Question

A 32.6mL sample of hydrogen gas is collected over water using water displacement. The temperature of the gas is 295K. The total pressure of the mixture is 785.2 torr. At 295K the vapor pressure of water is 19.8 torr.Find the pressure of hydrogen gas alone, using the volume and temperature as listed above, calculate moles of hydrogen and moles of water vapor. Show work

Answer 1

We know that total pressure of mixture of gases = sum of parital pressure of individual gases

So P_total = P_{water vapour} + P_H2

785.2 Torr = 19.8 + P_H2

So parital pressure of hydrogen gas = 765.4 torr

Volume of hydrogen gas = 32.6 moles

temperature = 295 K

Calcualte number of moles of hydrogen gas:

We know that

PV=nRT

R = 62.363 L?Torr?K^?1? ?mol^?1

number of moles = PV / RT = 765.4 X 0.0326 / 62.363 X 295 = 0.0013 moles

Similarly for water vapour

PV = nRT

n = PV / RT = 19.8 X 0.0326 / 295 X 62.363 = 3.5 X 10^-5 moles