# Question #48160

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to use the **ideal gas law** equation and the **molar mass** of nitrogen gas, *mass* of a sample of nitrogen gas and the *volume* it occupies at **STP**.

Now, **STP conditions** will most likely be given to you as a pressure of

The ideal gas law equation looks like this

#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

As you know, a compound's **molar mass** tells you the mass of **one mole** of that substance. IN your case, nitrogen gas has a molar mass of

#M_M = "28.0134 g mol"^(-1)#

This means that **every mole** of nitrogen gas has a mass of

You can thus replace the *number of moles*o f nitrogen gas, **molar mass** of the gas

#n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange to get

#PV *M_M = m * RT#

This will be equivalent to

#P * M_M = m/V * RT#

But since **density**,

#m/V = (P * M_M)/(RT)#

and therefore

#color(purple)(|bar(ul(color(white)(a/a)color(black)(rho = (P * M_M)/(RT))color(white)(a/a)|)))#

Now all you have to do is use the STP values for pressure and temperature and the molar mass of nitrogen gas to get

#rho = (1 color(red)(cancel(color(black)("atm"))) * "28.0134 g" color(red)(cancel(color(black)("mol"^(-1)))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 273,15color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))#

**ALTERNATIVE APPROACH**

As you know, **one mole** of any ideal gas kept at **molar volume of a gas at STP**.

Moreover, you know that **one mole** of nitrogen gas has a mass of **per unit of volume**, all you have to do now is determine the *mass of one liter* of nitrogen gas at STP

#1 color(red)(cancel(color(black)("L N"_2))) * overbrace((1 color(red)(cancel(color(black)("mole N"_2))))/(22.4color(red)(cancel(color(black)("L N"_2)))))^(color(purple)("molar volume of a gas at STP")) * overbrace("28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar mass of N"_2)) = "1.25 g"#

Since this is how many grams you get **per liter** of nitrogen gas at STP, it follows that its density will be

#rho = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))#