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Q)

An aqueous solution of $6.3\;g$ oxalic acid dihydrate is made up to $250\;ml$ . The volume of $0.1\;N\; NaOH$ required to completely neutralize 10 ml of this solution is

$(a)\;40\;ml \\ (b)\;20\;ml \\(c)\;10\;ml \\(d)\;4\;ml $

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A)
Equivalents of $H_2C_2O_4.2H_2O$ in 10 ml
= Equivalents of $NaOH$
$\large\frac{6.3}{63} \times \frac{10}{250}$$=0.1 \times v(in\;litres)$
(Eq.wt of $H_2C_2O_4.2H_2O=63)$
or $v= 0.04\;l$ or $40\;ml$
Hence a is the correct answer.
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