# Question #a9643

##### 1 Answer

#### Explanation:

The idea here is that you need to use the **vapor density** of ammonia, **molar mass**, then use this to find how many **moles** of ammonia you have in that sample.

At this point, you need to use the definition of **NTP (Normal Temperature and Pressure)** and the **ideal gas law** equation to find the volume of the gas.

So, the *vapor density* of a gas is calculated by comparing the **density** of the gas with that of hydrogen gas,

In essence, the vapor density of a gas tells you the ratio that exists between a mass of the gas and the *mass of hydrogen gas* that **occupies the same volume** as the mass of the gas.

You can thus say that

#color(blue)(|bar(ul(color(white)(a/a)"vapor density" = "molar mass of the gas"/"molar mass of H"_2color(white)(a/a)|)))#

If you take the molar mass of hydrogen gas to be equal to

#M_ ("M NH"_ 3) = "vapor density" xx M_ ("M H"_2)#

#M_("M NH"_3) = 8.5 xx "2 g mol"^(-1) = "17 g mol"^(-1)#

Now, use the molar mass of ammonia to find how many moles you have in

#85 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17color(red)(cancel(color(black)("g")))) = "5 moles NH"_3#

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

**NTP** conditions are defined as a pressure of

Rearrange the equation to solve for

#PV = nRT implies V = (nRT)/P#

#V_(NH_3) = (5 color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 293.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#

#V_(NH_3) = color(green)(|bar(ul(color(white)(a/a)"120 L"color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.