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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2019 · 8 April · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

0.27g of a long chain fatty acid was dissolved in 100 cm ^ 3 of...

0.27g0.27g of a long chain fatty acid was dissolved in 100cm3100{cm}^{3} of hexane. 10mL10mL 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 10cm10cm . What is the height of the monolayer?
[Density of fatty acid =0.9gcm−3;π=3]= 0.9gc{m}^{- 3}; \pi = 3 ]

Options

  1. A10−4m{10}^{- 4}m
  2. B10−6m{10}^{- 6}m
  3. C10−8m{10}^{- 8}m
  4. D10−2m{10}^{- 2}m

Correct answer

B. 10−6m{10}^{- 6}m

Step-by-step solution

Mass of fatty acid =0.027g= 0.027g in 10ml10ml solution

Density of fatty acid 0.90.9 g/cc

Volume of fatty acid =0.0270.9=0.03cc= \dfrac{0.027}{0.9} = 0.03cc

Area of plate =πr2=3×102=300cm2= \pi{r}^{2} = 3 \times {10}^{2} = 300c{m}^{2}

Height of fatty acid layer =volumearea=0.03300=10−4cm=10−6m= \dfrac{volume}{area} = \dfrac{0.03}{300} = {10}^{- 4}cm={10}^{- 6}m

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