Question

Following limiting molar conductivities are given as
am
(H2SO4) = x S cm mol-1
(K2So4)= y S cm mol-1
(CH COOK)=z S cm mol-1

(in S cm? molt) for CH,COOH will be

(1) (x-y)+2z/2
(2) x - y + 2z
(3) x+y+z
(4) x - y + z​

Answer 1

The limiting molar conductivity of CH₃COOH is x/2 -y/2 + z S cm mol-1.

• According to Kaulrausch's law:

        Λ⁰ (CH₃COOH) = Λ⁰ (CH₃COO⁻) +  Λ⁰ (H⁺)

• From the given data:

        Λ⁰ (CH₃COOH) = Λ⁰ (H₂SO₄)/2 - Λ⁰ (K₂SO₄)/2 + Λ⁰ (CH₃COOK)

                                  = x/2 - y/2 + z Scm mol-1

Answer 2

The limiting molar conductivity of $CH_3COOH$ is $\frac{2z+x-y}{2}$

Explanation:

Kohlrausch law of limiting molar conductivity is defined as the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte.

Mathematically,

$\Lambda^o_m\text{ for }A_xB_y=x\lambda^o_++y\lambda^o_-$

We are given:

$\Lambda^o_{H_2SO_4}=x\text{ S.cm }mol^{-1}\\\Lambda^o_{K_2SO_4}=y\text{ S.cm }mol^{-1}\\\Lambda^o_{CH_3COOK}=z\text{ S.cm }mol^{-1}$

The equation used to calculate the limiting molar conductivity of $CH_3COOH$ follows:

$\Lambda^o_{CH_3COOH}=\frac{2\Lambda^o_{CH_3COO^-}+2\Lambda^o_{K^+}+2\Lambda^o_{H^+}+\Lambda^o_{SO_4^{2-}}-2\Lambda^o_{K^+}-\Lambda^o_{SO_4^{2-}}}{2}$

$\Lambda^o_{CH_3COOH}=\frac{2\Lambda^o_{CH_3COOK}+\Lambda^o_{H_2SO_4}-\Lambda^o_{K_2SO_4}}{2}$

$\Lambda^o_{CH_3COOH}=\frac{2z+x-y}{2}$

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