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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

`10 vol` solution

B

`20 vol` solution

C

`34 vol` solution

D

`32 vol` solution

Text Solution

Verified by Experts

The correct Answer is:
A
`N=(W_(2)xx1000)/(EW_(2)xxVmL)`
`=(34xx1000)/(17xx1120)=200/12`
`1 N of H_(2)O_(2)=5.6 ` vol strength
`200/12 N of H_(2)O_(2)=5.6xx200/12=10` vol strength
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