NTA Abhyas JEE Main2020ChemistrySome Basic Concepts of ChemistryPractice
A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a strandarized acid as follows : Co NH 3 x Cl y aq + HCl → NH 4 + aq + Co y + + aq + Cl − aq A 1.8 g complex required 20.00 mL 1.54 M HCl to reach the equivalence point. Also, if the reaction mixture at equivalence point is treated with excess of AgNO 3 solution, 7.735 g of AgCl precipitate was produced. What is the form
Options
- ACo NH 3 4 Cl 3
- BCo NH 4 4 Cl 3
- CCo NH 3 3 Cl 4
- DCo NH 4 3 Cl 4
Correct answer
A. Co NH 3 4 Cl 3
Step-by-step solution
Co ( NH 3 ) x Cl y 1 .8 M + xHCL 1 .8 M → x NH 4 + aq + Co y + + ( x + y ) Cl - 1 .8 ( x + r ) M 1 . 8 x M = 2 0 × 1 . 5 4 × 1 0 - 3 = 3 0 . 8 × 1 0 - 3 ........(I) 1 · 8 x + y M = 7 . 7 3 5 1 4 3 . 5 = 5 3 . 9 × 1 0 - 3 ........(II) II I ⇒ 5 3 . 9 × 1 0 - 3 3 0 . 8 × 1 0 - 3 = 1 . 7 5 = x + y x = 1 + y x y x = 0 . 7 5 Substituting in (I) ⇒ 1 . 8 x 5 9 + 1 7 x + 3 5 . 5 × 0 . 7 5 x = 3 0 . 8 × 1 0 - 3 ∴ x = 3.98 = 4 ∴ y = 3