1 L, 0.02 M solution of [Co(NH3)5 SO4]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 AgNO3 solution and BaCl2 solution respectively as shown below:1 L Solution (X) + AgNO3 solution (excess) → Y 1 L Solution (X) + BaCl2 solution (excess) → ZThe number of moles of Y and Z respectively are

on mixing both complexes become 0.01 M each in one litre solution.Each one Lt has 0.01 moles each of complexCoNH30.015SO4Br+AgNO3→AgBr0.01 mole ↓CoNH35BrSO4+BaCl2→BaSO4↓0.01 mole

1 L, 0.02 M solution of [Co(NH3)5 SO4]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 AgNO3 solution and BaCl2 solution respectively as shown below:1 L Solution (X) + AgNO3 solution (excess) → Y 1 L Solution (X) + BaCl2 solution (excess) → ZThe number of moles of Y and Z respectively are

on mixing both complexes become 0.01 M each in one litre solution.Each one Lt has 0.01 moles each of complexCoNH30.015SO4Br+AgNO3→AgBr0.01 mole ↓CoNH35BrSO4+BaCl2→BaSO4↓0.01 mole

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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 AgNO₃ 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

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0.02, 0.01

0.02, 0.02

0.01, 0.02

0.01, 0.01

answer is C.

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Detailed Solution

on mixing both complexes become 0.01 M each in one litre solution.
Each one Lt has 0.01 moles each of complex
$\begin{array}{l} {(\underset{0.01}{Co ({NH}_{3}})}_{5} {SO}_{4}) Br + {AgNO}_{3} \to \underset{0.01 mole}{AgBr} ↓ \\ (Co {({NH}_{3})}_{5} Br) {SO}_{4} + {BaCl}_{2} \to \underset{0.01 mole}{{BaSO}_{4} ↓} \end{array}$