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Weight of 112 ml of oxygen at NTP on liq...

Weight of `112 ml` of oxygen at `NTP` on liquefaction would be

A

0.32 g

B

0.64 g

C

0.16 g

D

0.96 g

Text Solution

Verified by Experts

The correct Answer is:
C
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The weight of 1 L sample 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. Calculate the molecular weight of ozone.

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Knowledge Check

  • The weight of 112 m L of oxgen at NTP is

    A
    `0.64g`
    B
    `0.96g`
    C
    `0.32g`
    D
    `0.16g`
  • 5.6 litres of oxygen at NTP is equivalent to

    A
    1 mole
    B
    `1/2` mole
    C
    `1/4` mole
    D
    `1/8` mole
  • 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
  • Similar Questions

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    Calculate the volume of oxygen at N.T.P. that would be required to convert 5.2 L of carbon monoxide to carbon dioxide.

    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)

    Find the number of molecules per unit volume of oxygen at NTP. [Mass of an oxygen molecule = 5.313 xx 10^(-26) kg, rms speed of oxygen molecles at NTP = 461.2 m/s]

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    The ratio of the weight of one litre of a gas to the weight of 1.0 L of oxygen gas both measured at STP is 2.22. The molecular weight of the gas would be :