JEE Advanced 2024 · Paper 2 · numerical answer

JEE Advanced 2024 Chemistry question on Surface Chemistry

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 P x 10−2310^{-23} m2^2 surface area on charcoal, the value of P is _____.
[Use given data: Surface area of charcoal = 1.5 x 10210^{2} m2^2g−1^{-1}; Avogadro's number (NA_A) = 6.0 x 102310^{23} mol−1^{-1}]
Show answer

Answer: 2500

[JEE Advanced 2024 · Paper 2]

Practise Surface Chemistry the way it is asked

A timed set of real JEE Advanced Surface Chemistry questions, graded as the exam grades. No sign-up.

Start a Surface Chemistry test

More: all JEE Advanced Surface Chemistry questions · JEE Advanced 2024 · Chemistry chapter list