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Arrhenius Acid

the molar con...

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the molar conductivity of 0.025mol/l methanoic acid is 46.1 scm^2/mol calculate the degree of dissociation constant and its degree of dissociation . given lim molar conductivity for h+ =349.6 and hcoo- =54.6

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Q. The molar conductivity of 0.0025 mol/lit methanoic acid is 46.1 s cm/mol. Calculate it's degree of dissociation and dissociation constant. Given

[H^{+}] = 349.6

[H C O O^{-}] = 54.6

Q. The molar conductivity of

0.025 m o l L^{- 1}

methanoic acid is

46.1 S c m^{2} m o l^{- 1}

Calculate its degree of dissociation and dissociation constant.

λ_{(H -)}^{o} = 349.6 S c m^{2} m o l^{- 1}

λ_{(H C O O^{-})}^{o} = 54.6 S c m^{2} m o l^{- 1}

The molar conductivity of 0.025 mol L−1 methanoic acid is

46.1 S cm2 mol⁻¹.

Calculate its degree of dissociation and dissociation constant. Given λ ^°(H⁺)

= 349.6 S cm² mol⁻¹ and λ ^°(HCOO−) = 54.6 S cm² mol

Q. Molar conductivity of

0.025

L^{- 1}

methanoic acid is

46.1

c m^{2}

m o l^{- 1}

, the degree of dissociation and dissociation constant will be :
(Given:

λ_{H^{+}}^{o} = 349.6

c m^{2}

m o l^{- 1}

λ_{H C O O^{-}}^{o} = 54.6

S . c m^{2}

m o l^{- 1}

)

Q. The degree of dissociation of 2.5 × 10^–2 M methanoic acid solution having molar conductivity 46.1 S cm² mol^–1 will be (Given : λ°(H⁺) = 349.6 S cm² mol^–1 and λ°(HCOO^–) = 54.6 S cm² mol^–1)

मोलर चालकता 46.1 S cm² mol^–1 वाले 2.5 × 10^–2 M मेथनॉइक अम्ल विलयन के वियोजन की मात्रा होगी (दिया है: λ°(H⁺) = 349.6 S cm² mol^–1 तथा λ°(HCOO^–) = 54.6 S cm² mol^–1)

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