JEE Main 2019ChemistrySome Basic Concepts of ChemistryHardMCQ

JEE Main 2019 — Some Basic Concepts of Chemistry Question with Solution

JEE Main 2019 (08 Apr Shift 2)

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;π=3] 

Choose an option

Show full solutionCorrect option: B
Correct answer
B10-6 m

Step-by-step explanation

Mass of fatty acid =0.027 g in 10 ml solution

Density of fatty acid 0.9  g/cc

Volume of fatty acid =0.0270.9=0.03 cc

Area of plate =πr2=3×102=300  cm2

Height of fatty acid layer =volumearea=0.03300=10-4 cm = 10-6m

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About this question

This is a previous-year question from JEE Main 2019, covering the Some Basic Concepts of Chemistry 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.