0.27 g of a long chain fatty acid was dissolved... | Surface Chemistry

JEE MAIN 2019

08-04-2019 S2

Question

0.27 g of a long chain fatty acid was dissolved in $100 \mathrm{~cm}^3$ 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 $\left.=0.9 \mathrm{~g} \mathrm{~cm}^{-3} ; \pi=3\right]$

Select the correct option:

$10^{-8} \mathrm{~m}$

$10^{-4} \mathrm{~m}$

$10^{-2} \mathrm{~m}$

$10^{-6} \mathrm{~m}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution