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2.52 g of oxalic acid dehydrate was ...

`2.52 g` of oxalic acid dehydrate was dissolved in 100 ml of water, 10 mL of this solution was diluted to 500 mL. The normality of the final solution and the amount of oxalic acid (mg/mL) in the solution are respectively-

A

`0.16 N, 5.04`

B

`0.08 N, 3.60`

C

`0.04 N, 3.60`

D

`0.02 N, 10.08`

Text Solution

Verified by Experts

The correct Answer is:
C
Initial Normality `N = (2.52 xx 1000)/(63 xx 100) = 0.4`
`{:(underset(|)(C)OOH),(COOH):}.2H_(2)O`
`:.N_(1)V_(1)=N_(2)V_(2)`
`0.4 xx 10 = N_(2) xx 500`
`N_(2) = (0.4)/(50) = 0.08 N`
Then final weight `N = (w xx 100)/(E xx V_(ml))`
`0.08 = (w xx 1000)/(63 xx 500)`
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