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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

Water undergo...

Question

Water undergoes electrolysis to produce hydrogen and oxygen gas at $14$ , with the products collected above water. The vapour pressure of water at $14$ is $12$ mm Hg. The total pressure is $740$ mm Hg.
What is the pressure due to the electrolysis products $H_{2} (g) + O_{2} (g)$ ?

728

mm Hg

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628

mm Hg

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828

mm Hg

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528

mm Hg

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Solution

The correct option is A $728$ mm Hg
The vapour pressure of water at 14 $^{o} C$ is $12$ mmHg. The total pressure is $740$ mm Hg.
The pressure due to the electrolysis products $H_{2} (g) + O_{2} (g)$ is $740 - 12 = 728$ 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

14^{0} C

12 m m H g

. The total pressure is

740 m m H g

.
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

17^{o}

C and

750

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

[The water vapour pressure at

17^{o}

C is

14

mm Hg.]

$50 c m^{3}$ 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.

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____.

50 c m^{3}

of hydrogen is collected over water at

17^{\circ} C

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

17^{\circ} C

is 14 mm Hg.

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Water undergoes electrolysis to produce hydrogen and oxygen gas at 14, with the products collected above water. The vapour pressure of water at 14 is 12 mm Hg. The total pressure is 740 mm Hg.What is the pressure due to the electrolysis products H2(g)+O2(g)?

The correct option is A 728 mm HgThe vapour pressure of water at 14oC is 12 mmHg. The total pressure is 740 mm Hg.The pressure due to the electrolysis products H2(g)+O2(g) is 740−12=728 mm Hg .

Water undergoes electrolysis to produce hydrogen and oxygen gas at $14$ , with the products collected above water. The vapour pressure of water at $14$ is $12$ mm Hg. The total pressure is $740$ mm Hg.
What is the pressure due to the electrolysis products $H_{2} (g) + O_{2} (g)$ ?

The correct option is A $728$ mm Hg
The vapour pressure of water at 14 $^{o} C$ is $12$ mmHg. The total pressure is $740$ mm Hg.
The pressure due to the electrolysis products $H_{2} (g) + O_{2} (g)$ is $740 - 12 = 728$ mm Hg .