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Methods of expressing the concentration of a solution and n factor

A solution containing 2.675 {~g} of {CoCl}_{3} 6 {NH}_{3} (mo

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A solution containing $2.675 g$ of $CoCl_{3} \cdot 6 NH_{3}$ (molar mass $= 267.5 g mol^{- 1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $AgNO_{3}$ to give $4.78 g$ of AgCl (molar mass $= 143.5 g mol^{- 1}$ ). The formula of the complex is (Atomic mass of $Ag = 108 u)$

Name: Video solution 1: 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=108u)
Uploaded: 2022-11-09T18:30:14.000Z
Duration: 15 min
Description: Video solution 1: 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=108u)

[Co (NH_{3})_{6}] Cl_{3}

[CoCl_{2} (NH_{3})_{4}] Cl

[CoCl_{3} (NH_{3})_{3}]

[CoCl (NH_{3})_{5}] Cl_{2}

Views: 5,290 students

Updated on: Nov 9, 2022

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2010

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Asher Discussed

A solution containing

2.675 g

CoCl_{3} \cdot 6 NH_{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

AgNO_{3}

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)

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Mole of

CoCl_{3} \cdot 6 NH_{3} = \frac{2.675}{267.5} = 0.01 mol

AgNO_{3} (aq) + Cl^{-} (aq) ⟶ AgCl ↓

(white)

+ NO_{3}^{-} (aq)

Moles of

AgCl = \frac{4.78}{143.5} = 0.03 mol

0.01 mol CoCl 3 \cdot 6 NH_{3}

gives

= 0.03 mol AgCl

∴ 1 mol CoCl 3 \cdot 6 NH_{3}

ionises to give

= 3 mol Cl^{-}

Hence, the formula of compound is

[Co (NH_{3})_{6}] Cl_{3}

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Question Text	A solution containing $2.675 g$ of $CoCl_{3} \cdot 6 NH_{3}$ (molar mass $= 267.5 g mol^{- 1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $AgNO_{3}$ to give $4.78 g$ of AgCl (molar mass $= 143.5 g mol^{- 1}$ ). The formula of the complex is (Atomic mass of $Ag = 108 u)$
Updated On	Nov 9, 2022
Topic	Solutions
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	211
Avg. Video Duration	15 min

Text solutionVerified

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Text solutionVerified

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