To a 25 mL `H_(2)O_(2)` solution excess of an acidified solution of potassium iodide was added. The iodine liberated required 20 " mL of " 0.3 N sodium thiosulphate solution Calculate the volume strength of `H_(2)O_(2)` solution.

To a 25 ml H_2O_2 solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H_2O_2 solution.

To a 25 ml H_(2)O_(2) solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 ml of 0.3 N sodium thiosulphate solution. The volume strength of H_(2)O_(2) solution is

To a 25 ml H_(2)O_(2) solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 ml of 0.3 N sodium thiosulphate solution. The volume strength of H_(2)O_(2) solution is

25 mL of H_(2)O_(2) solution were added to excess of acidified solution of KI . The iodine so liberated required 20 mL of 0.1N Na_(2)S_(2)O_(3) for titration Calculate the strength of H_(2)O_(2) in terms of normalility, percentage and volumes. (b) To a 25 mL H_(2)O_(2) solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3N sodium thiosulphate solution. Calculate the volume strength of H_(2)O_(2) solution.

To a 25 mL of H_(2)O_(2) solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3 N Na_(2)S_(2)O_(3) solution. Calculate the volume strength of H_(2)O_2 solution. Strategy : Volume strength of H_(2)O_(2) solution is related to its normality by the following relation Volume strength (V)=5.6xx"Normality" (N) where, Normality =((meq)H_(2)O_(2))/V_(mL) According to the law of equivalence (meq)_(Na_(2)S_(2)O_(3))=(meq)_(I_(2))=(meq)_(H_(2)O_(2))

To a 25 mL H_2 O_2 solution, excess of acidified solution of KI was added. The iodine liberated required 20.0 mL of 0.3 N Na_2 S_2 O_3 solution. Calculate the volume strength of H_2 O_2 solution.