The conductivity of 0.25 M solution of monovalent electrolyte AB is 0.0125 ohm-1 cm-1 . The value of ∧OM of AB is 500 ohm-1 mol-1. Calculate the value of dissociation constant.

∧c=1000XkM=1000X0.01250.25=50α=∧c∧0=50500=0.1K=Cα2=0.25(0.1)2=2.5X10−3

The conductivity of 0.25 M solution of monovalent electrolyte AB is 0.0125 ohm-1 cm-1 . The value of ∧OM of AB is 500 ohm-1 mol-1. Calculate the value of dissociation constant.

∧c=1000XkM=1000X0.01250.25=50α=∧c∧0=50500=0.1K=Cα2=0.25(0.1)2=2.5X10−3

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The conductivity of 0.25 M solution of monovalent electrolyte AB is 0.0125 ohm^-1 cm^-1 . The value of ${\overset{O}{\land}}_{M}$ of AB is 500 ohm^-1 mol^-1. Calculate the value of dissociation constant.

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$2.5 X 10^{- 3}$

$2.5$

$2.5 X 10^{- 1}$

$0.00025$

answer is A.

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$\begin{array}{l} \land_{c} = \frac{1000 X k}{M} \\ = \frac{1000 X 0.0125}{0.25} \\ = 50 \\ α = \frac{\land_{c}}{\land_{0}} = \frac{50}{500} \\ = 0.1 \\ K = C α^{2} \\ = 0.25 {(0.1)}^{2} \\ = 2.5 X 10^{- 3} \end{array}$

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The conductivity of 0.25 M solution of monovalent electrolyte AB is 0.0125 ohm-1 cm-1 . The value of ∧OM of AB is 500 ohm-1 mol-1. Calculate the value of dissociation constant.

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The conductivity of 0.25 M solution of monovalent electrolyte AB is 0.0125 ohm^-1 cm^-1 . The value of ${\overset{O}{\land}}_{M}$ of AB is 500 ohm^-1 mol^-1. Calculate the value of dissociation constant.