If a sample of Cl2 gas in a 1000 mL container is exerting a pressure of 14.2 atm, how many grams of chlorine are present? The temperature of the container is 256 K.

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Mass of chlorine present

In this case, you know:

P= 14.2 atm
V= 1000 mL= 1 L
T= 256 K
R= 0.082 [tex]\frac{atmL}{molK}[/tex]
n= ?

Replacing in the ideal gas law:

14.2 atm× 1 L = n× 0.082 [tex]\frac{atmL}{molK}[/tex]× 256 K

Solving:

(14.2 atm× 1 L)÷ (0.082 [tex]\frac{atmL}{molK}[/tex]× 256 K)= n

0.676 moles= n

Being the molar mass of chlorine Cl₂ 70.9 g/mole, the mass of the compound can be calculated as:

0.676 moles× 70.9 g/mole= 47.9284 grams.

Finally, the mass of chlorine present is 47.9284 grams.

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