# Ideal Gas Law Practice Problems with Molar Mass

now we're going to look at some

## ideal

gas law### problems

where we also have to use#### molar

## mass

to convert between grams and moles of a gas take this problem for instance calculate the volume that 12.5 grams of co2 gas will occupy at a temperature of 40 degrees Celsius and a pressure of 1.0 atm I've gone ahead and filled in a number of the variables that we're going to be using as you as you can see here volume is what we're going to be solving for when we do this the first thingthat I always check is to make sure that the units of the variables match the units on our okay so we have ATM ATM that's great temperature as you know needs to be in Kelvin to work with gases so I've gone ahead and already done that because I'm sure you guys are probably familiar with how we add 273 to the degrees Celsius to get Calvin so I'm going to change 40 degrees Celsius here to 313 Kelvin the next thing that we have to do though is look here at the amount of gas we have n

right now it's reported in grams 12.5 grams but to use it in the

## ideal

gas law it needs to be in moles so this means that we're going to have to use the#### molar

## mass

of co2 to convert from grams of co2 into moles of co2 so the first thing we have to do is determine the#### molar

## mass

of carbon dioxide well it has one carbon and two oxygens so we take the## mass

of carbon plus two times the## mass

of oxygen add these together and it gives us 44.0 grams for each mole all right so now what we'regoing to do is we're going to take 12.5 grams of co2 times I want to cancel out grams so I'm going to flip this fraction one mole divided by forty 4.0 gram and the math that I'm going to do is 12.5 times 1 divided by 44 I'm going to round that to 3 significant figures 1 mole this is an exact counting number so we don't worry about significant figures for this 3 here 3 here so I'm going to round this to 3 significant figures and I'm going to 0.28 4 moles of co2 so we

can cancel out grams and show that we've changed this into 0.28 four moles now everything that I have in terms of my variables matches with a value that I have on our so I'm going to take PV equals NRT the

## ideal

gas law and calculate and rearrange this to get V by itself so all I have to do is divide both sides by P then the piece cancel out and I get V equals n R T divided by P so I've gone ahead and plugged most of these values into the equation I have n I have T and I have P asbefore I leave our until the end and here it is now let's go and cancel the unit's ATM's here ATM's up here moles here moles here kelvin there kelvin there which leaves me with liters which makes sense since I'm solving for volume now the math I'm going to do I'm going to do this times this times this all together then divided by one point two zero atm and I'm going to get six point oh eight one six nine how many significant figures draw around it to I round it to

three so keep the six I keep the zero and I keep the eight I look whether I round it up or down I keep it the same so I do my final answer is going to be six point oh eight what are my units the units are what I was left with after canceling six point oh eight litres and that's my final answer for this okay here's another problem a certain gas has a

#### molar

## mass

of twenty-eight point zero grams per mole how many grams of this gas would fit in a three point zero zero liter container at 182kPa and forty seven point two degrees Celsius I filled in my variables here and how many grams of this gas how much gas that's what we're solving for as before I'm going to look through at these variables and make sure that they match the units on our Celsius is always a problem when we're dealing with gases I think as you know by now we want to take any Celsius temperature with gases and add 273 to that to get Kelvin here I've had to also round using my addition rules for

significant figures so I get 320 Kelvin is going to be the new temperature that I'm going to use so now kelvins match I have liters here but check this out I have kilo Pascal's and I have atm so my pressure units don't match there are two things I could do one of these things that I could do is I could use a new R I said earlier that you want that you can use a different R so that your pressure units match so I could use 8.3 one which has kPa instead of atm that would be totally fine

if you want to do that what I'm going to do though is I'm going to stick with the R that I'm using right now and instead I'm going to convert kPa into ATM so that the unit's match 180 M is equal to 101.3 kPa so I can do this math kPa up here kPa down there they cancel out and I'm going to get rounded to three significant figures one point eight zero ATM so this expressed and atmospheres is 1.80 so now everything that I have matches the units on our and I'm ready to go

about solving this so PV equals NRT I'm solving for n divide both sides by RT now those guys cancel out I get PV divided by RT equals n or if you want to flip it I can have N equals PV divided by RT so here is what I get when I plug these values in I have pressure and I have volume I have temperature and I've left R till the end because R is on the bottom I take the value that I have here and I flip it so I have Kelvin moles then the number liters atm okay cancel the unit's and

I'm left with moles and when I do out this math this times this divided by this times this final answer that I get rounded to three significant figures is zero point two oh six and there's my unit's moles but we're not done with a problem yet because if you see here we're solving for grams not for moles so how do we go from moles to grams we use

#### molar

## mass

we know that this gas has a#### molar

## mass

of 28 grams per mole so what I can do is I can take my moles and multiply thisnumber by 28.0 grams divided by one mole moles up here moles down here so they cancel out the final answer that I'm going to get rounded again to three significant figures one mole doesn't count because it's a counting number is going to give me five point seven seven grams and that is how we solve