At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:
- A$52.4 \times 10^{-4}$
- B$524 \times 10^{-4}$
- C$402 \times 10^{-4}$
- D$262 \times 10^{-4}$
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Similar Questions
At $25^o C$,the molar conductivity of a $0.1 \, N \, CH_3COOH$ solution is $80 \, S \, cm^2 \, mol^{-1}$ and the molar conductivity at infinite dilution is $400 \, S \, cm^2 \, mol^{-1}$. What is the degree of dissociation $(\alpha)$ of $CH_3COOH$?
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View SolutionWhat is the molar conductivity of $CH_3COOH$ at infinite dilution if the molar conductivities of $H_2SO_4$, $K_2SO_4$, and $CH_3COOK$ at infinite dilution are respectively $x$, $y$, and $z$ $\Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$?
DifficultMHT CET 2026
View SolutionCalculate the molar conductivity of $NH_4OH$ at infinite dilution by using the following data:
$[\Lambda _m^o(NH_4Cl) = 129.8, \Lambda _m^o(KOH) = 248.0$ and $\Lambda _m^o(KCl) = 126 \ S \ cm^2 \ mol^{-1}]$
$[\Lambda _m^o(NH_4Cl) = 129.8, \Lambda _m^o(KOH) = 248.0$ and $\Lambda _m^o(KCl) = 126 \ S \ cm^2 \ mol^{-1}]$
Medium
View SolutionThe specific conductivity of $N/10$ $KCl$ solution at $20 \, ^oC$ is $0.012 \, \Omega^{-1} \, cm^{-1}$ and the resistance of the solution in the cell at $20 \, ^oC$ is $56 \, \Omega$. The cell constant is ........... $cm^{-1}$.
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View SolutionWhat is the molar conductivity of $0.1 \ M$ $NaCl$ if its conductivity is $1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$?
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