on treatment of 100 ml of 0.1 M CoCl3.6H2O with excess of Agno3 1.2x10^22 ions are precipitated. the formula of the complex is
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Hey dear,
● Answer -
[Co(H2O)4Cl]Cl2.H2O
● Explaination -
No of moles of CoCl3.6H2O in solution are -
n = M × V
n = 0.1 × 0.1
n = 10^-2
Total no of ions of Cl- in supposed complex -
n' = 3×10^-2
No of Cl- ions precipitated -
Np = (1.2×10^22) / (6.022×10^23)
Np = 2×10^-2
No of Cl- ions unprecipated -
Nu = 3×10^-2 - 2×10^-3
Nu = 10^-2
That is precipitated ions are double that of unprecipitated ions. This means out of 3, 2 Cl- ions are out of co-ordination sphere.
Therefore, formula for given complex is [Co(H2O)4Cl]Cl2.H2O .
Hope this help you...
● Answer -
[Co(H2O)4Cl]Cl2.H2O
● Explaination -
No of moles of CoCl3.6H2O in solution are -
n = M × V
n = 0.1 × 0.1
n = 10^-2
Total no of ions of Cl- in supposed complex -
n' = 3×10^-2
No of Cl- ions precipitated -
Np = (1.2×10^22) / (6.022×10^23)
Np = 2×10^-2
No of Cl- ions unprecipated -
Nu = 3×10^-2 - 2×10^-3
Nu = 10^-2
That is precipitated ions are double that of unprecipitated ions. This means out of 3, 2 Cl- ions are out of co-ordination sphere.
Therefore, formula for given complex is [Co(H2O)4Cl]Cl2.H2O .
Hope this help you...
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