Verified Solution • 2025–26 Session moderate

QJEEM19Ch387e: JEE Main 2019 Chemistry - Class 12 JEEM19Ch387e Solution: 0.27 g of a long chain fatty acid was dissolved in 100 cm 3 of…

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]

Multiple Choice Options

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

Step-by-Step Verified Solution

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 = 0.0270.9

As here Area of fatty acid layer = Area of round plate = πr2

∴ Volume of fatty acid layer = πr2 × h

∴ πr2 × h = 0.0270.9

⇒ 3 × (10)2 × h = 0.0270.9

⇒ h = 10-4 cm = 10-6 m

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