Chemistry · 2019
JEE Main · 8 April 2019, Shift 2 · Q60
27 g of a long chain fatty acid was dissolved in 100 cm^3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round…
$\displaystyle 27$ g of a long chain fatty acid was dissolved in $\displaystyle 100 \mathrm{~cm}^3$ of hexane. $\displaystyle 10$ 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 $\displaystyle 10$ cm. What is the height of the monolayer ?
[Density of fatty acid $\displaystyle =0.9 \mathrm{~g} \mathrm{~cm}^{-3} ; \pi=3$ ]
Official answer
(2)
$\displaystyle 10^{-6} \mathrm{~m}$
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JEE Main 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.