1.

The strength of 10 volumeof H_(2)O_(2) solution is

Answer»

10
68
60.7
30.36

Solution :10 vol means that 1 ml of `H_(2)O_(2)` GIVES
10 ml of `O_(2)` at NTP
`2xx34 22400 ml`
=68 g
now from aboveequation it is clear that 22400 ml`O_(2)` at NTP is obtained from68 g of `H_(2)O_(2)`
`THEREFORE ml` of `O_(2)` at NTP is obtained from
`=(68)/(22400)xx10`
`=0.03035 g H_(2)O_(2)`
1000 ml will contain =`0.03035 xx1000 g`
`=30.35 g H_(2)O_(2)`


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