A solution of palmitic acid (Molar mass = 256 g/mol) in benzene contains 5.12 g of acid per dm³. When this solution is dropped on a water surface the benzene evaporates and the palmitic acid forms a m

Question

A solution of palmitic acid (Molar mass = 256 g/mol) in benzene contains 5.12 g of acid per dm³. When this solution is dropped on a water surface the benzene evaporates and the palmitic acid forms a monomolecular film of the solid type. If we wish to cover an area of 480 cm² with a monolayer, what volume (in mm³) of palmitic acid solution should be used? The area covered by one palmitic acid molecule may be taken to be 0.2 nm².

Prashant Kotian · Accepted Answer

Correct answer: 0020. Area to be covered = 480 cm² = 4.8 × 10¹⁶ nm². Area covered by one molecule = 0.2 nm². Total number of molecules required = (4.8 × 10¹⁶) / 0.2 = 2.4 × 10¹⁷ molecules. Moles of palmitic acid = (2.4 × 10¹⁷) / (6 × 10²³) = 4 × 10⁻⁷ mol (Assuming N_A = 6 × 10²³ for simple calculation consistent with standard problem conventions). Mass of palmitic acid required = Moles × Molar Mass = 4 × 10⁻⁷ mol × 256 g/mol = 1.024 × 10⁻⁴ g. The concentration of the solution is 5.12 g/dm³ = 5.12 × 10⁻³ g/cm³ = 5.12 × 10⁻⁶ g/mm³. Volume of solution required = Mass / Concentration = (1.024 × 10⁻⁴ g) / (5.12 × 10⁻⁶ g/mm³) = 20 mm³. Therefore, correct answer is 0020.

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