Elevation of boiling point -Numerical part 3

Elevation of boiling point

source : HyperPhysics Concepts

Elevation of boiling point-Numerical

Question 1 ) What will be the boiling point of  solution when 174.5 mg of octa atomic sulphur is added to 78 gm of bromine . K_b for bromine is 5.2 K Kg /mole. Boiling point of bromine is 332.15 K.

Solution )

K_b =  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 

ΔT_b =1000 K_b. w /W.m

ΔT_b = 1000 x  5.2 x  0.1745 /  78 x  256

ΔT_b = 0.045 

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

boiling point of solvent ( benzene ) = 332.15 K

boiling point of solution =   ?

 0.045 = boiling point  of solution – 332.15

boiling point  of solution =332.195 K Ans.

Question 2) The molal elevation constant for water is 0.510 K Kg/mole. Calculate boiling point  of solution made by dissolving 6 gm urea in 200 gm water ?

Solution  )

K_b =  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 

ΔT_b =1000 K_b. w /W.m

ΔT_b = 1000 x  0.510 x 6.0 /  200 x  60

ΔT_b = 0.255 

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

boiling point of solvent ( water) = 100⁰C

boiling point of solution =   ?

 0.255 = boiling point  of solution – 100

boiling point  of solution = 100.255 ⁰C Ans.

Question 3) A solution of an organic compound freezes at 272.33 K and boils at 373.187 K. Calculate molal elevation constant for water ? If molal depression constant for water is 1.86 K Kg/mole.

Solution )

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

freezing point of solution = 272.33 K

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

ΔT_f  = 273 – 272.33

ΔT_f = 0.67 

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

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

boiling point of solution = 373.187 K

 ΔT_b = 373.187 – 373 

ΔT_b = 0.187

ΔT_b /ΔT_f  = K_b/K_f

K_f = 1.86 K Kg /mole

0.187 / 0.67 = K_b / 1.86

K_b = 0.187 x 1.86 /0.67

K_b = 0.519 = 0.52 K Kg /mole Ans.

Question 4 ) . An aqueous solution of urea has a boiling point  100.18⁰C. Calculate its freezing point ? Molal constants of water K_f and K_b are 1.86 & 0.512 K Kg/mole respectively

Solution )

freezing point of pure solvent (water ) = 0⁰C

freezing point of solution = ?

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

boiling point of solvent ( water) = 100 ⁰C 

boiling point of solution = 100.18 ⁰C

 ΔT_b = 100.18 –  100

ΔT_b = 0.18

ΔT_b /ΔT_f  = K_b /K_f

K_f = 1.86 K Kg /mole

K_b = 0.512 K Kg /mole Ans.

0.18 / ΔT_f  = 0.512 / 1.86

ΔT_f  = 0.654

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

0.654 = 0  – freezing point of  solution

freezing point of  solution = 0 – 0.654

freezing point of  solution = – 0.654 Ans