Question

A cylinder of compressed gas rolls off a boat and falls to the bottom of a lake. Eventually it rusts and the gas bubbles to the surface. A chemist collects a sample of the gas with the idea of trying to identify the gas. The wet gas collected occupies a volume of 281 mL at a pressure of 803 torr and temperature of 28.0^oC. The vapor pressure of water at 28.0^oC is 0.0372 atm.

1. Calculate the volume (L) that the gas occupies after it is dried (the water vapor removed) and stored at STP.

The mass of the dry gas is 972 mg. A fragment of the tank indicates that the gas is a monoatomic element.

2. Write out the full name of the gas.

Answer 1

1. Pressure of the gas = Pressure of the wet gas - vapor pressure of water. This is the Dalton's Law of partial pressure.

803 torr = 1.056 atm

Pressure of the gas = 1.05-0.0372 = 1.019 atm

PV=nRT

1.019*0.281=n*0.0821*301

n=0.0116 mol

Volume at STP = nRT/P

V = (0.0116*0.0821*273)/1

V = 0.26 L = 260 mL

No. of moles = mass/molar mass

0.0116 = 0.972/molar mass

molar mass = 83.79 g/mol

2. Since the gas is monoatomic element, the molar mass corresponds to Krypton gas.