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Calculate the strength of 5 volume `H_(2)O_(2)` solution. |
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Answer» By definiation , 5 volume `H_(2)O_(2)` solutions means that 1 L of 5 volume `H_(2)O_(2)` solution on decomposition produces 5 L of `O_(2)` at NTP. Consider the decompositon reaction, `2H_(2)O_(2) to 2H_(2)O + O_(2)` `2xx34g to 22.7 ` L at NTP Now 22.7 L `O_(2)` at NTP will be obtained from `H_(2)O_(2)` =68 g `therefore 5 `L of `O_(2)` at NTP will be obtained from `H_(2)O_(2)=(68xx5)/(22.7) g` `" "=14.98g =15g ` But 5 L of `O_(2)` at NTP is produced from 1 L of 5 volume `H_(2)O_(2)` `therefore` Strength of `H_(2)O_(2)` solution =` 15 g L^(-1)` or percentage strength of `H_(2)O_(2)` solution `=(15)/(1000)xx100=1.5%` |
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