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10 ml of H(2)O(2) could release 224 ml o...

10 ml of `H_(2)O_(2)` could release 224 ml of `O_(2)` at 273K and 2 atm pressure. What is the molarity of that `H_(2)O_(2)`

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The correct Answer is:
4
`n = (PV)/(RT) n = (2 xx 224 xx 10^(-3))/(0.0821 xx 273)`
`n =(2xx 224 xx 10^(-3))/(22.4) rArr n = 2xx 10^(-2)` moles of `O_2`
`2H_2O_2 to 2H_2O + O_2`
`4 xx 10^(-2) ` mole ` 2xx 10^(-2)` moles
`M =n/(V"lit") = (4xx 10^(-2) xx 1000)/(10) , M =4`
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