250 ml of a solution contains 6.3 grams of oxalic acid(mol. wt. =126). What is the volume (in litres) of water to be added to this solution to make it a 0.1 N solution?

750

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7.5

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0.075

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0.75

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Solution

The correct option is D $0.75$
$N o r m a l i t y = \frac{number of equivalent of solute}{1 liter of solution}$
$= \frac{W e i g h t}{e q . w e i g h t} \times \frac{1000}{V (m l)}$
By putting all the given values, we get-

$0 \cdot 1 = \frac{6 \cdot 3}{\frac{126}{2}} \times \frac{1000}{V}$
$V = 1000 m l = 1 l i t r e$
$Volume of water added = Total volume - initial volume of oxalic acid solution$
$V = 1 - 0 \cdot 25$
$V = 0 \cdot 75 l i t r e$

So, the correct option is $D$

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