Commercial sample of `H_(2)O_(2)` is labeled as 10V. Its % strength is nearly-

Options

A. 3
B. 6
C. 9
D. 12

Explanation

10V (10 volume) `H_(2)O_(2)` means that 1mL of `H_(2)O_(2)` solution will produce 10mL of `O_(2)` at STP.
Now, decomposition of `H_(2)O_(2)` is as below:
`underset(underset(" =68g ")(2 xx 34)g)(2H_(2)O_(2)) rarr 2H_(2) O + underset(underset(" at STP ")" 22400 mL ")(O_(2))`
`therefore` At STP, 10mL of `O_(2)` is obtained from 1mL of `H_(2)O_(2)`
`therefore` 22400mL of `O_(2)` is obtained from `(22400)/(10)` mL of `H_(2)O_(2)`
= 2240 mL of `H_(2)O_(2)`
`therefore` 22400 mL of `H_(2)O_(2)` solution contains 68g of `H_(2)O_(2)`
`therefore` 100 mL of `H_(2)O_(2)` solution contains `(68 xx 100)/(2240)` = 3.03g
of `H_(2)O_(2) cong 3g of H_(2)O_(2)`
`therefore` The percent strength of 10V `H_(2)O_(2)` is 3.