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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)xxV_(mL))`
`N=(34xx1000)/(17xx1120)=(200)/(112)`
`1N of H_(2)O_(2)=5.6` volume strength.
`(200)/(112)N of H_(2)O_(2)=5.6xx(200)/(112)=10V`
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