Question #eac7c
1 Answer
Explanation:
This is a great example of how you can use a variation of the ideal gas law equation to find the molar mass without finding the number of moles first.
As you know, a compound's molar mass tells you the mass of one mole of that substance. This means that you can find a compound's molar mass by diving the mass of a sample of said compound by the total number of moles present in that sample
#color(blue)("molar mass" = M_M = "mass"/"no. of moles" = m/n = ["grams"/"mol"])#
This of course implies that you can find the number of moles contained in a sample of a given compound by dividing the mass of the sample by the molar mass of the compound
#M_M = m/n implies n = m/M_M#
Now, the most common form of the ideal gas law equation looks like this
#color(blue)(PV = nRT)" "# , where
Notice that you can replace the number of moles by using the mass and the molar mass of the gas
#PV = overbrace(m/M_M)^(color(red)(=n)) * RT#
This variation of the ideal gas law equation will allow you to find the molar mass of the gas without calculating the number of moles present in the sample.
So, rearrange to isolate
These conversion factors will thus come in handy
#"1 atm " = " 760 torr"#
#"1 g" = 10^3"mg"#
#"1 L" = 10^3"mL"#
Also, keep in mind that the molar mass is expressed in grams per mole!
#PV = m/M_M * RT implies M_m = (m * RT)/(PV)#
#M_M = (10.1 * 10^(-3) "g" * 0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 23)color(red)(cancel(color(black)("K"))))/(10/760color(red)(cancel(color(black)("atm"))) * 255 * 10^(-3)color(red)(cancel(color(black)("L"))))#
#M_M = "73.190 g/mol"#
Now, you should round this off to one sig fig, the number of sig figs you have for the pressure of the gas, but I'll leave it rounded to two sig figs, just for good measure
#M_M = color(green)("73 g/mol")#