A sample of hydrogen was collected by water displacement at 23.0°C and an atmospheric pressure of 735 mmHg. Its volume is 568 mL. After water vapor is removed, what volume would the hydrogen occupy at the same conditions of pressure and temperature? (The vapor pressure of water at 23.0°C is 21 mmHg.)

V = 552 mL or 0.552 L

Explanation:

First, we need to calculate the number of moles of H2 using the ideal gas equation which is:

PV = nRT

Solving for n:

n = PV / RT

Where:

P = Pressure

V = Volume

R = Gas constant (0.082 L atm / K mol)

T = Temperature in K

Let's convert first both pressure in atm, remember that 1 atm = 760 mmHg

P = 735 / 760 = 0.967 atm

Pwater = 21 / 760 = 0.028 atm

Finally temperature to Kelvin:

T = 23 + 273.15 = 296.15 K

Now, at first the hydrogen was collected by water displacement so pressure is:

P = 0.967 - 0.028 = 0.939 atm

Now the moles of hydrogen:

n = 0.939 * 0.568 / 0.082 * 296.15

n = 0.022 moles

Now that we have the moles, let's calculate the volume when the pressure is 735 mmHg

V = nRT/P

V = 0.022 * 0.082 * 296.15 / 0.967

V = 0.552 L or 552 mL

This is the volume that hydrogen occupies.