0.27 g of a long chain fatty acid was dissolved in `100cm^(3)` of hexane. `10ml` of this solution was added dropwise to the surface of water in a round watch glass . Hexane evaporates and a monolayeer is formed. The distance from edge to center of the watch glass is `10 cm` . What is the height of the monolayers?
`["Density of fatty acid "=0.9 g cm^(-3),pi=3]`

Options

A. `10^4 m`
B. `10^2 m`
C. `10^6 m`
D. `10^8 m`

Explanation

Mass of fatty acid=0.027g, Density =`0.9g//cc`
Volume of fatty acid =0.9
Area of plate =100
Volume of fatty acid= Area of plate `times` height of fatty acid layer
`10^6r`