0.4g of a gas has a volume of 227cm^3 at 27°C and at a pressure of 100kPa. Calculate the relative molecular mass of the gas (R=8.31 J/kmol)?

2 Answers
Sep 10, 2015

Answer:

#M_r=43.95#

Explanation:

#PV=nRT#

#n=(pV)/(RT)#

#=(100xx10^3xx227xx10^(-6))/(8.31xx300)#

#=0.0091#

#n=m/M_r#

#M_r=m/n=0.4/0.0091=43.95#

It looks like it could be #CO_2# but there are other possibilities.

Sep 10, 2015

Answer:

Molecular mass: #"44 g/mol"#.

Explanation:

To solve this problem you need to use the ideal gas law equation

#color(blue)(PV = nRT)" "#, where

#P# - the pressure of the gas;
#V# - the volume of th gas;
#n# - the number of moles of gas;
#R# - the universal gas constant, in your case given as #"8.31 J/mol K"#
#T# - the temperature of the gas - expressed in Kelvin!

The trick here is to realize that the units of #R# can be rewritten as

#R = "J"/("mol" * "K") = ("kg" * "m"^2)/"s"^2 * 1/("mol" * "K")#

#="kg"/("m" * "s"^2) * "m"^3 * 1/("mol" * "K")#

# = ("Pa" * "m"^3)/("mol" * "K")#

To find the molecular mass of the gas, you need to know how many moles of gas have a mass of 0.4 g. Rearrange the ideal gas law equation to solve for #n#

#n = (PV)/(RT)#

Next, convert the pressure from kPa to Pa and the volume from cubic centimeters to cubic meters

#100color(red)(cancel(color(black)("kPa"))) * "1000 Pa"/(1color(red)(cancel(color(black)("kPa")))) = 10^5"Pa"#

and

#227color(red)(cancel(color(black)("cm"^3))) * ("1 m"""^3)/(10^6color(red)(cancel(color(black)("cm"^3)))) = 227 * 10^(-6)"m"^3#

Plug in these values to find #n#, but do not forget to convert the temperature from degrees Celsiu to Kelvin!

#n = (10^5color(red)(cancel(color(black)("Pa"))) * 227 * 10^(-6)color(red)(cancel(color(black)("m"^3))))/(8.31(color(red)(cancel(color(black)("Pa"))) * color(red)(cancel(color(black)("m"^3))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 27)color(red)(cancel(color(black)("K")))) = "0.00910 moles"#

The molar mass of the gas will be

#M_M = m/n = "0.4 g"/"0.00910 moles" = color(green)("44 g/mol")#

SIDE NOTE I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the gas.