JEE MAIN - Chemistry (2010 - No. 15)
A solution containing 2.675g of CoCl3. 6NH3 (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)
[Co(NH3)6]Cl3
[CoCl2(NH3)4]Cl
[CoCl3(NH3)3]
[CoCl(NH3)5]Cl2
Explanation
$$\mathop {CoC{l_3}.6N{H_3}}\limits_{2.675g} \buildrel \, \over
\longrightarrow xC{l^ - }$$
$$xC{l^ - } + AgN{O_3}\buildrel \, \over \longrightarrow \mathop {x\,AgCl \downarrow }\limits_{4.78g} $$
Number of moles of the complex
$$ = {{2.675} \over {267.5}} = 0.01$$ moles
Number of moles of $$AgCl$$ obtained
$$ = {{4.78} \over {143.5}} = 0.03$$ moles
$$\therefore$$ $$\,\,\,$$ Number of moles of $$AgCl$$ obtained
$$ = 3 \times \,\,$$ No. of moles of complex
$$\therefore$$ $$\,\,\,$$ $$n = {{0.03} \over {0.01}} = 3$$
$$xC{l^ - } + AgN{O_3}\buildrel \, \over \longrightarrow \mathop {x\,AgCl \downarrow }\limits_{4.78g} $$
Number of moles of the complex
$$ = {{2.675} \over {267.5}} = 0.01$$ moles
Number of moles of $$AgCl$$ obtained
$$ = {{4.78} \over {143.5}} = 0.03$$ moles
$$\therefore$$ $$\,\,\,$$ Number of moles of $$AgCl$$ obtained
$$ = 3 \times \,\,$$ No. of moles of complex
$$\therefore$$ $$\,\,\,$$ $$n = {{0.03} \over {0.01}} = 3$$
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