Question

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)

Answer 1

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 :

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