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The mass of one litre sample of ozonised...

The mass of one litre sample of ozonised oxygen at `NTP` was found to be `1.5 g`. When `100 mL` of this mixture at `NTP` were treated with terpentine oil, the volume was reduced to `90 mL`. Hence calculate the molecular mass of ozone.
(Terpentine oil absorbs ozone)

Text Solution

Verified by Experts

As ozone is absorbed by turpentine oil, therefore, volume of ozone in 100mL of the mixture
`=100-90=10mL`
`therefore" "O_(2)" in the mixture"=100-10=90mL`
As 1 L of the mixture weigh = 1.5 g, therefore, average molar mass of the mixture (mass of 22.4 L at S.T.P.)`=1.5xx22.4=33.6"g mol"^(-1)`
`"Ratio of ozone " : "oxygen in the mixture "=10:90`
It m is the molecular mass of ozone, then
`"Average mol. mass of mixture "=(10xxm+90xx32)/(100)="33.6 (calculated above)"`
`"or "m+288=336 " or "m=48`
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Knowledge Check

  • 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 is

    A
    49
    B
    47
    C
    46
    D
    47.9

  • The weight of 1 litre of ozonised at STP was found to be 1.5 gm. 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 is

    A
    49
    B
    47
    C
    46
    D
    47.9

  • 600 mL of ozonised oxygen at STP were found to weigh one gram. What is the volume of ozone in the ozonised oxygen?

    A
    `200 mL`
    B
    `150 mL`
    C
    `100 mL`
    D
    `50 mL`

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