Home
Class 12
CHEMISTRY
20% surface sites have adsorbed N 2 On h...

20% surface sites have adsorbed `N _2` On heating N, gas is evolved from sites and were collected at 0.001 atm and 298 K in a container of volume `2.46 cm^3`. Density of surface sites is `6.02 xx 10^14 cm^(-2)` and surface area is 1000 `cm^2` Find out the number of surface sites occupied per molecule of `N_2` .

Text Solution

Verified by Experts

The correct Answer is:
2
For adsorbed `N_2` on surface sites
`P_(N_2) = 0.001 atm , V = 2.46 cm^3 = 2.46 xx 10^(-3)` litre, T = 298 K
`therefore n_(N_2) = (PV)/(RT) = (0.001 xx 2.46 xx 10^(-3))/(0.0821 xx 298) = 1.0 xx 10^(-7)` mol
`therefore` Molecules of adsorbed `N_2 = 1.0 xx 10^(-7) xx 6.023 xx 10^(23) = 6.023 xx 10^(16)`
Total surface sites available = Number of sites per `cm^2 xx "Area" = 6.023 xx 10^(14) xx 1000 = 6.023 xx 10^(17)`
Surface sites on which `N_2` is adsorbed = 20 %` xx` Available sites ` = (20)/(100) xx 6.023 xx 10^(17) = 12.046 xx 10^(16)`
`therefore` Number of sites adsorbed per molecules of `N_2 = (12.046 xx 10^(16))/(6.023 xx 10^(16)) = 2`
Promotional Banner

Similar Questions

Explore conceptually related problems

10% sites of catalyst bed have adsorbed by H_2 . On heating H_2 gas is evolved from sites and collectedat 0.03 atm and 300 K in a small vessel of 2.46 cm^3 . No. of sites available is 5.4 xx 10^(16) per cm^2 and surfacearea is 1000 cm^2 . Find out the no. of surface sites occupied per molecule of H_2 : ["Given" N_A = 6 xx 10^(23)]

Slant height of a square pyramid is 30 cm and surface area 3456cm^2 . Find the volume of the pyramid?

4 g of N_2 is allowed to be adsorbed on a 2g solid surface at a 300 K and 1 atm, find the volume of gas adsorbed per gm of solid surface at 1 atm

One of the perpendicular side of a right angled triangle has length 4 cm more than twice of the other side.It has surface area , 80cm^2 .Find out the lengths of the perpendicular sides.

A uniformly charged conducting spheres of 2.4m diameter has a surface charge density of 80.0mu Cm^-2 ?. a) Find the charge on the sphere.

The radius of a sphere is 12.4 ± 0.1 cm.Calculate the percentage error in surface area of the sphere (A = 4pi r^2) .

In a solution of 100 mL 0.5 M acetic acid, one g of active charcoal is added, which adsorbs acetic acid. It is found that the concentration of acetic acid becomes 0.49 M if surface area of charcoal is 3.01 xx 10^(2) m^2 , calculate the area occupied by single acetic acid molecule on surface of charcoal.