The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. If λ(H+) = 349.6 S cm2 mol-1 and λ(HCOO-) = 54.6 S cm2 mol-1 its degree of dissociation is

Question

The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. If λ(H+) = 349.6 S cm2 mol-1 and λ(HCOO-) = 54.6 S cm2 mol-1 its degree of dissociation is (A) 34% (B) 25% (C) 5% (D) 11.4%

Tardigrade Admin · Accepted Answer

Lambda°(HCOOH) = λ°(H+) + λ°(HCOO-) =349.6 + 54.6 = 404 .2 S cm2 mol-1 α = ( Lambdam/ Lambda°m) = (46.1/404.2) = 0.114 ⇒ α = 11.4 %

Q. The molar conductivity of $0.025 m o l L^{- 1}$ methanoic acid is $46.1 S c m^{2} m o l^{- 1}$ . If $λ (H^{+}) = 349.6 S c m^{2} m o l^{- 1}$ and $λ (H CO O^{-}) = 54.6 S c m^{2} m o l^{- 1}$ its degree of dissociation is

34%

25%

5%

11.4%

$Λ_{(H COO H)}^{\circ} = λ_{(H^{+})}^{\circ} + λ_{(H CO O^{-})}^{\circ}$
$= 349.6 + 54.6 = 404.2 S c m^{2} m o l^{- 1}$
$α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{46.1}{404.2} = 0.114$
$\Rightarrow α = 11.4%$

Q. The molar conductivity of 0.025molL−1 methanoic acid is 46.1Scm2mol−1. If λ(H+)=349.6Scm2mol−1 and λ(HCOO−)=54.6Scm2mol−1 its degree of dissociation is

34%

25%

5%

11.4%

Λ(HCOOH)∘​=λ(H+)∘​+λ(HCOO−)∘​ =349.6+54.6=404.2Scm2mol−1 α=Λm∘​Λm​​=404.246.1​=0.114 ⇒α=11.4%

Q. The molar conductivity of $0.025 m o l L^{- 1}$ methanoic acid is $46.1 S c m^{2} m o l^{- 1}$ . If $λ (H^{+}) = 349.6 S c m^{2} m o l^{- 1}$ and $λ (H CO O^{-}) = 54.6 S c m^{2} m o l^{- 1}$ its degree of dissociation is

$Λ_{(H COO H)}^{\circ} = λ_{(H^{+})}^{\circ} + λ_{(H CO O^{-})}^{\circ}$
$= 349.6 + 54.6 = 404.2 S c m^{2} m o l^{- 1}$
$α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{46.1}{404.2} = 0.114$
$\Rightarrow α = 11.4%$