What is the molar mass of a gas that has a density of 1.02 g/L at 0.990 atm pressure and 37°C?

1 Answer
May 30, 2016

Answer:

#"26.3 g mol"^(-1)#

Explanation:

Your starting point here will be the ideal gas law equation

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#, 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

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas, #rho#, under those conditions for pressure and temperature, and its molar mass, #M_M#.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass #m# of this gas, you can express its molar mass as the ratio between #m# and #n#, the number of moles it contains

#color(blue)(|bar(ul(color(white)(a/a)color(black)(M_M = m/n)color(white)(a/a)|)))" " " "color(orange)((1))#

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass #m# of this gas, you can express its density as the ratio between #m# and the volume it occupies

#color(blue)(|bar(ul(color(white)(a/a)color(black)(rho = m/V)color(white)(a/a)|)))" " " "color(orange)((2))#

Plug equation #color(orange)((1))# into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange to get

#PV * M_M = m * RT#

#P * M_M = m/V * RT#

#M_M = m/V * (RT)/P#

Finally, use equation #color(orange)((2))# to write

#M_M = rho * (RT)/P#

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

#M_M = 1.02 "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 + 37)color(red)(cancel(color(black)("K"))))/(0.990color(red)(cancel(color(black)("atm"))))#

#M_M = color(green)(|bar(ul(color(white)(a/a)color(black)("26.3 g mol"^(-1))color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.