A chemist puts 200.0 mL of water into a 2.50 L container, and adds enough gaseous H2S to give a pressure of 180.2 torr at a temperature of 13.0˚C. Some of the H2S then dissolves in the water, causing the pressure in the container to drop to 155.9 torr. The temperature remains constant throughout this experiment. What is the molar concentration of H2S in the water at the end of the experiment?

0.016 mol/L

Explanation:

First, we need to know the number of moles of the gas at the beginning of the experiment.

Transforming the unites, 760 torr = 1 atm, so 180.2 torr = 0.237 atm (180.2/760), and 155.9 torr = 0.205 atm (155.9/760). The temperature must be em Kelvin so T = 13.0 + 273.15 = 286.15 K.

The constant of the gases R = 0.082 atm. L / (mol. K), and the volume of the gas, will be the total volume less the water volume, so:

V = 2500 - 200 = 2300 mL = 2.3 L

PV = nRT (ideal gas equation)

0.237x2.3 = nx0.082x286.15

0.5451 = 23.4643n

n = 0.0232 moles

At the final of the experiment, for the same volume and temperature, and P = 0.205 atm

0.205x2.3 = nx0.082x286.15

0.4715 = 23.4643n

n = 0.0200 moles

So, the number of moles that were dissolved in the water, is the initial less the final in the gas:

n = 0.0232 - 0.0200 = 0.0032 moles

The molar concentration (M) is the number of moles divided by the volume, so for 200 mL (0.2 L) of water

M = 0.0032/0.2

M = 0.016 mol/L