# 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 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; = 3] (a) 10−8 (b) 10−4 (c) 10−2 (d) 10−6

## Question ID - 52489 | SaraNextGen Top Answer 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 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; = 3] (a) 10−8 (b) 10−4 (c) 10−2 (d) 10−6

Answer Key / Explanation : (d) -

0.27 gm in 100 ml of hexane

∴ in 10 ml of aqueous solution only 0.027 gm acid is present

Volume of 0.027g acid = ml

r2h =  (given r = 10 cm,  = 3)

∴ h = 10−4cm

= 10−6m