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Calculate the amount of H(2)O(2) present...

Calculate the amount of `H_(2)O_(2)` present is 600 mL of 10 volume `H_(2)O_(2)` solution.

Text Solution

Verified by Experts

10 volume `H_(2)O_(2)` mean that 1 mL of `H_(2)O_(2)` solution will
produce 10 mL of `O_(2)` at STP .
`2H_(2)O_(2) rarr 2H_(2)O + O_(2)`
`" "(2 '34)`g = 68 g `" "` 22400 mL (at STP)
At STP, 10 mL `O_(2)` is obtained from 1 mL of 10vol `H_(2)O_(2)` solution.
`therefore` 22400 mL `O_(2)` is obtained from `(22400)/(10)` mL of 10 vol
`H_(2)O_(2)` solution = 2240 mL of `H_(2)O_(2)`
2240 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 xx100)/(2240)` = 3.03g of `H_(2)O_(2)`.
So, 600 mL of `H_(2)O_(2)` solution contains = `(3.03 xx 600)/(100)` g of `H_(2)O_(2)` = 18.18g of `H_(2)O_(2) cong ` 18.2g of `H_(2)O_(2)`
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