InterviewSolution
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InterviewSolution
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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 beA. 1.34 mlB. 1.44 mlC. 1.60 mlD. 2.42 ml |
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Answer» 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`. |
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