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
Radhika Batra Grade: 11
        A 20.0 cm3 mixture of CO, CH4 and He gases is exploded by an electric discharge at room temperature with excess of further contraction is found to be 13.0 cm3.  A further contraction of 14.0 cm3 occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.
3 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
										Sol. The reaction involved in the explosion process is :

CO(g) + 1/2 O2 (g) → CO2 (g)
x mL x/2 mL x mL
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l)
y mL 2 y mL y mL
The first step volume contraction can be calculated as :
(x + x/2 + y + 2 y)-(x + y) = 13
⇒ x + 4 y = 26
The second volume contraction is due to absorption of CO2.
Hence, x + y = 14 …. (ii)
Now, solving equations (i) and (ii),
X = 10 mL, y = 4 mL and volume of He = 20 – 14 = 6 mL
⇒ Vol % of CO = 10/20 × 100 = 50%
Vol of CH4¬ = 4/20 × 100 = 20%
Vol % of He = 30%
3 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