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 containing 10% w/w non reacting impurities is diluted to 100 mL. 40mL of this diluted solution reacted with impure sample of anhydrous oxalic acid. The weight of impure sample required is:

Options

A. `0.15 g`
B. `0.135 g `
C. ` 0.59 g`
D. `0.38 g`

Answer

A

Explanation

`M _(NaOH)` resultant `= ( N _(1) V _(1) + N _(2) V _(2))/( V _(1) + V _(2)) = ( 20 xx 0.2 xx 35 + 0.1)/( 100) = 0. 0 75M`
Milli-equivalent of NaOH = milli - equivalent of `H _(2) C_(2) O _(4)`
Let wt of impure sample is .x. grom then `40 xx 0.075 = ( x xx 0.90)/(90) xx 2 xx 1000 implies x =0. 15` gram