The cell constant of a cell is $0.5 \ cm^{-1}$ and the resistance $R = 50 \ \Omega$. If the normality is $1 \ N$,then the equivalent conductivity of the electrochemical cell is ...... $\Omega^{-1} \ cm^2 \ (g \ eq)^{-1}$.
- A$10$
- B$20$
- C$300$
- D$100$
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Similar Questions
Which among the following statements is true for conductivity?
The resistance of a $1 \, N$ solution of a salt is $50 \, \Omega$. The two platinum electrodes in the solution are $2.1 \, cm$ apart and each has an area of $4.2 \, cm^2$. Calculate the equivalent conductivity of the solution.
Easy
View SolutionMatch the following:
List-$I$ (Symbol of electrical property) List-$II$ (Units) $A.$ $\wedge_m$ $I.$ $S\,cm^2\,mol^{-1}$ $B.$ $G$ $II.$ $S$ $C.$ $\kappa$ $III.$ $S\,cm^{-1}$ $D.$ $G^*$ $IV.$ $cm^{-1}$
The correct answer is
| List-$I$ (Symbol of electrical property) | List-$II$ (Units) |
|---|---|
| $A.$ $\wedge_m$ | $I.$ $S\,cm^2\,mol^{-1}$ |
| $B.$ $G$ | $II.$ $S$ |
| $C.$ $\kappa$ | $III.$ $S\,cm^{-1}$ |
| $D.$ $G^*$ | $IV.$ $cm^{-1}$ |
The correct answer is
If $ᴧ^{0} CH_{3}COONa = 91 \ S \ cm^{2} \ mol^{-1}$ and $ᴧ^{0} HCl = 462.2 \ S \ cm^{2} \ mol^{-1}$,then what value is required to calculate $ᴧ^{0} CH_{3}COOH$?
Medium
View SolutionCalculate the molar conductivity of a solution having a concentration of $0.08 \ M$ and a resistivity of $5 \times 10^{-3} \ \Omega \ cm$.
Medium
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