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Q. N2 is absorbed in 20% of the surface sites. N2 gas evolved on heating was collected at 0.001atm and 298 K in a container of volume 2.46cm3. Find out the no. of surface sites occupied per molecule of N2. If the density of surface sites is 6.023 x 1014 / cm2 and surface area is 1000 cm2.[IIT 2005][NOTE : Please the one who wants to post answer to this question can give me the steps of calculation so that it would be clear for me to understand.]– This is a numeric response type question

hiphop aadhi2001 , 7 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 7 Years ago

Dear student
 
Partial pressure of N2 = 0.001 atm,T = 298 K,V = 2.46 dm3.From ideal gas law : pV = nRTn= pV/RT(N ) 0.001 *2.462/ 0.082 *298 = 10-7No. of molecules of N2 = 6.023 ´ 10^23 ´* 10-7= 6.023 ´ *1016Now, total surface sites available= 6.023 ´* 10^14 *´ 1000= 6.023 * 1017Surface sites used in adsorption = 20/100*6.023 ´* 10^17= 2 ´ *6.023 ´ 10^16Sites occupied per molecules = Number of sites/Number of molecules= 2* 6.023* 10^16/6.023 10^16 = 2
 
 
Regards
Arun (askIITians forum expert)

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