Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A mixture of argon and nitrogen has a density of 1.40gL-1 at STP. What is mole fraction of nitrogen.
At STP conditions 1mole of any gas occupies 22.4L of volume, so x moles of gas occupies 22.4x L of volume.
Let us take there are ''A'' moles of Argon and N moles of Nitrogen in the mixture.
Then the total mass of the mixture = (40A+28N) gm. (since molecular weights of Argon and Nitrogen are 40 & 28)
Total Volume of the mixture = 22.4(A+N) Liters.
So, density of the mixture = (40A+28N)/22.4(A+N) gm/Liter
Given density of the mixture = 1.4 gm/Liter
Equating them we get A = (8/27)N
Adding N on both sides, we get
A+N =(35/27)N => [N/(A+N)] = 27/35
So, mole fraction of Nitrogen in the above mixture is 27/35
let no. of moles of nitrogen per be x and that of argon be y.
density=1.4=(no. of moles of nitrogen*molar mass of nitrogen +no. of moles of argon*molar mass of argon)/volume
d=(28x + 40y)/[22.4*(x+y)]=1.4
3.36x=8.64y
x=2.57y
mole fraction of nitrogen=x/(x+y)
=2.57y/(2.57y+y)
=0.72
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !