A sample of gas weighing 0.0286 g occupies a volume of 50 #cm^3# at 76 cmHg. and 25 degree centigrade. What is the molar mass of the gas?

1 Answer
Jun 26, 2017

Answer:

#MM = 14.0# #"g/mol"#

Explanation:

We can find the molar mass #MM# of this gas using the equation

#MM = (dRT)/P#

where

  • #d# is the density of the gas, in #"g/L"#

  • #R# is the universal gas constant, equal to #0.082057("L"•"atm")/("mol"•"K")#

  • #T# is the absolute temperature of the gas, in #"K"#

  • #P# is the pressure of the gas, in #"atm"#

We need to convert some units to get where we need to be. Let's do the density first.

#1# #"L"# is equal to #1# #"dm"^3#, which is equal to #10^3# #"cm"^3#, so the density is

#((0.0286color(white)(l)"g")/(50cancel("cm"^3)))((10^3cancel("cm"^3))/(1cancel("dm"^3)))((1cancel("dm"^3))/(1color(white)(l)"L")) =color(red)(0.572# #color(red)("g/L"#

Degrees centigrade is the same as degrees Celsius, so let's convert this to #"K"# by adding #273#:

#T = 25^"o""C" + 273 = color(green)(298# #color(green)("K"#

Lastly, let's convert our pressure, which is in centimeters of mercury:

#76cancel("cm Hg")((10cancel("mm Hg"))/(1cancel("cm Hg")))((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(purple)(1# #color(purple)("atm"#

Now that we have all our values, let's plug them into the equation:

#MM = ((color(red)(0.572"g"/"L"))(0.082057("L"•"atm")/("mol"•"K"))(color(green)(298"K")))/(color(purple)(1"atm"))#

# = color(blue)(14.0# #color(blue)("g/mol"#