`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-

Options

A. `0.16 N, 5.04`
B. `0.08 N, 3.60`
C. `0.04 N, 3.60`
D. `0.02 N, 10.08`

Answer

C

Explanation

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)`