0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?[Density of fatty acid = 0.9 g cm-3, π = 3]

0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane.

1.	0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?[Density of fatty acid = 0.9 g cm-3, π = 3]
A.	10-6 m
B.	10-8 m
C.	10-2 m
D.	10 -4 m
Answer» B. 10-8 m