1 L, 0.02 M solution of [Co(NH3)5SO4] Br is mixed with 1 L, 0.02 M solution of [Co(NH3)5Br]SO4. The resulting solution is divided into two equal parts (X) and treated with excess of AgNO4 solution and

Question

1 L, 0.02 M solution of [Co(NH3)5SO4] Br is mixed with 1 L, 0.02 M solution of [Co(NH3)5Br]SO4. The resulting solution is divided into two equal parts (X) and treated with excess of AgNO4 solution and BaCl2 solution respectively as shown below: 1 L Solution (X) + AgNO3 solution (excess) Y 1 L Solution (X) + BaCl2 solution (excess) Z The number of moles of Y and Z respectively are

(A) 0.02, 0.02
(B) 0.02, 0.01
(C) 0.01, 0.01
(D) 0.01, 0.02 | JEE(Main) 2023 | DATE : 30-01-2023 (SHIFT-2) | PAPER-1 | CHEMISTRY PAGE # 6

Prashant Kotian · Accepted Answer

Correct answer: C. ution respectively as shown below: 1 L Solution (X) + AgNO3 solution (excess)  Y 1 L Solution (X) + BaCl2 solution (excess)  Z The number of moles of Y and Z respectively are (1) 0.02, 0.02 (2) 0.02, 0.01 (3) 0.01, 0.01 (4) 0.01, 0.02 | JEE(Main) 2023 | DATE : 30-01-2023 (SHIFT-2) | PAPER-1 | CHEMISTRY PAGE # 6 NTA. (3) RESO. (3) Sol. On mixing equal volume concentration is Half. [Co(NH3)5SO4]Br, [Co(NH3)5Br]SO4 0.01 M 0.01 M (i) X [Co(NH3)5SO4]Br + AgNO3  AgBr       02 . mole 0.01 mole ‘Y’ (ii) X [Co(NH3)5Br]SO4 + BaCl2  BaSO4       02 . mole 0.01 mole ‘Y’

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