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A solution containing 2.675 g of {C}{o}{C}{l}_{{{3}}}. 6 {N}{H

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A solution containing 2.675 g of $C o C l_{3}$ . 6 $N H_{3}$ (molar mass = 267.5 g $mol^{- 1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $A g N O_{3}$ 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)

Name: Video solution 1: A solution containing 2.675 g of CoCl3. 6 NH3 (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)
Uploaded: 2021-01-01T18:41:02.000Z
Duration: 2 min
Description: Video solution 1: A solution containing 2.675 g of CoCl3. 6 NH3 (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)

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A solution containing 2.675 g of

C o C l_{3}

. 6

N H_{3}

(molar mass = 267.5 g

mol^{- 1}

) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of

A g N O_{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)

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Question Text	A solution containing 2.675 g of $C o C l_{3}$ . 6 $N H_{3}$ (molar mass = 267.5 g $mol^{- 1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $A g N O_{3}$ 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)
Answer Type	Video solution: 1
Upvotes	150

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