Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12th pass
Pls help me to solve this.this is from gaseous state answer given is 8 but i am getting 6.
one year ago

Answers : (2)

Aastha Singh
askIITians Faculty
204 Points
							dear stude
one year ago
Vikas TU
10040 Points
Using equation PV=nRT
The number of moles of N2, CO2 and N2 in A
P=60 atm
V=12.315 litres = 12.315 X 10-3 M3
R= 8.314JK-1mol
T = 27+ 273= 300K
n =PV/RT = 2.962 x 10^-4
Number of moles CO2 and N2 after liquefaction of NH3
P=17-1=16 atm  due to vapour pressure of ammonia
V= 12.315x2 litres= 24.63x10-3 M3
T= -73+273 = 200K
n= PV/RT gives 2.37 x10-4
Moles of N2
Take P=4, V=12.315x 3 litres=36.945x10-3 M3  T=-173=273= 100
Then substitute the values in the eaquation n,  it gives 1.778x10-4
Moles of ammonia =(2.962-2.37)x10-4 =0.592x10-4
Moles of N= (2.37-1.778)x10-4 = 0.593x10-4
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details