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Byju's Answer

Standard XII

Chemistry

Raoult's Law

What is the t...

Question

What is the total pressure of the dry gas collected if the vapor pressure due to water at $25^{o} C$ is $24 m m$ $H g$ ?

24 m m

H g

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240 m m

H g

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736 m m

H g

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760 m m

H g

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Solution

The correct option is C $736 m m$ $H g$
Since, the total vapour pressure = atmospheric pressure = 760 mm Hg
therefore, the total pressure of dry gas = 760 - vapour pressure of water
the total pressure of dry gas
$= 760 - 24 = 736$ mm Hg

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Similar questions

Q. Water undergoes electrolysis to produce hydrogen and oxygen gas at

14^{0} C

, with the products collected above water. The vapor pressure of water at

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12 m m H g

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740 m m H g

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If the partial pressure of oxygen is

242.7 m m H g

, what is the mole fraction of hydrogen gas?

50 c m^{3}

of hydrogen is collected over water at

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and 750 Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at

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50

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of hydrogen is collected over water at

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mm Hg pressure. Calculate the volume of dry gas at S.T.P.

[The water vapour pressure at

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14

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$50 c m^{3}$ of hydrogen is collected over water at $17^{\circ} C$ and 750 mm Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at $17^{\circ} C$ is 14 mm Hg.

Q. Vapor pressure of pure water at

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What is the total pressure of the dry gas collected if the vapor pressure due to water at 25oC is 24 mm Hg?

The correct option is C 736 mm HgSince, the total vapour pressure = atmospheric pressure = 760 mm Hgtherefore, the total pressure of dry gas = 760 - vapour pressure of water the total pressure of dry gas =760−24=736 mm Hg

What is the total pressure of the dry gas collected if the vapor pressure due to water at $25^{o} C$ is $24 m m$ $H g$ ?

The correct option is C $736 m m$ $H g$
Since, the total vapour pressure = atmospheric pressure = 760 mm Hg
therefore, the total pressure of dry gas = 760 - vapour pressure of water
the total pressure of dry gas
$= 760 - 24 = 736$ mm Hg