v37
v37
A
`30%`
B
`3%`
C
`1%`
D
`10%`
Text Solution
Verified by Experts
The correct Answer is:
B
“10 volume `H_(2)O_(2)`” means 1 mL of its solution on decomposition at NTP give 10 mL oxygen gas. Volume of oxygen formed from 100 mL of solution at NTP = 1000 mL
`undersetunderset(2xx34g)("2 mol")(2H_(2)O_(2))rarr2H_(2)O+undersetunderset("22400mL")("1 mol")(O_(2))`
`because 22400 mL O_(2)` formed at NTP by decomposition of 68 g `H_(2)O_(2)`.
`therefore 1 mL O_(2)` formed at NTP from `(68)/(22400)gH_(2)O_(2)`
`therefore 1000 mL O_(2)` formed at NTP from `(68xx1000)/(22400)gH_(2)O_(2)`
`=3.035gH_(2)O_(2)`
So ..10 volume `H_(2)O_(2)`.. = 3.0% approximately.
`undersetunderset(2xx34g)("2 mol")(2H_(2)O_(2))rarr2H_(2)O+undersetunderset("22400mL")("1 mol")(O_(2))`
`because 22400 mL O_(2)` formed at NTP by decomposition of 68 g `H_(2)O_(2)`.
`therefore 1 mL O_(2)` formed at NTP from `(68)/(22400)gH_(2)O_(2)`
`therefore 1000 mL O_(2)` formed at NTP from `(68xx1000)/(22400)gH_(2)O_(2)`
`=3.035gH_(2)O_(2)`
So ..10 volume `H_(2)O_(2)`.. = 3.0% approximately.
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