1 gm of charcoal adsorbs 100 mL of 0.5 M CH 3 COOH to form mono layer and there by the molarity of CH 3 COOH reduces to 0.49 M. Calculate the surface area of the charcoal adsorbed by each molecule of CH 3 COOH. Surface area of charcoal = 3.01 × 10 2 m 2 /g.
Text Solution
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5 × 10 –19 m 2
Sol. 100 ml of 0.5 CH 3 COOH contains = 0.05 mole
After adsorption, CH 3 COOH present = 0.049 mole
Acetic acid adsorbed by 1 gm charcoal = 0.05 – 0.049 = 0.001 mole = 6.023 × 10 20 molecule
Surface area of 1 gm charcoal = 3.01 × 10 2
Surface area of charcoal adsorbed by each molecule = 3.01 × 10 2 / 6.023 × 10 20 = 5 × 10 –19 m 2 .
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