1.

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

Answer»

`[Co(NH_(3))_(3)Cl_(3)]`
`[Co(NH_(3))_(5)Cl]Cl_(2)`
`[Co(NH_(3))_(6)]Cl_(3)`
`[Co(NH_(3))_(4)Cl_(2)]NO_(3)`

SOLUTION :4.305 g AgCl = `(4.305)/(143.5)` mole = 0.03 mole
As 0.01 mole of the COMPLEX gives 0.03 mole of AgCl, this shows that there are 3 ionizable Cl. Hence, FORMULA is (c ).


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