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What is the specific surface area (in m²/g) of a sample of activated carbon that adsorbs 20 ml of nitrogen gas at STP, assuming monolayer formation and a nitrogen cross-sectional area of 0.162 nm²? The mass of the activated carbon sample is 0.5 g.

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To calculate the specific surface area of the activated carbon, we need to determine the total surface area occupied by the adsorbed nitrogen gas and then divide it by the mass of the activated carbon sample.First, we need to find the number of nitrogen molecules adsorbed on the activated carbon. At STP  standard temperature and pressure , 1 mole of any gas occupies 22.4 L  22,400 mL . Given that the activated carbon adsorbs 20 mL of nitrogen gas, we can calculate the number of moles of nitrogen gas adsorbed:Number of moles of N2 =  Volume of N2 adsorbed  /  Volume occupied by 1 mole of gas at STP Number of moles of N2 = 20 mL / 22,400 mL/molNumber of moles of N2 = 0.000892857 molNow, we can calculate the number of nitrogen molecules using Avogadro's number  6.022 x 10^23 molecules/mol :Number of N2 molecules = Number of moles of N2  Avogadro's numberNumber of N2 molecules = 0.000892857 mol  6.022 x 10^23 molecules/molNumber of N2 molecules  5.37 x 10^20 moleculesNext, we can calculate the total surface area occupied by the adsorbed nitrogen gas using the given nitrogen cross-sectional area  0.162 nm :Total surface area = Number of N2 molecules  Cross-sectional area of N2Total surface area = 5.37 x 10^20 molecules  0.162 nm/moleculeTotal surface area  8.70 x 10^20 nmTo convert the surface area from nm to m, we use the conversion factor 1 nm = 10^-9 m:Total surface area = 8.70 x 10^20 nm   10^-9 m/nm Total surface area  8.70 x 10^2 mFinally, we can calculate the specific surface area by dividing the total surface area by the mass of the activated carbon sample  0.5 g :Specific surface area = Total surface area / Mass of activated carbonSpecific surface area = 8.70 x 10^2 m / 0.5 gSpecific surface area  1.74 x 10^3 m/gSo, the specific surface area of the activated carbon sample is approximately 1,740 m/g.

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