1 g of CHARCOAL HAVING SURFACE AREA 3.02×10^2 IS MIXED IN 100 ML OF 0.5 M ACETIC ACID SOLUTION ,AFTER FEW TIMES CONCENTRATIONS OF SOLUTION BECOMES 0.49 M WHAT WILL BE SURFACE AREA OCCUPIED ONE MOLECULE OF ACETIC ACID?
Answers
given,
molarity of acetic acid=0.5M
volume=100ml ie,0.1L
no of moles=molarity X volume
Therefore no of moles =0.05
similarly moles left after adsorption=0.49M X 0.1L =0.049
hence no of moles of adsorbed=0.05-0.049=0.001 mole
0.001 mole corresponds to 6.02 X 10^20 molecules
surface area occupied by each molecule=total surface area of charcoal/number of molecules of acetic acid
=3.02 X 10^2/6.02 X 10^20
=5 X 10^-19 m^2
Answer: 5×10^-19 m2
Explanation:
Number of moles of acetic in 100 ml before adding charcoal = 0.05
Number of moles of acetic acid in 100 ml after adding charcoal = 0.049
Number of moles of acetic acid adsorbed on the surface of charcoal = 0.001
Number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 * 6.02 * 1023 = 6.02 * 1020
Surface area of charcoal = 3.01 * 102 m2(given)
Area occupied by single acetic acid molecule on the surface
of charcoal 3.01 *102/6.02 * 1020 = 5 *10-19m2