Question

Following limiting molar conductivities are given as $\lambda^{o}_{m}\left(H_{2}SO_{4}\right) \,= \,x S\, cm^{2} mol^{-1}$ $\lambda^{o}_{m}\left(K_{2}SO_{4}\right) = y S\,cm^{2} mol^{-1}$ $\lambda^{o}_{m}\left(CH_{3}COOK\right) =z S\, cm^{2} mol^{-1}$ $\lambda^{o}_{m}\left(in \,S \,cm^{2} mol^{-1}\right)$ for $CH_3COOH$ will be-

• A. $x - y + 2 z$
• B. $x + y - z$
• C. $x - y + z$
• D. $\frac{\left(x -y\right)}{2}+z$

Answer 1

Answer: (D)

$\frac{\left(x -y\right)}{2}+z$

Answer 2

$CH_3COOH {->} CH_3COO^- + H^+\,...(1)$
$H_2SO_4 {->} 2H^+ + SO_4^{-2} \,...(2)$
$K_2SO_4 {->} 2K^+ + SO_4^{-2} \,...(3)$
$CH_3COOK {->} CH_3COO^- + K^+ \,...(4)$
According to Kohlrausch's law-
$\lambda^{\circ}_{CH_3COOH} = \lambda^{\circ}_{CH_3COO^-} + \lambda^{\circ}_{H^+}$
$e(1) = eq .(4) + eq . \frac{(2)}{2} - e \frac{(3)}{2}$
$\therefore \lambda^{\circ}_{CH_3COOH} = z + \frac{X}{2} - \frac{Y}{2}$
$\lambda^{\circ}_{CH_3COOH} = \frac{(X - Y)}{2} + z (S \times cm^2 mol^{-1})$