Question

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 the water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from the 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]

Options

• 10⁻⁶ m

• 10⁻⁸ m

• 10⁻² m

• 10⁻⁴ m

Answer 1

10⁻⁶ m

Explanation:

Given: 0.27 g is present in 100 cm³ of hexane.

∴ 10 mL of aqueous solution contains only 0.027 g of acid.

Volume of 0.027 g of acid = `0.027/0.9` mL

∴ πr²h = `0.027/0.9` ...(given r = 10 cm, π = 3)

∴ h = 10⁻⁴ cm = 10⁻⁶ m