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1.22 g of a gas measured over water at 1...

1.22 g of a gas measured over water at 15°C and a pressure of 775 mm of Hg Occupied 900 ml. Calculate the volume of dry gas at vapour pressure of water at `15^(@)C` is 14 mm of Hg

A

372.21 mL

B

854.24 mL

C

869.96 mL

D

917.76 mL

Text Solution

Verified by Experts

The correct Answer is:
B
Pressure of dry gas = Pressure of moist gas - Vapour pressure of water = 775 - 14 = 761 mm
From , `(p_(1) V_(1))/(T_(1)) = (p_(2) V_(2))/(T_(2)) implies V_(2) = (p_(1) V_(1) T_(2))/(T_(1) p_(2)) = (761 xx 900 xx 273)/(288 xx 760) = 854.24 `mL
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Knowledge Check

  • 1.22 g of a gas measured over water at 15^(@)C and a pressure of 775 mm of mercury occupied 900 mL. Calculate the volume of dry gas at NTP (vapour pressure of water at 15^(@)C is 14 mm).

    A
    `372.21 mL`
    B
    `854.24 mL`
    C
    `869.96 mL`
    D
    `917.76 mL`

  • The volume of oxygen collected by the decomposition of potassium chlorate at 24^(@) C and atmospheric pressure of 760 mm Hg is 128 mL. Calculate the mass of oxygen gas obtained. The pressure of the water vapour at 24^(@)C is 22.4 mm Hg.

    A
    0.123 g
    B
    0.163 g
    C
    0.352 g
    D
    1.526 g

  • The volume of oxygen collected by the decomposition of potassium chlorate at 24^(@)C and atmospheric pressure of 760 mm Hg is 128 mL. Calculate the mass of oxygen gas obtained. The pressure of the water vapour at 24^(@)C" is "22.4 mm Hg .

    A
    `0.123 g`
    B
    `0.163 g`
    C
    `0.352 g`
    D
    `1.526 g`

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