# 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

#### 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

*universal gas constant*, usually given as

**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, **molar mass**,

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 *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 *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

#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

#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**.