JEE Advanced 2024 Paper 2 · Q08 · Solutions & Colligative Properties
To form a complete monolayer of acetic acid on 1 g 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 $\times 10^{-23}$ $m^2$ surface area on charcoal, the value of P is _____. [Use given data: Surface area of charcoal = $1.5 \times 10^2$ $m^2 g^{-1}$; Avogadro's number ($N_A$) = $6.0 \times 10^{23}$ $mol^{-1}$]
Reveal answer + step-by-step solution
Correct answer:2500
Solution
Moles of NaOH used = $40 \times 10^{-3} \times 1 = 4 \times 10^{-2}$ mol = moles of unadsorbed acetic acid. Total moles of acetic acid taken = $100 \times 10^{-3} \times 0.5 = 5 \times 10^{-2}$ mol. Moles adsorbed = $5\times 10^{-2} - 4\times 10^{-2} = 1 \times 10^{-2}$ mol. Number of molecules adsorbed = $1 \times 10^{-2} \times 6 \times 10^{23} = 6 \times 10^{21}$. Total surface area available on 1 g charcoal = $1.5 \times 10^{2}\ m^2$. Area per molecule $= \frac{1.5 \times 10^{2}}{6 \times 10^{21}} = 2500 \times 10^{-23}\ m^2$. So $P = 2500$.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →