Use the equation for the ideal gas law to determine moles of the gas. Then divide the mass of gas by the moles to get molar mass.
#PV=nRT#,
where:
#P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins.
Since the given value for #R# is in #"L atm"/"mol K"#, you need to convert the pressure in #"mmHg"# to #"atm"#.
#"1 atm"##=##"760.0 mmHg"#
#836color(red)cancel(color(black)("mmHg"))xx("1 atm")/(760.0color(red)cancel(color(black)("mmHg")))="1.10 atm"#
The given temperature is #25^@"C"#, so you need to convert it to Kelvins.
#25^@"C + 273.15"="298 K"#
Organize the data:
Known/Given
#P="1.10 atm"#
#V="2.00 L"#
#R="0.0821 L atm K"^(-1) "mol"^(-1)"#
#T="298 K"#
Unknown
#n#
Solution
Rearrange the equation to isolate #n#. Plug in the known values and solve.
#n=(PV)/(RT)#
#n=((1.10color(red)cancel(color(black)("atm")))xx(2.00color(red)cancel(color(black)("L"))))/((0.0821 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1))xx(298color(red)cancel(color(black)("K"))))="0.0899 mol"#
Molar mass of gas
#M=("15.0 g")/("0.0899 mol")="167 g/mol"# (rounded to three significant figures)
Molecular weight is numerically equal to the molar mass.
#"M"_r="167 u"#