A powdered solid adsorbs a monolayer of $5.6 c m^{3}$ (at STP) of nitrogen per gram. Each $N X_{2}$ molecule covers $0.16 n m^{2}$ . Taking molar volume at STP $= 22400 c m^{3} mo l^{- 1}$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , calculate the surface area of the solid in $m^{2} g^{- 1}$ (one decimal place).

Numerical answer type, answer in m^2/g, tolerance 0.2.

Answer: 24.088 m^2/g

Moles

= 5.6/22400 = 2.50 \times 1 0^{- 4}

. Molecules

= 2.50 \times 1 0^{- 4} \times 6.022 \times 1 0^{23} = 1.506 e + 20

. Area

= 1.506 e + 20 \times 0.16 \times 1 0^{- 18} m^{2} = 24.1 m^{2}

per gram.