InterviewSolution
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InterviewSolution
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Two metals X and Y from the salts `XSO_(4)` and `Y_(2)SO_(4)` respectively. The solution of salt `XSO_(4)` is blue colour whereas that of `Y_(2)SO_(4)` is colourless. When barium chloride solution is added to `XSO_(4)` solution, then a white precipitate Z is formed along with 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 metals 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 precipitate Z ? (e) Write the name and formula fo the salt which turns the solution green in the first case. |
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Answer» (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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