To form a complete monolayer of acetic acid on 1g of charcoal, 100 mL of 0.5M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M

Question

To form a complete monolayer of acetic acid on 1g of charcoal, 100 mL of 0.5M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of 1 M NaOH solution was required. If each molecule of acetic acid occupies  surface area on charcoal, the value of P is .

[Use given data: Surface area of charcoal = ; Avogadro’s number () = ]

Canvas Classes · Accepted Answer

The acid that disappeared from solution is the acid stuck to the charcoal. Find it by subtracting the leftover (titrated by ) from what you started with; then spread those molecules over the known surface area to get the footprint of one molecule.

Acid taken initially

Unadsorbed acid  meq of

Adsorbed acid

Molecules adsorbed

Total surface area  (since )

Area per molecule

So .

Watch out: the unit jump  is , not  — that is a squared length. Miss it and the power of ten is off by six.

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