1.

Calculate the strengthof 20 V of H_(2)O_(2) in terms of (i) normality

Answer»

Solution :The strength of `H_(2)O_(2)`as . 20 V. means 1 volume of `H_(2)O_(2)` decomposition gives 20 volumes of oxygen at NTP or 1 LITRE of `H_(2)O_(2)` gives 20 litresof oxygen at NTP .
(i) `2H_(2)O_(2) to 2H_(2)O + O_(2)`
`1 lit "" 20 " lit at NTP "` .
` 1 lit"" 20/(5.6) eq . ""...(Eqn.4ii)`
`{:(( :. " 1 mole oxygen (32 g) occupies a vol. of 22 . li at NTP"),( :. " 1 eq of oxygen(8 g) shall occupy "(22.4)/4" lit . at NtP = 5.6 lit")):}`
`:. `equivalent in 1 lit . of `H_(2)O_(2)`= eq . of oxygen produced
` 20/(5.6) = 3.58`
` :. ` equivalent per litre REPRESENTS normality
` :. ` normality of 20 V `H_(2)O_(2) = 3.58 N `


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