MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
amulya bijjam Grade:
        

4 grams of H2 (hydrogen) is diffused through a small hole in 5 sec at 1 atm pressure.The amount of oxygen diffused in the same interval of time at 1.5 atm and same temperatuure will be_____

5 years ago

Answers : (1)

Pallav Agarwal
33 Points
										

Let gas 1 be H2. Gas 2 is O2


 


(R1/R2)=(P1/P2)*(sqroot(m2/m1))=(n1/n2)


 


=> (P1/P2)*(sqroot(m2/m1))=(n1/n2)


=> (1/1.5)*(sqroot(32/2))=(2/n2)


=> (2/3)*(4)=(2/n2)


=> (n2)= 8/6   moles = 4/3 moles


=> Weight = 32*4/3 = 128/3 grams


 


 


R1 - Rate of diffusion of gas 1.


M1 - Molecular Weight of gas 1.


n1 - No. of moles of gas 1.


 


I am new to this forum. Please approve if you find this helpful.

5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 2,385 off
USE CODE: COUPON10
Get extra Rs. 413 off
USE CODE: COUPON10

Ask Experts

Have any Question? Ask Experts

Post Question

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