Question

Question #a9643

Answer 1

`"120 L"`

Explanation:

The idea here is that you need to use the vapor density of ammonia, `"NH"_3`, to determine its 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, `"H"_2`, kept under the same conditions for pressure and temperature.

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 `"2 g mol"^(-1)`, you can say that the molar mass of ammonia will be

`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 g"` of the compound

`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

`P` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

NTP conditions are defined as a pressure of `"1 atm"` and a temperature of `20^@"C" = "293.15 K"`.

Rearrange the equation to solve for `V` and plug in your values to find

`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.