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  • A solution contains 2.675 g of (molar mass = 267.5 g mol-1) is passed through a cation exchanger. The ch

A solution contains 2.675 g of (molar mass = 267.5 g mol-1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 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)

Option: 1

\left [ CoCl\left ( NH_{3} \right ) _{5}\right ]Cl_{2}


Option: 2

\left [ Co\left ( NH_{3} \right ) _{6}\right ]Cl_{3}


Option: 3

\left [ CoCl_{2}\left ( NH_{3} \right ) _{4}\right ]Cl


Option: 4

\left [ CoCl_{3}\left ( NH_{3} \right ) _{3}\right ]


Answers (1)

best_answer

Some Cl ion are outside of the complex sphere, and that will react with AgNO3 and form AgCl.

if Cl has 1 mol is there then 1 mol of AgCl form.

so for that we need to find mol--

check below.

 (COCl_{3}.6NH_{3})+AgNO_{3}\rightarrow AgCl\\ \frac{2.675}{267.5} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: Excess\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \frac{4.78}{143.5}=0.03

From given data 0.01 Mole COCl3.6NH3 produce 0.03-mole AgCl which indicates that compound has three ionizable Cl-  hence the formula of the compound is \left [ CO(NH_{3})_{6} \right ]Cl_{3}

Posted by

manish

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