JEE Main 2019 — Solid State Question with Solution
From: JEE Main 2019 (Online) 8th April Evening Slot
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 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 = 0.9 g cm–3, = 3]
[Density of fatty acid = 0.9 g cm–3, = 3]
Choose an option
Show full solutionCorrect option: D
Correct answer
D10–6 m
Step-by-step explanation
In 100 ml of hexane solution contains 0.27 g of fatty acid.
In 10 ml of hexane solution contains 0.027 g of fatty acid.
Volume of fatty acid present on the round glass =
As here Area of fatty acid layer = Area of round plate =
Volume of fatty acid layer = h
h =
3 h =
h = 10-4 cm = 10-6 m
In 10 ml of hexane solution contains 0.027 g of fatty acid.
Volume of fatty acid present on the round glass =
As here Area of fatty acid layer = Area of round plate =
Volume of fatty acid layer = h
h =
3 h =
h = 10-4 cm = 10-6 m
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This is a previous-year question from JEE Main 2019, covering the Solid State chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.