Question

Volume of n2 at ntp to form a mono layer on the surface of iron catalyst is 8.15ml/gram of the absorbent. What will be the surface area of the adsorbent per gm if each nitrogen molecules occupies 16*10^-22 m2.

Answer 1

Answer : The total surface area of absorbent per gram is $0.350m^2/g$

Explanation:

First we have to calculate the moles of $N_2$ at NTP.

Using ideal gas equation:

$PV=nRT\\\\n=\frac{PV}{RT}$

where,

P = pressure of gas at NTP = 1 atm

V = volume of gas at NTP = $8.15mL/g=8.15\times 10^{-3}L/g$

T = temperature of gas at NTP = 273 K

R = gas constant = 0.0821 L.atm/mole.K

n = number of moles = ?

Now put all the given values in this formula, we get:

$n=\frac{(1atm)\times (8.15\times 10^{-3}L/g)}{(0.0821L.atm/mole.K)\times (273K)}=3.64\times 10^{-4}mole/g$

Now we have to calculate the number of molecules of $N_2$ required.

As, 1 mole of $N_2$ required $6.022\times 10^{23}$ number of molecules

So, $3.64\times 10^{-4}mole/g$ mole of $N_2$ required $3.64\times 10^{-4}\times 6.022\times 10^{23}=2.19\times 10^{20}$ number of molecules per gram.

Now we have to calculate the total surface area of absorbent.

Total surface area of absorbent = $(16\times 10^{-22})\times (2.19\times 10^{20})=0.350m^2/g$

Therefor, the total surface area of absorbent per gram is $0.350m^2/g$