Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a strandarized acid as follows :
 CoNH3xClyaq+HClNH4+aq+Coy++aq+Claq  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 AgNO3  solution, 7.735 g of AgCl precipitate was produced. What is the formula of this complex?
[Given: atomic weight of CO = 59 gmol-1]
ChemistrySome Basic Concepts of ChemistryJEE Main
Options:
  • A Co NH 3 4 Cl 3
  • B Co NH 4 4 Cl 3
  • C Co NH 3 3 Cl 4
  • D Co NH 4 3 Cl 4
Solution:
2293 Upvotes Verified Answer
The correct answer is: Co NH 3 4 Cl 3
Co(NH3)xCly1.8M+xHCL1.8MxNH4+aq+Coy++(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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.