At 25^{circ} C,the molar conductance at infin | vedclass

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(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the mol

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$524 	imes 10^{-4}$

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(C) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.$\lambda_{m}^{o} (Ba(OH)_{2}) = \lambda_{m}^{o} (Ba^{2+}) + 2\lambda_{m}^{o} (OH^{-})$Given values:$\lambda_{m}^{o} (NaOH) = \lambda_{m}^{o} (Na^{+}) + \lambda_{m}^{o} (OH^{-}) = 248 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$$\lambda_{m}^{o} (NaCl) = \lambda_{m}^{o} (Na^{+}) + \lambda_{m}^{o} (Cl^{-}) = 126 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$$\lambda_{m}^{o} (BaCl_{2}) = \lambda_{m}^{o} (Ba^{2+}) + 2\lambda_{m}^{o} (Cl^{-}) = 280 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$To obtain $\lambda_{m}^{o} (Ba(OH)_{2})$,we perform the operation:$\lambda_{m}^{o} (Ba(OH)_{2}) = \lambda_{m}^{o} (BaCl_{2}) + 2\lambda_{m}^{o} (NaOH) - 2\lambda_{m}^{o} (NaCl)$Substituting the values:$\lambda_{m}^{o} (Ba(OH)_{2}) = 280 	imes 10^{-4} + 2(248 	imes 10^{-4}) - 2(126 	imes 10^{-4})$$= (280 + 496 - 252) 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$$= 524 	imes 10^{-4} \ S \ m^{2} \ mol^{-1}$

At $25^{\circ} C$,the molar conductance 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. $\lambda_{m}^{o} Ba(OH)_{2}$ in $S \ m^{2} \ mol^{-1}$ is

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