Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393600681
Author: Gilbert
Publisher: W. W. Norton & Company
Question
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Chapter 10, Problem 10.121QA
Interpretation Introduction

To identify:

The element X from the reaction of ammonia and unknown acid HX by using 1.0 m glass tube.

Expert Solution & Answer
Check Mark

Answer to Problem 10.121QA

Solution:

The Element X is Bromine (Br).

Explanation of Solution

1) Concept:

Graham’s law of diffusion states that the diffusion rate of a gas is inversely proportional to the square root of its molar mass.We can use the equation of Graham’s law of diffusion to find the atomic mass of element X and from its molar mass we can identify it the element X.

In this reaction, ammonia and HX are 1.00 m away from each other in a glass tube. When ammonia reacts with HX acid it forms a product at 68.5 m.

2) Formula:

effusion rateNH3effusion rateHX= molar massHXmolar massNH3

3) Given:

i) lenght of glass tube=1.00 m=100 cm.

ii) Diffusion of NH3=68.5cm.

iii) M of NH3=17.0gmol

4) Calculations:

The molar mass of the ammonia is 17.0 g/mol. So, the diffusion rate of ammonia is greater than HX acid. So, the distance travelled by the ammonia gas is 68.5 cm.

Calculate the distance travelled of the HX  acid in the tube.

Diffusion of HX=100 cm-68.5 cm=31.5 cm

effusion rateNH3effusion rateHX= molar massHXmolar massNH3

68.5 cm31.5 cm=molar massHX17.03 gmol

2.1746=molar massHX17.03 gmol

Now take the square of both sides.

4.7289=molar massHX17.03 gmol

molar massHX=4.7289×17.03 gmol=80.5329gmol

molar massHX=atomic mass of H+atomic mass of X

80.5329gmol=1.0 gmol+atomic mass of X

atomic mass of X=80.5329gmol-1.0gmol=79.53gmol

79.53gmol is the atomic mass of Bromine.

Therefore, the unknown element X in this reaction is Bromine (Br).

Conclusion:

Using diffusions of ammonia and HX acid, we got the atomic mass of X. From the atomic mass, we identified the element is Bromine.

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Chapter 10 Solutions

Chemistry: An Atoms-Focused Approach

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