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0.27 g of a long chain fatty acid was dissolved in 100 cm 3 of hexane, 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaportates and a monolayer is formed. The distance from edge to centre of the watch glassf is 10 cm. What is the height of the monolayer ? [Density of fatty acid = 0.9 g cm –3 , π = 3]

1. 10 –2 m
2. 10 –4 m
3. 10 –8 m
4. 10 –6 m

Correct answer: 10 –6 m

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