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?

Correct option is B. 10⁻⁶ m
Solution:- (B) 10⁻⁶ m
Radius of watchglass = 10 cm ⇒ surface area = πr² = 3 × (10 cm)² = 300 cm²
Mass of fatty acid in 10 ml solution = 10 × 0.27/100 = 0.027 gm
volume of fatty acid = 0.027 g / 0.9 g/ml = 0.03 cm³
⇒ Height = volume of fatty acid / surface area of watch glass = 0.03 cm³ / 300 cm² = 0.0001 cm = 10⁻⁶ m