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Question

0.02 mole of [Co(NH3)5Br]Cl2 and 0.02 mole of [Co(NH3)5Cl]SO4 are present in 200 cc of a solution X. The number of moles of the preceipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively :

A
0.02,0.02
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B
0.01,0.02
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C
0.02,0.04
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D
0.04,0.02
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Solution

The correct option is D 0.04,0.02
0.02 mole of [Co(NH3)5Br]Cl2 will dissociate to give 2×0.02=0.04 moles of chloride ions. On reaction with silver nitrate, 0.04 moles of silver chloride precipitate will be obtained.
[Co(NH3)5Br]Cl2[Co(NH3)5Br]2++2Cl
0.02 mole of [Co(NH3)5Cl]SO4 will dissociate to give 0.02 moles of sulphate ions. On reaction with barium sulphate, 0.02 moles of barium sulphate precipitate will be obtained.
[Co(NH3)5Cl]SO4[Co(NH3)5Cl]2++SO24

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