Question

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)?

Answer 1

`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.

Answer 2

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.