Consider a mixtm-e 'X' which is made by dissolving 0.4 mol of $$[Co(NH_{3})_{5}SO_{4}]Br$$ and 0.4 mol of $$[Co(NH_{3})_{5}Br]SO_{4}$$ in water to make 4 L of solution. When 2 L of mixture 'X' is allowed to react with excess of $$AgNO_{3}$$,it fonns precipitate 'Y'. The rest 2 L of mixture 'X' reacts with excess $$BaCl_{2}$$ to fonn precipitate 'Z'. Which of the following statements is CORRECT?

A mixture X contains 0.4 mol [Co(NH$$_3$$)$$_5$$SO$$_4$$]Br and 0.4 mol [Co(NH$$_3$$)$$_5$$Br]SO$$_4$$ in 4 L of water.

[Co(NH$$_3$$)$$_5$$SO$$_4$$]Br ionizes as: [Co(NH$$_3$$)$$_5$$SO$$_4$$]$$^+$$ + Br$$^-$$. Similarly, [Co(NH$$_3$$)$$_5$$Br]SO$$_4$$ ionizes as: [Co(NH$$_3$$)$$_5$$Br]$$^{2+}$$ + SO$$_4^{2-}$$. In 4 L total there are 0.4 mol Br$$^-$$ and 0.4 mol SO$$_4^{2-}$$.

When 2 L of this mixture is treated with excess AgNO$$_3$$, the 0.2 mol Br$$^-$$ ions precipitate as AgBr (Y) according to AgNO$$_3$$ + Br$$^-$$ $$\to$$ AgBr, so moles of Y = 0.2 mol.

Treating 2 L with excess BaCl$$_2$$ precipitates the 0.2 mol SO$$_4^{2-}$$ ions as BaSO$$_4$ (Z) according to BaCl$$_2$$ + SO$$_4^{2-}$$ $$\to$$ BaSO$$_4$$, giving moles of Z = 0.2 mol.

• Option A: 0.1 mol of Y — Incorrect (0.2 mol formed).
• Option B: 0.2 mol of Z — Correct (0.2 mol BaSO$$_4$$).
• Option C: 0.4 mol of Z — Incorrect.
• Option D: Y is BaSO$$_4$$ and Z is AgBr — Incorrect (Y is AgBr, Z is BaSO$$_4$$).

The correct answer is Option B.