trong>Exercise 13.10 Consider a sample of hydrogen gas collected over water at 25 ° C where the vapor pressure of water is 24 torr . The volume occupied by the gaseous mixture is 0. 5 00  L , and the total pressure is 0. 95 0  atm . Calculate the partial pressure of H 3 and the number of moles of H 2 present.

Interpretation:

The partial pressure of H₂ and the number of moles of H₂ present in the sample should be determined.

Concept Introduction:

The problem can be solved using ideal gas law and Dalton’s gas law. The total number of moles (moles of hydrogen gas plus moles of water vapor) can be calculated using ideal gas law as follows

PV=nRTn=PVRT

Where, P is the total pressure of the mixture

V is the volume of the mixture

R is the gas constant

T is the absolute temperature and,

n is the total number of moles of hydrogen gas and water vapor.

Now,

n = nH2+ nH2O

nH2=n - nH2O

nH2O can be determined using the Daltons law as follows:

PH2O= xH2O.Ptotal = (nH2Ontotal). Ptotal

Once nH2 is calculated, one can use Dalton’s law to find out the partial pressure of hydrogen gas in the mixture as follows:

PH2 = xH2 ?P_{t o t a l}

Where PH2 is the partial pressure of hydrogen gas in the mixture

xH2 is the mole fraction of hydrogen gas and,

P_{t o t a l} is the total pressure of the mixture.

Here given,

Vapor pressure of water = 24 torr

Total pressure of the mixture = 0.950 atm = 0.950 ?760 torr = 722 torr

Hence, mole fraction of water xH2O = (24 torr/722 torr) = 0.033

Now the total number of moles of hydrogen and water vapor can be determined using ideal gas equation as follows:

PV=nRTn=PVRT=0.950 atm×0.500 L0.082 L.atm/mol.K×298 Kn = 0.0194 moles

Now, mole fraction of water vapor is given as,

xH2O= nH2On

nH2O = 0