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Volume of 2m HCl required to neutralize the solution containing 1 mole of $NH_4Cl$ and $1$ mole of NaOH is

$(a)\;1litre\qquad(b)\;2litre\qquad(c)\;3litre\qquad(d)\;\large\frac{1}{2}$$litre$

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$NH_4Cl+NaOH+HCl\rightarrow NaCl+H_2O$
Milli-mole of NaOH=milli-mole of HCl
$1000=2\times V$ (in ml)
$\therefore V=\large\frac{1000}{2}$
$\Rightarrow 500ml$
Hence (d) is the correct answer.
answered Oct 29, 2013 by sreemathi.v
 

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