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The volume strength of 1*5 N H(2)O(2) so...

The volume strength of `1*5` N `H_(2)O_(2)` solution is

Text Solution

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Normality of 10 volume `H_(2)O_(2)` = 1.78 N
Thus, volume strength of 1.78 N solution = 10
Volume strength of 1.5 N `H_(2)O_(2) = 1.5 xx 10/1.78 = 15/1.78 = 8.4`
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