20 " mL of " `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2)O_(7)` solution 40 " mL of " `K_(2)Cr_(3)O_(7)` solution was required to oxidised the `H_(2)O_(2)` completely. Also, 2.0 " mL of " the same `K_(2)Cr_(2)O_(7)` solution required 5.0 " mL of " a 1.0 M `H_(2)C_(2)O_(4)` solution to reach equivalence point. Which of the following statements is/are correct?

Options

A. The `H_(2)O_(2)` solution is 5 M.
B. The volume strength of `H_(2)O_(2)` is 56V.
C. The volume strength of `H_(2)O_(2)` is 112V.
D. If 40 " mL of " `(5M)/(8H_(2)O_(2))` is further added to the 10 " mL of " above `H_(2)O_(2)` solution the volume strength of the resulting solution is changed to 16.8 V.

Answer

A, B, D

Explanation

(i). `mEq H_(2)O_(2)-=m" Eq of "Cr_(2)O_(7)^(2-)`
`(n=2)(n=6)`
`20mLxxN_(1)=40mL xxN_(2)`
(a). `m" Eq of "Cr_(2)O_(7)^(2-)-=m" Eq of "C_(2)O_(4)^(2-)(n=2)`
`2.0xxN_(2)-=5.0xx1.0xx2`
`N_(2)(Cr_(2)O_(7)^(2-))=5` ..(ii)
Therefore substituting the `N_(2)` in equation (i) `20 mLxxN_(1)=40mLxx5`
`N_(1)(H_(2)O_(2))=10`
`M_(1)(H_(2)O_(2))=(10)/(2)=5M`
(b). `1 N H_(2)O_(2)=5.6V`
`therefore10 N H_(2)O_(2)=56V`
(d). `N_(1)V_(1)+N_(2)V_(2)=N_(3)V_(3)(V_(3)=10+40=50mL)`
`10xx10+40xx(5)/(8)xx2=N_(3)xx50`
`N_(3)` (final) `H_(2)O_(2)=3N`
The volume strength of `H_(2)O_(2)=5.6xx3=16.8V`