10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm² on a polar substrate. If the polar head is approximated as a cube, what is its edge length?
- A.2.0 pm
- B.0.1 nm
- C.1.0 pm
- D.2.0 nm
Show answer and solution
Correct option: A
Solution
Total moles of surfactant = 1×10⁻³ mol/L × 10×10⁻³ L = 10⁻⁵ mol.
Number of molecules = 10⁻⁵ × 6.022×10²³ ≈ 6×10¹⁸ molecules.
Area per molecule = 0.24 cm² / 6×10¹⁸ = 4×10⁻²⁰ cm² = 4×10⁻²⁴ m².
Since the polar head is a cube sitting on the substrate, its face area = edge² , so edge = √(4×10⁻²⁴) = 2×10⁻¹² m = 2.0 pm.
Number of molecules = 10⁻⁵ × 6.022×10²³ ≈ 6×10¹⁸ molecules.
Area per molecule = 0.24 cm² / 6×10¹⁸ = 4×10⁻²⁰ cm² = 4×10⁻²⁴ m².
Since the polar head is a cube sitting on the substrate, its face area = edge² , so edge = √(4×10⁻²⁴) = 2×10⁻¹² m = 2.0 pm.
[JEE Main 2019 · 9 Apr · Shift II]