Question #66004
1 Answer
Explanation:
The idea here is that you need to use the ideal gas law equation to find a relationship between the density of the gas and its molar mass.
As you know, the ideal gas law equation looks like this
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where
Now, the number of moles can be expressed using the mass,
#color(purple)(|bar(ul(color(white)(a/a)color(black)(n = m/M_M)color(white)(a/a)|)))#
Plug this into the ideal gas law equation to get
#PV = m/M_M * RT" " " "color(red)("(*)")#
Density is defined as mass per unit of volume. If you take
#color(purple)(|bar(ul(color(white)(a/a)color(black)(rho = m/V)color(white)(a/a)|)))#
Notice that you can rearrange equation
#PV * M_M = m/color(red)(cancel(color(black)(M_M))) * RT * color(red)(cancel(color(black)(M_M)))#
Now divide both sides by
#(P * color(red)(cancel(color(black)(V))) * M_M)/color(red)(cancel(color(black)(V))) = m/V * RT#
Finally, isolate
#color(purple)(|bar(ul(color(white)(a/a)color(black)(M_M = rho * (RT)/P)color(white)(a/a)|)))#
Now, STP conditions are defined as a pressure of
#"1 atm " = " 101.325 kPa"#
Plug in your values to get
#M_M = 1.5"g"/color(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 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))#
#M_M = color(green)(|bar(ul(color(white)(a/a)"34 g mol"^(-1)color(white)(a/a)|))) -># rounded to two sig figs
SIDE NOTE Many sources still use STP conditions as a pressure of *
If this is the definition of STP given to you, simply redo the calculations using these values for pressure and temperature. In this particular case, the answer will be the same because it must be rounded to two sig figs.