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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)`

Text Solution

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The correct Answer is:
4
`n=(PV)/(RT)= n(2xx224xx10^(-3))/(0.0821xx273)`
`n=(2xx224xx10^(-3))/(22.4) rArr n=2xx10^(-2) "moles" of O_(2)`
`2H_(2)O_(2) to 2H_(2)O+O_(2)`
`4xx10^(-2)` moles `2xx10^(-2)` moles
`M=(n)/(Vlit)=(4xx10^(-2)xx1000)/(10), M=4`
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