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(D) The number of milli-equivalents of $NaOH$ is given by $N 	imes V = 0.1 	imes 20 = 2 \ mEq$. Since $1 \ mEq = 10^{-3} \ Eq$,the number of equival

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$63$

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(D) The number of milli-equivalents of $NaOH$ is given by $N 	imes V = 0.1 	imes 20 = 2 \ mEq$. Since $1 \ mEq = 10^{-3} \ Eq$,the number of equivalents is $2 	imes 10^{-3} = 0.002 \ Eq$. At the point of neutralisation,the number of equivalents of acid equals the number of equivalents of base. Therefore,$0.002 = \frac{	ext{mass of acid}}{	ext{equivalent weight}} = \frac{0.126}{E}$. Solving for $E$,we get $E = \frac{0.126}{0.002} = 63$.

Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

$0.126 \ g$ of an acid is needed to completely neutralise $20 \ mL$ of $0.1 \ N$ $NaOH$ solution. The equivalent weight of the acid is:

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Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$

(in $u$)

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