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Question

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 $$280\times 10^{-4} Sm^{2} mol^{-1}$$ respectively, $$\lambda_{m}^{\circ}Ba(OH)_{2}$$ in $$Sm^{2} mol^{-1}$$ is


A
52.4×104
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B
524×104
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C
4.2×104
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D
262×104
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Solution

The correct option is D $$524\times 10^{4}$$
$$BaCl_{2} + 2NaOH\rightarrow Ba(OH)_{2} + 2NaCl$$
$$\lambda^{\infty}_{m\ Ba(OH)_{2}} = \lambda^{\infty}_{m\ BaCl_{2}} + 2\lambda^{\infty}_{m\ NaOH} - 2\lambda^{\infty}_{m\ NaCl}$$
$$= 280 \times 10^{-4} + 2\times 248\times 10^{-4} - 2 \times 126\times 10^{-4}$$
$$= (280 + 496 - 252) \times 10^{-4}$$
$$= 524\times 10^{-4} Sm^{2} mol^{-1}$$.

Chemistry

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