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Ammonia and potassium iodide solutions are added to an aqueous solution of Cr(NO 3 ) 3 . A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess 0 2, 0.0203 g CrO 3 was formed. ii. In a second experiment it took 32.93 mL of 0.100 M HCI to titrate completely the NH 3 present in 0.341 g compound A. iii. Compound A was found to contain 73.53% iodine by mass. iv. The freezing point of water was lowered by 0.64°C when 0.601 g compound A was dissolved in 10.00 g H 2 O ( K f =1.86°C·kg/mol). What is the formula of the compound? What is the structure of the complex ion present? ( Hints: Cr 3+ is expected to be sixcoordinate, with NH 3 and possibly I − as ligands. The I − ions will be the counterions if needed.)

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 20, Problem 99IP
Textbook Problem
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Ammonia and potassium iodide solutions are added to an aqueous solution of Cr(NO3)3. A solid is isolated (compound A), and the following data are collected:

i. When 0.105 g of compound A was strongly heated in excess 0 2, 0.0203 g CrO3 was formed.

ii. In a second experiment it took 32.93 mL of 0.100 M HCI to titrate completely the NH3 present in 0.341 g compound A.

iii. Compound A was found to contain 73.53% iodine by mass.

iv. The freezing point of water was lowered by 0.64°C when 0.601 g compound A was dissolved in 10.00 g H2O (Kf =1.86°C·kg/mol).

What is the formula of the compound? What is the structure of the complex ion present? (Hints: Cr3+ is expected to be sixcoordinate, with NH3 and possibly I as ligands. The I ions will be the counterions if needed.)

Interpretation Introduction

Interpretation: The formation of compound A by the addition of ammonia and potassium iodide solutions to an aqueous solution of Cr(NO3)3 is given. The formula of the compound and the structure of complex ion present on the basis of given information is to be stated.

Concept introduction: Electronic configuration is used to describe the distribution of the electrons in the orbitals of an atom. Structure of an atom can be defined by its electronic configuration. It can also be used to denote an atom which is ionized to a cation or anion formed by the loss or gain of electrons in their respective orbitals.

To determine: The formula of the compound A and the structure of complex ion present on the basis of given information.

Explanation of Solution

Explanation

The mass percent of Cr is 10.1%Cr_ .

The molar mass of Cr is 51.99g .

The molar mass of CrO3 is 99.99g .

In 99.99g CrO3 , the mass of Cr present is 51.99g .

In 0.0203g CrO3 , the mass of Cr present =51.99×0.020399.99=0.0106gCr

Mass percent of Cr is calculated by the formula,

Masspercent=MassofCrMassofcompoundA×100%

The mass of compound A is 0.105g .

The mass of Cr is 0.0106gCr .

Substitute the value of mass of compound A and MassofCr in the above formula.

Masspercent=0.0106g0.105g×100%=10.1%Cr_

The mass percent of NH3 is 16.5%NH3_ .

One mL HCl equals to 0.100HCl .

Therefore, 32.93mLHCl is equal to 32.93×0.100mLHCl

Molar mass of NH3 is 17.04g .

One mL of HCl titrate 17.04mgNH3 .

Therefore, the mass of NH3 titrated by 32.93×0.100mLHCl is calculated as,

17.04×32.93×0.100=56.1gNH3

Mass percent of NH3 is calculated by the formula,

Masspercent=MassofNH3MassofcompoundA×100%

The mass of compound A is this case is 0.341g .

The mass of NH3 is 56.1g .

Substitute the value of mass of compound A and mass of NH3 in the above formula.

Masspercent=56.1g0.341g×100%=16.5%NH3_

Except chromium, ammonia and iodine, no other element is needed in compound A.

Given

The mass percent of iodine is 73.53% .

To check whether any other element is needed or not, mass percent of Cr , NH3 and iodine are added shown in the formula given below.

MasspercentofCr+MasspercentofNH3+MasspercentofI

Substitute the values of mass percent in the above formula.

74.53%+16.5%+10.1%=100%

Therefore, no other element is required in the compound A

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Chapter 20 Solutions

Chemistry: An Atoms First Approach
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