JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

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]

A
10^–8 m
B
10^–4 m
C
10^–2 m
D
10^–6 m

Solution

0.27 gm in 100 ml of hexane

$∴$ in 10 ml of aqueous solution only 0.027 gm acid is present

volume of 0.027 g acid $= \frac{0.027}{0.9} ml$

$∴ π r^{2} h = \frac{0.027}{0.9} (given r = 10 cm, π = 3)$

$∴ h = 1 0^{- 4} cm$

$= 1 0^{- 6} m$

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