InterviewSolution
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InterviewSolution
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A solution containing 2.675 g of `CoCl_(3).6NH_(3)` (molar mass = 267.5 g `mol^(-1)` is passed through a cation exchanger. The chloride ions obtained in solution are treated with excess of `AgNO_(3)` 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 ) .A. `[CoCl(NH_(3))_(5)]Cl_(2)`B. `[Co(NH_(3))_(6)]Cl_(3)`C. `[CoCl_(2)(NH_(3))_(4)]Cl`D. `[CoCl_(3)(NH_(3))_(3)]` |
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Answer» Correct Answer - B Moles of `CoCl_(3).6NH_(3) = ( 2.675)/(267.5) = (1)/(100) `mole `= 0.01` mole Mole of `AgCl= (4.78 )/(143.5) = (3)/(100)` mole `=0.03` mole 0.01 mole of compound gives moles of `AgCl= 0.03` mole So, 1 mole of compound gives moles of `AgCl = (0.03)/(0.01) = 3` mole. So, structural formula of compound having `3Cl^(-)` ions out side of coordination sphere, so formula is `[Co(NH_(3))_(6)]Cl_(3)` |
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