A sample of hydrogen gas is mixed with water vapor. The mixture has a total pressure of 755 torr and the water vapor has a partial pressure of 24 torr. What amount (in moles) of hydrogen gas is contained in 1.55 L of this mixture at 298 K?

Question

Bronson Noble

Chemistry

10 April, 10:52

A sample of hydrogen gas is mixed with water vapor. The mixture has a total pressure of 755 torr and the water vapor has a partial pressure of 24 torr. What amount (in moles) of hydrogen gas is contained in 1.55 L of this mixture at 298 K?

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Answers (1)

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Luciano Leblanc · Answer 1

The moles of hydrogen that contained in 1.55 L calculated using the ideal gas equation

that is PV = nRT where n is number of moles

by making n the subject of formula n = Pv/RT

P = total pressure - partial pressure of water vapor = 755-24 = 731 torr

V=1.55 l

T=298 k

R (gas Constant) = 62.364 l. torr/mol. k

n = (731 torr x 1.55 L) / (62.364 l. torr/mol. k x298 K) = 0.06 moles of hydrogen