When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is

A coordination complex compound of cobalt has molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution. On reacting this solution with excess of silver nitrate solution, two moles of AgCl get precipitated. The formula of this compound would be

When 1 mol CrCl_3 .6H_2O is treated with excess of AgNO_3 , 3 3 mol of AgCl are obtained. The formula of the complex is:

The impure 6g of NaCl is dissolved in water and then treated with excess of silver nitrate solution. The mass of p[recipitate of silver chloride is found to be 14g. The % purity of NaCl solution would be:

Chlorobenzene gives a white precipitate with alcoholic silver nitrate solution.

On reaction with silver nitrate solution, the complex CoCl_3 .4NH_3 gives precipitates corresponding to one/two chloride ions.

When a strip of copper is placed in the solution of silver nitrate, the solution become blue. Why ?

If the freezing point of a 0.01 molal aqueous solution of a cobalt chloride ammonia complex is -0.0558^@C , the number of chloride in the coordination sphere of the complex is [K_f of water =1.86K kg mol^-1]

Chloroform is a chlorine compound but it does not give white precipitate with silver nitrate solution. Give reason.

A solution containing 2.675 g of CoCl_3 .6NH_3 (molar mass = 267.5 g mol^-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 mol^-1 ). The formula of the complex is : (At. mass of Ag = 108 u)

When an aqueous solution of CuSO_(4) is sstirred with a silver spoon then :