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100 mL of ozone at STP was passed throug...

100 mL of ozone at STP was passed through 100 mL of 10 volume `H_(2)O_(2)` , solution. What is the volume strength of `H_(2)O_(2)`, after the reaction?

A

`9.5`

B

`9.0`

C

`4.75`

D

`4.5`

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(34g)(H_(2)O_(2)) + underset(48 g)(O_(3)) rarr H_(2) O + underset(224L)(O_(2))`, 100 mL 10 Volume `H_(2) O_(2) = (68xx10)/(22.4) xx (100)/(1000) = 3.0357g`
`22.4 L O_(3) = 48 g O_(3) implies` 100 ml `O_(3) = (48xx100 mL)/(22400 mL) =0.2143 g`
48 g `O_(3)= 34g H_(2) O_(2) , :. 0.2143 g O_(3) = ((34xx0.2143)/(48))= 0.1518 g H_(2) O_(2)`
Quantity of `H_(2) O_(2) ` remaining after passing 100 mL `O_(3)=3.0357-0.1518 = 2.8839 g = 28.839 gL^(-1)`
Volume strength (x) of `H_(2) O_(2)` remaining `= (68xx x)/(22.4)= 28.839 x = (22.4xx 28.839)/(68 ) =9.5 `
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