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Two metals X and Y form the salts XSO(4)...

Two metals X and Y form the salts `XSO_(4)` and `Y_(2)SO_(4)` respectively.The solution of salt `XSO_(4)` is blue in colour whereas that of `Y_(2)SO_(4)` is formed alongiwth a salt Which turns the solution green. And when barium chloride solution is added to `Y_(2)SO_(4)` solution, then the same white precipitate Z is formed alongwith colourless common salt solution.
(a) What could the metal X and Y be ?
(b) Write the name and formula of salt `XSO_(4)`
(c) Write the name and formula of salt `Y_(2)SO_(4)`
(d)What is the name and formula of white precipate Z ?
(e) Write the name and formula of the salt which turns the solution green in the first case.

Text Solution

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(a) The available information suggests that metals X and Y are copper (Cu) and sodium (Na) respectively.
(b) The salt X is `CuSO_(4)` and its aqueous solution in blue in colour.
(c) The salt Y is `Na_(2)SO_(4)` and its aqueous solution is colourless.
(d) When `BaCl_(2)` solution is added to `CuSO_(4)` solution, a white precipitate of `BaSO_(4)` (Z) is formed. The solution acquires greenish colour due to formation of
`underset((X))(CuSO_(4)) + BaCl_(2) to underset("White ppt (z)")(BaSO_(4)) + underset("Greenish solution")(CuCl_(2))`
(e) The green solution as stated above is of cupric chloride.
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