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Calculate the volume in `mL` hydrogen peroxide labelled `10` volume required to liberate `600 mL` of oxygen at `27^(@)C` and `760 mm`. |
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Answer» `10V` of `H_(2)O_(2)` `1V` of `H_(2)O` gives `10V` of `O_(2)` at `STP`. `1 mL` of `10 V H_(2)O_(2)` gives `10 ml` of `O_(2)`. Volume of `O_(2)` required at `NTP` `{:(V_(1)= 600, P_(1)=760 mm),(V_(2)=?, P_(2)=760 mm),(T_(1)=300K, T_(2)=273K):}` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))=600xx(273)/(300)=546 mL` `10 V` of `H_(2)O_(2)` produces `10 mL` of `O_(2)` at `STP` for `1 mL` of its volume. Hence, for `546 mL` of `O_(2)` at `NTP`, the volume of `10 V` of `H_(2)O_(2)` required `=(546xx1)/(10)=54.6 mL` (Alternative method) Volume of `O_(2)` at `STP=?` `600 mL` of `O_(2)` at `27^(@)C(273+27=300 K)` and `760 mm` or `1 atm` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `{:(P_(1)= 1 atm, P_(2)=1 atm),(V_(1)=600 mL, V_(2)=?),(T_(1)=300K, T_(2)=273K):}` `(1xx600)/(300)=(1xxV_(2))/(273)` `V_(2)=(600xx273)/(300)=546 mL` Volume strength of `H_(2)O_(2)xx` Volume of `H_(2)O_(2)` `=` volume of `O_(2)` at `STP` `10 Vxx x=546 mL` `x=(546)/(10)=54.6 mL` Volume of `H_(2)O_(2)=54.6 mL` |
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