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

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

A

`10 vol`

B

`20 vol`

C

`34 vol`

D

`32 vol`

Text Solution

Verified by Experts

The correct Answer is:
A
`N=(W_(2)xx1000)/(EW_(2)xx"vol of sol(in mL)")`
`=(34xx1000)/(17xx1120)=200/112`
`1N H_(2)O_(2)=5.6` volumes strength
`200/112 NH_(2)O_(2)=5.6xx200/112 =10 V`
Hence, the correct option is (a)
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