View Full Version : Mole Concept question
Pranav-Arora
Jun9-11, 12:03 AM
1. The problem statement, all variables and given/known data
Hello Everyone!!
I am solving a lots and lots of questions on Mole Concept but i am getting stuck in this:-
Calculate the volume occupied by 5.25g of nitrogen at 26\circ C and 74.2 cm of pressure.
3. The attempt at a solution
I actually dont know from where should i start. I have done questions when the conditions are at STP but i have never done a question like this.
Thanks!! :smile:
piyushkumar
Jun9-11, 12:19 AM
start by the fact that one mole of any gas at 273K and 1 bar pressure occupies 22.7L volume
Pranav-Arora
Jun9-11, 02:38 AM
start by the fact that one mole of any gas at 273K and 1 bar pressure occupies 22.7L volume
Thanks for your reply Piyush!!
I thought of starting by this but i am not able to think of the next step. :confused:
This is direct application of ideal gas equation, PV=nRT. The only problem is with units of the pressure - is it cm (which doesn't make sense to me), or cm Hg?
Pranav-Arora
Jun9-11, 04:34 AM
This is direct application of ideal gas equation, PV=nRT. The only problem is with units of the pressure - is it cm (which doesn't make sense to me), or cm Hg?
So what should be the unit? :confused:
cm Hg most likely. Close relative of mmHg, or torr.
Pranav-Arora
Jun9-11, 09:41 AM
cm Hg most likely. Close relative of mmHg, or torr.
Thanks for your help Borek!! :smile:
I am having many questions to solve in mole concept (almost a full book). If i get any doubts, can i post them in this thread one by one? :smile:
In general you should start a new thread with each one. But there is a catch - if you start several at once, that's wrong as well.
Let's hope you will be able to deal with most of the questions on your own :wink:
Pranav-Arora
Jun9-11, 10:17 AM
In general you should start a new thread with each one. But there is a catch - if you start several at once, that's wrong as well.
Let's hope you will be able to deal with most of the questions on your own :wink:
Yes i will try to solve them by my own :wink:
But they are really tough....:uhh:
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