Equivalent conductances of $N a C l, H C l$ and $C H_{3} C O O N a$ at infinite dilution are $126.45, 426.16$ and $91 o h m^{- 1} c m^{2} e q^{- 1}$ respectively. The equivalent conductance of $C H_{3} C O O H$ at infinite dilution would be:

101.38 o h m^{- 1} c m^{2} e q^{- 1}

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253.62 o h m^{- 1} c m^{2} e q^{- 1}

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389.71 o h m^{- 1} c m^{2} e q^{- 1}

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678.0 o h m^{- 1} c m^{2} e q^{- 1}

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Solution

The correct option is D $389.71 o h m^{- 1} c m^{2} e q^{- 1}$
Given that,
$_{\infty} N a C l = 126.45 o h m^{- 1} c m^{2} e q^{- 1}$

$_{\infty} H C l = 425.16 o h m^{- 1} c m^{2} e q^{- 1}$

$_{\infty} C H_{3} C O O N a = 91 o h m^{- 1} c m^{2} e q^{- 1}$

$_{\infty} N a C l =_{\infty} N a^{+} +_{\infty} C l^{-} = 126.45 \dots \dots (i)$

$_{\infty} H C l =_{\infty} H^{+} +_{\infty} C l^{-} = 425.16 \dots \dots (i i)$

$_{\infty} C H_{3} C O O N a =_{\infty} C H_{3} C O O^{-} +_{\infty} N a^{+} = 91 \dots \dots (i i i)$

$_{\infty} C H_{3} C O O H =_{\infty} C H_{3} C O O^{-} +_{\infty} H^{+}$

$\Rightarrow (i i i) + (i i) - (i)$
$\Rightarrow 425.16 + 91 - 126.45$
$= 389.71 o h m^{- 1} c m^{2} e q^{- 1}$

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