20 ml of `0.2 M NaOH` (aq) solution is mixed with 35 ml of 0.1 M NaOH (aq) solution and the resultant solution is diluted to 100 ml. 40 ml of this diluted solution reacted with 10% impure sample of oxalic acid `(H_(2)C_(2)O_(4))`. The weight of the impure sample is

Options

A. 0.15 gram
B. 0.135 gram
C. 0.59 gram
D. None of these

Answer

A

Explanation

`M_(NaOH)` resultant `=(20xx0.2+35xx0.1)/(100)=0.075M`
Milli-equivalent of NaOH = Milli equivalent of `H_(2)C_(2)O_(4)`
Let the weight of the impure sample be x gram
`40xx0.075=(x xx0.90)/(90)xx2xx1000`
`x=0.15` gram