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34 g of H(2)O(2) is present in 1120 " mL...

34 g of `H_(2)O_(2)` is present in 1120 " mL of " solution. This solution is called

A

10volume

B

20volume

C

30 volume

D

32volume

Text Solution

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The correct Answer is:
A
`34gH_(2)O_(2)` inj 1.12 L solution
1 mole `H_(2)O_(2)` in 1.12 L solution `implies (1)/(1.2)` mole in 1 L solution
`n_(O_(2))=(1)/(2.24)` mole =`(V_(O_(2)))/(22.4) implies V_(O_(2))=10`
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