First slide

Kohlraush law

Question

Following limiting molar conductivities are given as
$λ_{m}^{\circ} (H_{2} {SO}_{4}) = x {Scm}^{2} {mol}^{- 1} λ_{m}^{\circ} (K_{2} {SO}_{4}) = y {Scm}^{2} {mol}^{- 1} λ_{m}^{\circ} ({CH}_{3} COOK) = z {Scm}^{2} {mol}^{- 1} λ_{m}^{\circ} (in {Scm}^{2} {mol}^{- 1}) for {CH}_{3} COOH willbe$

Moderate

Solution

According to Kohlrausch's law
$λ_{m}^{o} (AB) = λ_{m}^{o} (A^{+}) + λ_{m}^{o} (B^{-})$
$So, λ_{m}^{o} ({CH}_{3} COOH) = λ_{m}^{\circ} ({CH}_{3} {COO}^{-}) + λ_{m}^{o} (H^{+})$
$So λ_{m}^{o} ({CH}_{3} COOH)$
$= λ_{m}^{o} ({CH}_{3} COOK) + \frac{1}{2} λ_{m}^{o} (H_{2} {SO}_{4}) - \frac{1}{2} λ_{m}^{\circ} (K_{2} {SO}_{4})$
$\begin{array}{l} = z + \frac{x}{2} - \frac{y}{2} \\ = z + (\frac{x - y}{2}) \end{array}$

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