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20% surface sites have adsorbed N2. On heating N2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume is 2.46 cm3. Density of surface sites is 6.023×1014 / cm2and surface area is 1000 cm2, find out the no. of surface sites occupied per molecule of N2.

20% surface sites have adsorbed N2. On heating N2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume is 2.46 cm3. Density of surface sites is 6.023×1014 / cm2and surface area is 1000 cm2, find out the no. of surface sites occupied per molecule of N2. 


Grade:11

1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
Partial pressure of N2 = 0.001 atm,
T = 298 K,V = 2.46 dm3.
From ideal gas law : pV = nRT
n= pV/RT
(N ) 0.001 *2.462/ 0.082 *298 = 10-7
No. of molecules of N2 = 6.023 ´ 10^23 ´* 10-7
= 6.023 ´ *1016
Now, total surface sites available
= 6.023 ´* 10^14 *´ 1000
= 6.023 * 1017
Surface sites used in adsorption = 20/100*6.023 ´* 10^17
= 2 ´ *6.023 ´ 10^16
Sites occupied per molecules = Number of sites/Number of molecules
= 2* 6.023* 10^16/6.023 10^16 = 2

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