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What is the mass of hydrogen peroxide present in 1 litre of 2M solution ? Calculate the volume of oxygen at STP liberated upon complete decomposition of 100 cm^(3) of the above solution. |
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Answer» Solution :Step 1. To calculate the mass of `H_(2)O_(2)` in 1 litre of 2M solution. Molecular mass of `H_(2)O_(2)=2xx1+2xx16=34` u. By definition, 1 litre of 1 M `H_(2)O_(2)` contains 34 g of `H_(2)O_(2)` `therefore` 1 litres of 2M `H_(2)O_(2)` will contain `H_(2)O_(2)=34xx2=68` Step 2. To calculatethe volume of `O_(2)` LIBERATED at STP from `100 cm^(3)` of 2M solution. 1 litre of 2M `H_(2)O_(2)` solution will contain `H_(2)O_(2)=(68)/(1000)xx100=6.8g` The equation representing the DECOMPOSITION of `H_(2)O_(2) ` is `underset(2xx34=68 g)(2H_(2)O_(2))to 2H_(2)O + underset(22400 cm^(3) at "STP")(O_(2))` Now 68 g of `H_(2)O_(2)` at STP give `O_(2)=22400 cm^(3)` `therefore 6.8 ` g of `H_(2)O_(2)` at STP will evolve `O_(2)=(22400)/(68)xx6.8=2240 cm^(3)=2.24 litres.` |
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