When 0.01 mole of a cobalt complex is treated with excess of silver nitrate solution 4.035 g of silver chloride is precipitated. The formula of the complex is :

Options

A. `[Co(NH_(3))_(3)Cl_(3)]`
B. `[Co(NH_(3))_(5)Cl]Cl_(2)`
C. `[Co(NH_(3))_(6)]Cl_(3)`
D. `[Co(NH_(3))_(4)Cl_(2)]NO_(3)`

Answer

C

Explanation

Molar mass of AgCl = 143.5 g`mol^(-)`
No . Of moles of AgCl precipitated
= (4.305 g) / (143.4 g`mol^(-)`) = 0.03 mol
0.01 mole of complex precipitates AgCl = 0.03 mol
1.0 mole of complex precipitates AgCl = 3 mol
This means that the complex provides three `Cl^(-)` ions in solution `therefore` Formula of complex = `[C0(NH_3)_6]Cl_3`