Failed to connect to MySQL: Access denied for user 'examnext_online'@'localhost' (using password: YES) A solution containing 2.675 g of `CoCl_3 . 6 NH_3` (molar mass=267.5 g mo`l^(-1)`) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of `AgNO_3` to give 4.78 g of AgCl (molar mass=143.5 g mo`l^(-1)`). The formula of the complex is (At mass of Ag= 108 u) - Sarthaks eConnect

A solution containing 2.675 g of `CoCl_3 . 6 NH_3` (molar mass=267.5 g mo`l^(-1)`) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of `AgNO_3` to give 4.78 g of AgCl (molar mass=143.5 g mo`l^(-1)`). The formula of the complex is (At mass of Ag= 108 u)

A solution containing 2.675 g of `CoCl_3 . 6 NH_3` (molar mass=267.5 g mo`l^(-1)`) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of `AgNO_3` to give 4.78 g of AgCl (molar mass=143.5 g mo`l^(-1)`). The formula of the complex is (At mass of Ag= 108 u)

A.

`[Co(NH_3)_6]Cl_3`

B.

`CoCl_2(NH_3)_4]Cl`

C.

`[CoCl_3(NH_3)_3]`

D.

`[CoCl(NH_3)_5]Cl_2`

Solution

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Best answer

0.01 mole complex give 0.03 moles of AgCl