A solution containing $$2.675$$ grams of $$CoCl3$$ . $$6NH3$$ (molar$$ mass $$=$$ $$267.6$$ gr . mole−$$1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $$AgNO3$$ to give $$4.78$$ grams of AgCl (Molar$$ $$mass=$$ $$143.5$$ gr mole−$$1). The formula of the complex is

Question

A solution containing $$2.675$$ grams of $$CoCl3$$ . $$6NH3$$ (molar$$ mass $$=$$ $$267.6$$ gr . mole−$$1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $$AgNO3$$ to give $$4.78$$ grams of AgCl (Molar$$ $$mass=$$  $$143.5$$ gr mole−$$1). The formula of the complex is

Infinity Learn · Accepted Answer

WAgCl  $$precipitated=4.78gnAgCl$$  $$precipitated=4.78143.5=0.0333Wcomplex=2.675$$ gncomplex $$=$$ $$2.675267.5=0.010.01$$ moles of complex gave $$0.033$$ moles of $$AgCl1$$ mole of complex gives $$3$$ moles of AgClso one mole complex must have three moles of ionizable chloride ions.The complex is [$$Co(NH3)6$$]$$Cl3$$ .

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A solution containing 2.675 grams of CoCl3 . 6NH3 (molar mass = 267.6 gr . mole−1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 grams of AgCl (Molar mass= 143.5 gr mole−1). The formula of the complex is

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