Nitrogen adsorption on a solid follows x/m=kp^1/n. The values are x/m=6.0 mg g^-1 at 0.20 atm and 12.0 mg g^-1 at 0.80 atm. Find the number of N2 molecules (in units of 10^22, two decimals) adsorbed by 50 g of the solid at 5.0 atm. Take N2=28 g mol^-1 and N_A=6.022x10^23 mol^-1.

JEE Advanced levelNumerical answerDifficulty 3 of 3sc049

Nitrogen adsorption on a solid follows $x / m = k p^{1/ n}$ . The values are $x / m = 6.0 mg g^{- 1}$ at $0.20 atm$ and $12.0 mg g^{- 1}$ at $0.80 atm$ . Find the number of $N X_{2}$ molecules (in units of $1 0^{22}$ , two decimals) adsorbed by $50 g$ of the solid at $5.0 atm$ . Take $N X_{2} = 28 g mo l^{- 1}$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ .

Numerical answer type, answer in x 10^22 molecules, tolerance 0.03.

Pressure ratio

0.80/0.20 = 4

and

x / m

ratio

= 2

, so

1/ n = 0.50

and

k = 6.0/0.2 0^{0.5} = 13.416

. At

5.0 atm

x / m = 13.416 \times 5. 0^{0.5} = 30.0 mg g^{- 1}

. Mass on

50 g

= 1500 mg = 1.5 g = 0.0536 mol

, giving

3.23 \times 1 0^{22}

molecules.