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To a 25 mL H(2)O(2) solution excess of a...

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.

A

1.34 mL

B

1.44 mL

C

1.60 mL

D

2.42 mL

Text Solution

Verified by Experts

The correct Answer is:
A
20 mL of 0.3N `Na_(2)S_(2)O_(3)`
= 20mL of 3.0 2 N `I_(2)` solution
= 20 mL of 3.0 N `H_2 O_2` solution
= 25 mL of 08N `H_(2)O_(2)` solution
Mass of 100 mL `H_(2)O_(2)` solution ` = (0.80 xx 17 xx 100)/(1000)`
` = 0.136 g`
` % = 0.136`
68 g `H_(2)O_(2)` evolve oxygen at N.T.P. = 22400mL
0.00136 g `H_(2)O_(2)` evolve oxygen at N.T.P
` = (22400)/(68) xx 0.00136 = 0.448`
For 0.1N , the solution is of 0.448 volume.
`therefore ` 3N , volume = .0 448 × 3 = .1 344 mL
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