Question

To a 25 ml of `H_(2)O_(2)` solution, excess of acidified solution of KI was mixed. The liberated `I_(2)` require 20 ml of 0.3 M hypo solution for neutralization. The volume strength of `H_(2)O_(2)` will be
A. 1.34 ml
B. 1.44 ml
C. 1.60 ml
D. 2.42 ml

Answer 1

Correct Answer - A
20 ml of 0.3N `Na_(2)S_(2)O_(3)`
=20ml of 0.3 N `I_(2)` solution
=20ml of 0.3N `H_(2)O_(2)` solution
=25ml of 0.08N `H_(2)O_(2)` solution
Mass of 100 ml `H_(2)O_(2)` solution
Mass of 100 ml `H_(2)O_(2)` evolve oxygen at N.T.P.=22400 ml
0.00136 gm `H_(2)O_(2)` evolve oxygen at N.T.P.
`=(22400)/(68)xx0.00136=0.448`
For 0.1N, the solution is of 0.448 volume.
`therefore3N`, volume =0.448`xx3=1.344ml`.