1.

The weight of `1 L` of ozonised oxygen at `STP` was found to be `1.5 g`. When `100 mL` of this mixture at `STP` was treated with turpentine oil, the volume was reduced to `90 mL`. The molecular weight of ozone isA. 49B. 47C. 46D. 47.9

Answer» Correct Answer - D
Volume of `O_(3)` in `100 mL` of ozonised `O_(2)`
`= 100 - 90 = 10 mL` (dissolved in turpentine)
Volume of `O_(3)` in `1 L` of ozonised `O_(2)`
`= (10 x 1000)/(100) = 100 mL`
Volume of `O_(2)` in `1 L = 1000 - 100 = 900 mL`
Weight of `900 mL` of `O_(2)` at `STP = (900 xx 32)/(22400) = 1.286 g`
Weight of `100 mL "of" O_(3)` at `STP = 1.5 - 1.286`
`= 0.214 g`
Now `100 mL` of`O_(3)` at `STP = 0.214 g`
`22400 mL` of `O_(3)` at `STP` Weighs `= (0.214 xx 22400)/(100)`
`= 47.94 g`
Molecular weight of `O_(3) = 47.94`


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