A solution containing $$2.675$$ g 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 (Atomic$$ mass of Ag $$=$$ $$108$$ $$u)

Question

A solution containing $$2.675$$ g 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 (Atomic$$ mass of Ag $$=$$ $$108$$ $$u)

Infinity Learn · Accepted Answer

Mole of $$CoCl3.6NH3=2.675267.5=0.01$$ $$molAgNO3(aq)+Cl$$−(aq)→AgCl↓$$(white)+NO3$$−$$(aq)Mole$$ of $$AgCl=4.78143.5=0.03$$ $$mol0.01$$ mol $$CoCl3.6NH3$$ gives $$=$$ $$0.03$$ mol AgCl∴ $$1$$ mol $$CoCl3.6NH3$$ ionises to give $$=$$ $$3$$ mol Cl−Hence, the formula of compound is [$$Co(NH3)6$$]$$Cl3$$

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A solution containing 2.675 g 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 (Atomic mass of Ag = 108 u)

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