Volume of N 2 at NTP required to form a monolayer on the surface of iron catalyst is 8.15 mL / g of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16 × 10-22 m 2 area?

Question

Volume of N 2 at NTP required to form a monolayer on the surface of iron catalyst is 8.15 mL / g of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16 × 10-22 m 2 area? (A) 16 × 10-16 cm2 (B) 0.35 m 2 / g (C) 39 m 2 / g (D) 22400 cm 2

Tardigrade Admin · Accepted Answer

N 2=8.15 × 10-3 L of NO 2 Moles of N 2=(8.15 × 10-3/22.4)=0.3638 × 10-3 Molecules =0.3638 × 10-3 × 6.022 × 1023=2.19 × 1020 ∴ surface area covered =(16 × 10-29 m 2)(2.19 × 1020) =35.072 × 10-9 m 2 =0.35 m 2 / g

Q. Volume of $N_{2}$ at $NTP$ required to form a monolayer on the surface of iron catalyst is $8.15 m L / g$ of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies $16 \times 1 0^{- 22} m^{2}$ area?

$16 \times 1 0^{- 16} c m_{2}$

$0.35 m^{2} / g$

$39 m^{2} / g$

$22400 c m^{2}$

Q. Volume of N2​ at NTP required to form a monolayer on the surface of iron catalyst is 8.15mL/g of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16×10−22m2 area?

16×10−16cm2​

0.35m2/g

39m2/g

22400cm2

N2​=8.15×10−3L of NO2​ Moles of N2​=22.48.15×10−3​=0.3638×10−3 Molecules =0.3638×10−3×6.022×1023=2.19×1020 ∴ surface area covered =(16×10−29m2)(2.19×1020) =35.072×10−9m2 =0.35m2/g

Q. Volume of $N_{2}$ at $NTP$ required to form a monolayer on the surface of iron catalyst is $8.15 m L / g$ of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies $16 \times 1 0^{- 22} m^{2}$ area?

$16 \times 1 0^{- 16} c m_{2}$

$0.35 m^{2} / g$

$39 m^{2} / g$

$22400 c m^{2}$