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Question

20% surface sites have absorbed N2. On heating, N2 gas is evolved from the sites and is collected at 0.001 atm and 298 K in a container of volume 2.46 cm3. Density of surface sites is 6.023×1014 cm2 and surface area is 1000 cm2. The number of surface sites occupied per molecule of N2 is:

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Solution

For absorbed N2 on surface sites,
PN2=0.001 atm, V=2.46 cc =2.46×103 L, T=298 K

nN2=PVRT=0.001×2.46×1030.0821×298=1.0×107

Molecules of N2 absorbed =1.0×107×6.023×1023 =6.023×1016

Total surface sites available = Number of sites per cm2× Area =6.023×1014×1000=6.023×1017

Surface sites on which N2 is absorbed =20%× Available sites
=20100×6.023×1017 =12.046×1016

Number of sites absorbed per molecule of N2
=12.046×10166.023×1016=2

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