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Question

The radioactive decay, 83Bi211 81Tl207, takes place in 10 L closed container at 0C. Half - life of 83Bi211 is 2.15 min. Starting with 1 mole of 83Bi211, the pressure developed in the container after 4.30 min. will be


A

16.8 atm

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B

22.4 atm

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C

2.24 atm

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D

1.68 atm

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Solution

The correct option is D

1.68 atm


83Bi211 81Tl207+2He4
Fraction left =(12)n=(12)4.302.15=14
Moles of He produced = moles of Bi decayed
=114=0.75
Pressure developed due to He
=22.4×0.7510=1.68 atm


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