# What is the mole fraction of #NO# in a 55.0 L gas cylinder at 30.0°C which comes from a mixture of #N_2# and #NO# if you have 3.238 mol of #N_2# and the gas cylinder has a total pressure of 2.14 atm?

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to use the ideal gas law equation in order to find the **total number of moles** present in the mixture, then use the number of moles of nitrogen gas to determine the **mole fraction** of nitric oxide.

So, the ideal gas law equation looks like this

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

*universal gas constant*, usually given as

Plug in your values and solve for

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

#n = (2.14 color(red)(cancel(color(black)("atm"))) * 55.0color(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 + 30.0)color(red)(cancel(color(black)("K")))) = "4.729 moles"#

Since the mixture only contains nitric oxide and nitrogen gas, it follows that the number of moles of nitric acid will be equal to

#n_"total" = n_(NO) + n_(N_2)#

#n_(NO) = 4.729 - 3.238 = "1.491 moles NO"#

Now, the **mole fraction** of nitric oxide will be equal to the number of moles of nitric oxide divided by the *total number of moles* present in the mixture.

#chi_"NO" = n_"NO"/n_"total"#

In this case, you have

#chi_"NO" = (1.491 color(red)(cancel(color(black)("moles"))))/(4.729 color(red)(cancel(color(black)("moles")))) = color(green)(0.315)#

The answer is rounded to three sig figs.