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Help me answer: - Some basic concepts in chemistry - JEE Main-3

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 , \pi=3]

  • Option 1)

    10-6 m

  • Option 2)

    10-8 m

  • Option 3)

    10-2 m

  • Option 4)

    10-4 m

Answers (1)
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Density -

Density is a measure of mass per unit  volume. The average density of an object is equals to its total mass divided by its total volume. An object made from a comparatively dense material (such as iron) will have less volume than an object of equal mass made from some less dense substance (such as water).

To find the density of any object, we need to know the Mass (grams) of the object, and its Volume (measured in mL or cm³). Divide the mass by the volume in order to get an object's Density.

                                                   D = m/v.

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1 cm3 = 1 mL

\therefore 100 cm3 = 100 mL

Then 10 mL of hexanme contains=\frac{0.27 \times 10}{100}=0.027g

 

Distance from edge to centre of watch glass=10 cm

height of monolayer=?

The volume of fatty acid over glass plate=\frac{Mass}{Density}

                                                    =\frac{0.027g}{0.9g/cm^{3}}\\ =0.03 cm^{3}

Volume = Area X height

0.03 cm^{3}=\pi r^{2} cm^{2} \times H

0.03 cm^{3}=3.14 \times 10^{2} cm^{2}\times H

H=\frac{0.03 cm^{3}}{314 cm^{2}}=10^{-4} cm

So, height of monolayer=10-4 cm

                                       =10-4 X 10-2

                                       =10-6 m  

 


Option 1)

10-6 m

Option 2)

10-8 m

Option 3)

10-2 m

Option 4)

10-4 m

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