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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

A sample of ...

Question

A sample of $O_{2}$ gas is collected over water at $23^{o} C$ at a barometric pressure of $751 m m$ Hg (vapour pressure of water at $23^{o} C$ is $21 m m$ Hg). The partial pressure of $O_{2}$ gas in the sample collected is:

21 m m H g

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751 m m H g

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0.96 a t m

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1.02 a t m

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Solution

The correct option is C $0.96 a t m$
Using the Dalton's law
P (Total) = Poxygen +Pwater
The vapor pressure of water is 21mmHg
751mmHg = Poxygen + 21 mmHg
Poxygen = 751 mmHg - 21 mmHg
Poxygen =730 mmHg
Convert the partial pressure of oxygen in to atmosphere
(730 mmHg) (1 atm / 760 mmHg) = 0.96 atm

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Similar questions

A sample of $O_{2}$ gas is collected over water at $23^{\circ}$ C at a barometric pressure of 751 mm Hg (vapour pressure of water at $23^{\circ}$ C is 21 mm Hg). The partial pressure of $O_{2}$ gas in the sample collected is ......

Q. A

3.00 L

oxygen gas is collected over water at

27^{o} C

when the barometric pressure is

787 T o r r

. If vapour pressure of water is

27 T o r r

27^{o} C

, moles of

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Q. A sample of hydrogen is collected by water displacement in the laboratory .The volume of the sample of hydrogen is 25

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at 24.0

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and barometric pressure for the day of 758

m m H g

. Find the pressure of the dry gas at this temperature ?

(The vapor pressure of water at 24.0

^{\circ} C

is 22.4

m m H g

O_{2}

is collected over water a t 20C.The pressure inside shown by the gas is 750 mm Hg. What is the pressure due to

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it vapour pressure of

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O is 18 mm at 20'C?

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of hydrogen is collected over water at

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A sample of O2 gas is collected over water at 23oC at a barometric pressure of 751mm Hg (vapour pressure of water at 23oC is 21mm Hg). The partial pressure of O2 gas in the sample collected is:

The correct option is C 0.96 atmUsing the Dalton's law P (Total) = Poxygen +PwaterThe vapor pressure of water is 21mmHg751mmHg = Poxygen + 21 mmHgPoxygen = 751 mmHg - 21 mmHgPoxygen =730 mmHgConvert the partial pressure of oxygen in to atmosphere(730 mmHg) (1 atm / 760 mmHg) = 0.96 atm

A sample of $O_{2}$ gas is collected over water at $23^{o} C$ at a barometric pressure of $751 m m$ Hg (vapour pressure of water at $23^{o} C$ is $21 m m$ Hg). The partial pressure of $O_{2}$ gas in the sample collected is:

The correct option is C $0.96 a t m$
Using the Dalton's law
P (Total) = Poxygen +Pwater
The vapor pressure of water is 21mmHg
751mmHg = Poxygen + 21 mmHg
Poxygen = 751 mmHg - 21 mmHg
Poxygen =730 mmHg
Convert the partial pressure of oxygen in to atmosphere
(730 mmHg) (1 atm / 760 mmHg) = 0.96 atm