Question

250 ml of a solution contains 6.3 grams of
oxalic acid(mol. wt. =126). What is the volume(inlitres) of water to be added to this solution to make it a 0.1 N solution?
1) 750 237.5 3) 0.075 4) 0.75
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Answer 1

Answer:

your answer is 0.75

explanation:

Eq. Wt. of oxalic acid = 126/2 = 63

Normality of oxalic acid = 6.3 ×1000/(63× 250)

= 0.4N

Equation for dilution

V 2= V1N1/N2

= (0.250×0.4)/0.1

=1 lit

Volume of water added = 1-0.25 = 0.75 lit

HOPE..IT..HELPS..!!