1 gram of charcoal adsorbs 100 mL of 0.5 M CH_(2)COOH to from a monolayer and thereby the molarity of acetic acid isreduced to 0.49 M. Calculate the surface area of the charcool adsorbed by each molecule of acetic acid. Surface area of charcoal 3.01xx10^(2)m^(2)//gram.

1 gram of charcoal adsorbs 100 mL of 0.5 M CH_(2)COOH to from a monola

1.	1 gram of charcoal adsorbs 100 mL of 0.5 M CH_(2)COOH to from a monolayer and thereby the molarity of acetic acid isreduced to 0.49 M. Calculate the surface area of the charcool adsorbed by each molecule of acetic acid. Surface area of charcoal 3.01xx10^(2)m^(2)//gram.
Answer» Solution :Sept I. Calculation of number of molecules of `CH_(3)COOH` adsorbed. No. of moles of acetic acid initially present `=((0.5mol))/((1000mL))xx(100mL)=0.05 MOL.` No of moles of acetic acid left `=((0.49mol))/((1000mL))xx(100mL)=0.0049 mol.` No. of moles of acetic acid adsorbed = 0.05 - 0.049 = 0.001 mol. No. of molecules of acetic acid adsorbed `=0.001molxx6.022xx10^(23)mol^(-1)` `=6.022xx10^(20)` molecules. Sept II. Calculate of surface AREA of charcoal adsorbed by ONE molecule of acetic acid. According to available data, 1g of charcoal has area `=3.01xx10^(2)m^(2)` Thus, `6.022 xx 10^(20)` molecules of acetic acid get adsorbed on surface area `=3.01xx10^(2)m^(2)` `therefore` 1 molecule of acetic acid gets adsorbed on surface area `=((3.01xx10^(2)m^(2)))/((6.002xx10^(20)))=5.0xx10^(-19)m^(2)`