The equivalent conductivity of N/10 solution of acetic acid at 25°C is 14.3 ohm-1 cm2 equiv-1. What will be the degree of dissociation of acetic acid ? ( Lambda∞ CH3COOH=390.71 ohm-1 cm2 equiv-1)

Question

The equivalent conductivity of N/10 solution of acetic acid at 25°C is 14.3 ohm-1 cm2 equiv-1. What will be the degree of dissociation of acetic acid ? ( Lambda∞ CH3COOH=390.71 ohm-1 cm2 equiv-1) (A) 3.66 % (B) 3.9 % (C) 2.12 % (D) 0.008 %

Tardigrade Admin · Accepted Answer

Lambda∞m CH3COOH=390.71 ohm-1 cm2 equiv-1 LambdaCH3COOH=14.3 ohm-1 cm2 equiv-1 Degree of dissociation (α) =( Lambdacm/ Lambda∞m)=(14.3/390.71) =0.0366 i.e. 3.66 %

Q. The equivalent conductivity of $N /10$ solution of acetic acid at $2 5^{\circ} C$ is $14.3 o h m^{- 1} c m^{2} e q u i v^{- 1}$ . What will be the degree of dissociation of acetic acid ?
$(Λ_{\infty C H_{3} COO H} = 390.71 o h m^{- 1} c m^{2} e q u i v^{- 1})$

$3.66%$

$3.9%$

$2.12%$

$0.008%$

$Λ_{m C H_{3} COO H}^{\infty} = 390.71 o h m^{- 1} c m^{2} e q u i v^{- 1}$
$Λ_{C H_{3} COO H} = 14.3 o h m^{- 1} c m^{2} e q u i v^{- 1}$
Degree of dissociation $(α)$
$= \frac{Λ _{m}^{c}}{Λ _{m}^{\infty}} = \frac{14.3}{390.71}$
$= 0.0366$ i.e. $3.66%$

Q. The equivalent conductivity of N/10 solution of acetic acid at 25∘C is 14.3ohm−1cm2equiv−1. What will be the degree of dissociation of acetic acid ? (Λ∞CH3​COOH​=390.71ohm−1cm2equiv−1)

3.66%

3.9%

2.12%

0.008%