Question

If $20$ mL of $H_2{O}_2$ solution reacts with $80$ mL of $0.05M$ ${KMnO}_4$ in acidic medium, then what is the volume strength of $H_2{O}_2$?

• A. $2.8$
• B. $5.6$
• C. $11.2$
• D. None of the above

Answer 1

The correct option is B $5.6$

$2KMn{O}_4\left(aq\right)+5H_2{O}_2\left(aq\right)+3H_{2}S{O}_4\left(aq\right)\to K_{2}S{O}_4\left(aq\right)+2MnS{O}_4\left(aq\right)+5O_2\left(g\right)+8H_{2}O\left(aq\right)$
This is the balance chemical equation taking place :
This reaction suggests that for the for the same molarity ,suppose 0.05M , if KMnO4 has 80 mL , then H2O2 volume which will react completely is $5/2\ast 80=200mL$ .
As given is only 20 mL $H_2{O}_2$
so 0.05 M (200 mL ) = actual molarity * 20 mL
actual molarty = $0.05\ast 200/20=0.50M.$
As 1 M $H_2{O}_2$ has 11.2 M volume strenth , then 0.5 will has $5.6$ M, volume strength.