# A sample of gas occupies a volume of 272 mL at a pressure of .999 atm and a temp of 357K. If the sample weighs 1.03 g, what is the molecular weight of the gas?

##### 1 Answer

#### Answer:

#### Explanation:

You know that a gas occupies a volume of

Now, the **molar mass** of a substance tells you exactly what the mass of *one mole* of that substance is. This means that in order to be able to find the molar mass of the gas, you need to know exactly how many moles of gas you have in that sample.

To do that, use the ideal gas law equation

#PV = nRT" "# , where

*universal gas constant*, usually given as

Plug in your values and solve for *liters*!

#PV = nRT implies n = (PV)/(RT)#

#n = (0.999color(red)(cancel(color(black)("atm"))) * 272 * 10^(-3)color(red)(cancel(color(black)("mL"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 357color(red)(cancel(color(black)("K")))) = "0.009282 moles"#

This means that the molar mass of the gas will be

#M_"M" = m/n#

#M_"M" = "1.03 g"/"0.009282 moles" = "110.97 g/mol"#

Rounded to three sig figs, the answer will be

#M_"M" = color(green)("111 g/mol")#