$\wedge_{\mathrm{m}}^{\circ}$ for $\mathrm{NaCl}$, $\mathrm{HCl}$ and NaA are $126.4, 425.9$ and $100.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, respectively. If the conductivity of 0.001 M HA is $5 \times 10^{-5} \mathrm{Scm}^{-1}$, degree of dissociation of HA is :

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Redox Reactions and Electrochemistry

\land_{m}^{\circ}

for

NaCl

HCl

and NaA are

126.4, 425.9

and

100.5 S cm^{2} mol^{- 1}

, respectively. If the conductivity of 0.001 M HA is

5 \times 1 0^{- 5} Scm^{- 1}

, degree of dissociation of HA is :

Answer: (C)

By Kohlrausch's law,

Λ_{m}^{\circ} (HA) = Λ_{m}^{\circ} (HCl) + Λ_{m}^{\circ} (NaA) - Λ_{m}^{\circ} (NaCl) = 425.9 + 100.5 - 126.4 = 400 S c m^{2} mo l^{- 1}

. The molar conductivity at the given concentration is

Λ_{m} = \frac{κ \times 1000}{M} = \frac{5 \times 1 0 ^{- 5} \times 1000}{0.001} = 50 S c m^{2} mo l^{- 1}

. The degree of dissociation is

α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{50}{400} = 0.125

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option