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Concentration Terms

The molar con...

Question

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

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Solution

C = 0.025 mol L⁻¹

Now, degree of dissociation:

Thus, dissociation constant:

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Similar questions

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}

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 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 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)

Q. Conductivity of

2.5 \times 10^{- 4}

M methanoic acid is

5.25 \times 10^{- 5} S c m^{- 1}

. Calculate it molar conductivity and degree of dissociation.
(Given:

λ^{0} (H^{+}) = 349.5 s c m^{2} m o l^{- 1} a n d λ^{0} (H C O O^{.}) = 50.5 S c m^{2} m o l^{- 1}

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