Correct Answer is Option (C)

According to Kohlrausch law of independent migration of ions.

${\mathrm{\Lambda }}_m^o(C{H}_{3}COOH)$

$={\mathrm{\Lambda }}_m^o(C{H}_{3}COONa)+{\mathrm{\Lambda }}_m^o(HCl)-{\mathrm{\Lambda }}_m^o(NaCl)$

= 91.0 S cm² mol^$-$1 + 426.16 S cm² mol^$-$1 $-$ 126.45 S cm² mol^$-$1

= 390.71 S cm² mol^$-$1

Correct Answer is Option (D)

${\mathrm{\Lambda }}_m^{}$ = 20 S cm² mol^$-$1

According to Kohlrausch’s law,

${\mathrm{\Lambda }}_{mC{H}_{3}COOH}^o={\mathrm{\Lambda }}_{C{H}_{3}CO{O}^{-}}^o+{\mathrm{\Lambda }}_{m{H}^{+}}^o$

= 50 + 350 = 400 S cm² mol^$-$1

Degree of dissociation,

$\alpha =\frac{{\mathrm{\Lambda }}_m^{}}{{\mathrm{\Lambda }}_m^o}=\frac{20}{400}=\frac{1}{20}$


$K_a=\frac{C{\alpha }^2}{1-\alpha }\simeq C{\alpha }^2$

As $\alpha$ << 1 so 1 - $\alpha$ $\approx$ 1

$=7\times {10}^{-3}\times {\left(\frac{1}{20}\right)}^2$

$=7\times {10}^{-3}\times \frac{1}{4}\times {10}^{-2}$

$=1.75\times {10}^{-5}$ mol L^-1

Correct Answer is Option (D)

$\mathrm{\Lambda }$^o_CH3COOH = $\mathrm{\Lambda }$^o_CH3COONa + $\mathrm{\Lambda }$^o_HCl - $\mathrm{\Lambda }$^o_NaCl

= = 91 + 425.9 – 126.4 = 390.5

Correct Answer is Option (D)

According to Kohlrausch’s law, the molar conductivity of NH₄OH

${\mathrm{\Lambda }}_{m\left(N{H}_{4}OH\right)}^0$ = ${\mathrm{\Lambda }}_{m\left(N{H}_{4}Cl\right)}^0+{\mathrm{\Lambda }}_{m\left(NaOH\right)}^0-{\mathrm{\Lambda }}_{m\left(NaCl\right)}^0$

Correct Answer is Option (B)

Equivalent conductance of an electrolyte at infinite dilution is given by the sum of equivalent conductances of the respective ions at infinite dilution.