JEE Main 8 Apr 2019 Paper 2 Solved Paper

Section: Chemistry

Question : 36 of 90

Marks: +1, -0

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 = 0.9gcm−3;π=3 ]

[8 Apr 2019 Shift 2]

10−6m
10−2m
10−4m
10−8m

Validate

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