g What are the relative rates of diffusion of the three naturally occurring isotopes of krypton, 80Kr80Kr , 82Kr82Kr, and 83Kr83Kr. What are

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g What are the relative rates of diffusion of the three naturally occurring isotopes of krypton, 80Kr80Kr , 82Kr82Kr, and 83Kr83Kr. What are the relative rates of diffusion of the three naturally occurring isotopes of krypton, , , and . The relative rates of diffusion are: 82Kr(1.02)>82Kr(1.02)>83Kr(1.01)>80Kr(1.00)83Kr(1.01)>80Kr(1.00) The relative rates of diffusion are: 80Kr(1.02)>80Kr(1.02)>82Kr(1.01)>83Kr(1.00)82Kr(1.01)>83Kr(1.00) The relative rates of diffusion are: 82Kr(1.02)>82Kr(1.02)>80Kr(1.01)>83Kr(1.00)80Kr(1.01)>83Kr(1.00) The relative rates of diffusion are: 83Kr(1.02)>83Kr(1.02)>82Kr(1.01)>80Kr(1.00)82Kr(1.01)>80Kr(1.00)

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Hồng Cúc 3 years 2021-08-31T05:04:39+00:00 1 Answers 14 views 0

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  1. danthu
    0
    2021-08-31T05:05:39+00:00
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    Solution :

    According to the Graham’s law of diffusion, we know that, the rate of the diffusion varies inversely to the molar mass of the gas, i.e.

    Rate of diffusion, $r_d = \frac{a}{\sqrt M}$

    where, the ‘M’ is the molar mass of the gas.

    Now in the case of the isotopes of the Krypton,

    Atomic mass of $^{80}Kr$ = 80 AMU

    Atomic mass of $^{82}Kr$ = 82 AMU

    Atomic mass of $^{83}Kr$ = 83 AMU

    So the ratio of the rate of diffusion of the three isotopes are :

    $M_{d,^{80}Kr}:M_{d,^{82}Kr}:M_{d,^{83}Kr}$

    $=\frac{1}{\sqrt{80}}:\frac{1}{\sqrt{82}}:\frac{1}{\sqrt{83}}$

    $=0.1118:0.1104:0.10976$

    Dividing the above three with the smallest number among the three i.e. 0.10976, we get the relative rates of diffusion.

    ∴ $M_{d,^{80}Kr}:M_{d,^{82}Kr}:M_{d,^{83}Kr}$

     = 1.02 : 1.01 : 1

    Hence the relative rate of diffusion are :

    $^{80}Kr(1.02)>^{82}Kr(1.01)>^{83}Kr(1.00)$

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