September 23, 2020 ## Ideal Gas Law Practice Problems with Molar Mass

what is the molar mass of an ideal gas if a 0.622 g sample This is a topic that many people are looking for. voteyesons.org is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, voteyesons.org would like to introduce to you Ideal Gas Law Practice Problems with Molar Mass. Following along are instructions in the video below:
Were going to look at some ideal gas law problems. Where we also have have to use molar mass to convert between grams and moles of a gas take problem for. Instance calculate the volume that 125.
Grams. Of co2 gas will occupy at a temperature of 40. Degrees celsius and a pressure of 10.
Atm. Ive gone ahead and filled in a number of the variables that were going to be using as you as you can see here volume is what were going to be solving for when we do this the first thing that i always check is to make sure that the units of the variables match the units on our okay. So we have atm atm thats great temperature.
As you know needs to be in kelvin to work with gases. So ive gone ahead and already done that because im sure you guys are probably familiar with how we add 273 to the degrees celsius to get calvin. So im 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 its reported in grams. 125 grams.
But to use it in the ideal gas law. It needs to be in moles. So this means that were 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 440. Grams.
For each mole all right so. Now what were going to do is were going to take 125 grams of co2 times. I want to cancel out grams so im going to flip this fraction one mole divided by forty.
40. Gram and the math that im going to do is 125 times 1 divided by 44. Im going to round that to 3 significant figures 1 mole.
This is an exact counting number so we dont worry about significant figures for this 3 here. 3. Here so im going to round this to.
3 significant figures and im going to 028.

4. Moles of co2 so we can cancel out. Grams and show that weve changed this into 028.
Four moles now everything that i have in terms of my variables. Matches with a value that i have on our so im 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 ive gone ahead and plugged most of these values into the equation. I have n. I have t and i have p as before i leave our until the end and here.
It is now lets go and cancel the units atms here. Atms up here moles here moles here kelvin there kelvin there which leaves me with liters which makes sense since im solving for volume now the math. Im going to do im going to do this times this times this all together then divided by one point two zero atm and im 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 thats my final answer for this. Okay. Heres 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 182 kpa and forty seven point two degrees celsius i filled in my variables.
Here and how many grams of this gas. How much gas thats what were solving for as before im going to look through at these variables. And make sure that they match the units on our celsius is always a problem when were 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. Ive had to also round using my addition rules for significant figures. So i get 320 kelvin is going to be the new temperature that im going to use so now kelvins match.
I have liters here. But check this out i have kilo pascals and i have atm so my pressure units dont 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 83. One which has kpa instead of atm that would be totally fine.
If you want to do that what im going to do. Though is im going to stick with the r that im using right now and instead. Im going to convert kpa into.
Atm so that the units match 180. M is equal to. 1013.
Kpa. So i can do this math kpa up here kpa down there they cancel out and im going to get rounded to three significant. Figures one point eight zero.
Atm so this expressed and atmospheres is 180. So now everything that i have matches. The units on our and im ready to go about solving this so pv equals nrt.
Im 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 ive 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 units and im 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 theres my units moles.
But were not done with a problem yet because if you see here were 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 this number by 280 grams. Divided by one mole moles up here moles down here so they cancel out the final answer that im going to get rounded again to three significant figures.
One mole doesnt count because its a counting number is going to give me five point seven seven grams. And that is how we solve ideal gas law problems using molar mass for converting grams. And moles.
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