badge image

Enroll For Free Now & Improve your performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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
Grade: 12th pass

                        

0.5 g of bleaching powder liberates iodine from ki that reacts with 20 ml of 0.1n sodiumthiosulphate solution. what is the active oxygen content in bleaching powder?

3 years ago

Answers : (2)

Ameen Sha M
15 Points
							
CaOCl2 + 2KI ------  2KCl + I2 + oxygen + Ca
I2 + 2Na2S2O3 --------- Na2S4O6 + 2NaI
The above reaction should be remembered 
Moles of Na2S2O3 = NV/1000 = 20×0.1/1000 = 0.002 moles
1 mole of I2 react with 2 mole of Na2S2O3
Hence moles of Na2S2O3 = 0.002÷2 = 0.001
Number of moles of I2 = number of moles of O = 0.001 moles 
Mass of O = 0.001×16 = 0.016 g
%of oxygen in given sample = 0.016 × 100/ 0.5 = 3.2 %
5 months ago
Vikas TU
12119 Points
							
Dear student 
60mL of 0.5N Na2S2O3 = 60mL 0.5N I2 
Amt of Cl = 35.5*0.5*60/1000 = 1.065 g 
% of available cl = 1.065 /3.55 * 100 = 30%
5 months 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
Vouchers
To Win!!! Click Here for details