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What is the molar mass of 31.3 g of gas exerting a pressure of 5.71 atm on the walls of a 3.04 L container at 297 K?

2 Answers
Mar 9, 2018

Answer:

The molar mass of the gas is #43.946# grams per mole or #43.9# grams per mole using significant figures.

Explanation:

Use the ideal gas law #PV=nRT#.

#P# is the pressure in atmospheres (atm)

#V# is the volume in liters (L)

#n# is the number of moles

#R# is the universal gas constant (#.08206# # L*atm*K^(-1)*mol^(-1)#)

#T# is the temperature in kelvin

Notice that number of moles is #(grams)/(molar mass)# or #g/x# where x is the molar mass of the gas.

Plug it into the ideal gas law for #n#.

#PV=g/xRT#

Rearrange and isolate x.

#x=(gRT)/(PV)#

Plug everything in (excluding units)

#x= (31.3*.08206*297)/(5.71*3.04)#

#x= 43.946 g/(mol)#

(If you cancel out all the units, you will find that you will get #g/(mol)# which is the correct unit for molar mass.)

Mar 9, 2018

Answer:

The molar mass of the gas is #"43.9 g/mol"#.

Explanation:

Use the ideal gas law equation to determine moles. Then divide the given mass 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.

Organize the data:

Known

#P="5.71 atm"#

#V="3.04 L"#

#R="0.0820575 L atm K"^(-1) "mol"^(-1)"#

#T="297 K"#

Unknown

#n"#

Calculate moles

Rearrange the ideal gas law equation to isolate #n#. Plug in the known values and solve.

#n=(PV)/(RT)#

#n=((5.71color(red)cancel(color(black)("atm")))xx(3.04color(red)cancel(color(black)("L"))))/((0.0820575color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1))xx(297color(red)cancel(color(black)("K"))))="0.712254 mol"# (I am keeping some guard digits to reduce rounding errors.)

Determine molar mass

Divide the given mass by the calculated moles.

#M=("31.3 g")/("0.712254 mol")="43.9 g/mol"# (rounded to three significant figures)