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If the ration of the masses of SO(3) and...

If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would be

A

`5:2`

B

`2:5`

C

`2:1`

D

`1:2`

Text Solution

Verified by Experts

The correct Answer is:
B
`"Let mass of "SO_(3)=x," mass of "O_(2)=x`
`"Moles of "SO_(3)=(x)/(80)," Moles of "O_(2)=(x)/(32)`
`"Mole fraction of "SO_(3)=((x)/(80))/((x)/(80)+(x)/(32))`
`"Mole fraction of "O_(2)=((x)/(32))/((x)/(80)+(x)/(32))`
`"Ratio of mole fractions, "(x(SO_(3)))/(x(O_(2)))=(32)/(80)=(2)/(5)`
Since partial pressure are proportional to mole fraction
`therefore" Ratio of partial pressures "=2:5`
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Knowledge Check

  • If the ratio of the masses of SO_3 and O_2 gases confined in a vessel is 1 : 1, then the ratio of their partial pressures would be

    A
    `5:2`
    B
    `2:5`
    C
    `2:1`
    D
    `1:2`

  • Equal masses of He, O_(2) and SO_(2) are taken in a closed container. The ratio of the partial pressures of gases He, O_(2) and SO_(2) would be

    A
    `1:2:8`
    B
    `8:16:1`
    C
    `16:2:1`
    D
    `1:4:16`

  • A gaseous mixture of 2 moles of A, 3 moles of B, 5 moles of C and 10 moles of D is contained in a vessel. Assuming that the gases are ideal and the partial pressure of C is 1.5 atm, the total pressure is

    A
    3 atm
    B
    6 atm
    C
    9 atm
    D
    15 atm

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