The rise in the boiling point of a solution containing 1.8 gram of glucose in 100g of a solvent in 0.1⁰C. The molal elevation constant of the liquid is

Q. Assertion :When

0.1 M

weak diprotic acid

H_{2} A

dissociates with its dissociation constants

K_{a_{1}} = 10^{- 3}

and

K_{a_{2}} = 10^{- 8}

, then

[A^{2 -}]

is almost equal to

10^{- 3} M

Reason: Since

K_{a_{2}} ≪ K_{a_{1}}

for

0.1 M H_{2} A

, so

[A^{2 -}]

is negligible w.r.t.

[H A^{-}]

pH of 0.1 M weak acid solution is 4. What is the dissociation constant of the acid ?

Join BYJU'S Learning Program

When 0.1 mole CH3COOH is present in a litre solvent it shows elevation in boiling point of 0.75oC. Acid dissociation constant will be:(Kb=5Kkgmol−1) and (M≃m)

The correct option is A 5×10−2 CH3COOH⇌CH3COO⊝+H⊕at t=0 0.1 0 0at eqm 0.1−C C Cnow, i=0.1−C+C+C=0.1+CΔTb=iKf×m⇒0.75=(0.1+C)×5×0.1 [Since M≃m=0.1]∴ dissociation constant= 5×10−2

When $0.1$ mole $C H_{3} C O O H$ is present in a litre solvent it shows elevation in boiling point of ${0.75}^{o}$ C. Acid dissociation constant will be:
$(K_{b} = 5 K k g m o l^{- 1})$ and $(M ≃ m)$