JEE Advanced 2024 Chemistry question
To form a complete monolayer of acetic acid on 1g of charcoal, 100 mL of 0.5 M 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 ]
Answer: 2500
Solution
Acetic acid taken mmol. Unadsorbed acid NaOH used mmol. Adsorbed acid mmol mol, i.e. 10^{-2}10^{23}$ =
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