ΔTb = 100.18 –  100

ΔTb = 0.18

ΔTb /ΔTf  = K/Kf

Kf = 1.86 K Kg /mole

Kb = 0.512 K Kg /mole Ans.

0.18 / ΔTf  = 0.512 / 1.86

ΔTf  = 0.654

ΔTf  = freezing point of pure solvent – freezing point of  solution