A 283.3-g sample of #X_2(g)# has a volume of 30 L at 3.2 atm and 27°C. What is element #X#?

1 Answer
Feb 13, 2016

Answer:

Chlorine, #"Cl"#.

Explanation:

Your strategy here is to apply the ideal gas law equation to find the number of moles of gas present in your sample.

In order to be able to identify the unknown element, you need to know its molar mass. A substance's molar mass tells you the mass of one mole of that substance.

The important thing to notice here is that you're given a sample of #"X"_2# and asked for the identity of element #"X"#. Keep this in mind.

Since you know the mass of your sample, you can use the number of moles it contains to determine the element's molar mass.

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "#, where

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant, usually given as #0.0821 ("atm" * "L")/("mol" * "K")#
#T# - the absolute temperature of the gas

Rearrange the equation to solve for #n#

#PV = nRT implies n = (PV)/(RT)#

Plug in your values - do not forget to convert the temperature of the gas from degrees Celsius, to Kelvin

#n = (3.2 color(red)(cancel(color(black)("atm"))) * 30color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 27)color(red)(cancel(color(black)("K")))) = "3.896 moles"#

So, if #3.896# moles of this element have a mass of #"283.3 g"#, it follows that one mole will have a mass of

#1color(red)(cancel(color(black)("mole"))) * "283.3 g"/(3.896color(red)(cancel(color(black)("moles")))) = "72.7 g"#

Since one mole of #"X"_2# has a mass of #"72.7 g"#, it follows that its molar mass will be equal to #"72.7 g mol"^(-1)#.

Now, a molecule of #"X"_2# contains two atoms of #"X"#. This means that the molar mass of the element #"X"# will be half that of the molecule

#M_M = 1/2 * "72.7 g mol"^(-1) = "36.25 g mol"^(-1)#

The closest match to this value is the molar mass of chlorine, #"Cl"#, which is listed as #"36.453 g mol"^(-1)#. Therefore, element #"X"# is chlorine, #"Cl"#.