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If `100 mL` of acidified `2NH_(2)O_(2)` is allowed to react with `KMnO_(4)` solution till there is light tinge of purples colour, the volume of oxygen produced at `STP` is :A. `2.24 L`B. `1.12 L`C. `3.36 L`D. `4.48 L` |
Answer» Correct Answer - A Volume of `O_(2)` at `STP` `=100mLxx11.2mL` volume strength `=1120 mL` of `O_(2)at STP` Since `1N=5.6` volume strength of `H_(2)O_(2)` `2N=11.2` volume strength of `H_(2)O_(2)` Volume of `O_(2)` produced by `H_(2)O_(2)=1120 mL` Same volume of `O_(2)` will be produced by `KMnO_(4)`, i.e., `1120 mL` |
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