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Question: 0.27 g of a long chain fatty acid was dissolved in \(100c{m^3}\)of hexane. 10 mL of this solution wa...

0.27 g of a long chain fatty acid was dissolved in 100cm3100c{m^3}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−30.9gc{m^{ - 3}}, π=3\pi = 3]
A. 10−8m{10^{ - 8}}m
B. 10−6m{10^{ - 6}}m
C. 10−4m{10^{ - 4}}m
D. 10−2m{10^{ - 2}}m

Explanation

Solution

Hint: In this question we will use some basic concepts of chemistry. To find the density of any object, we need to know the Mass (grams) of the object, and its Volume (measured in mL or cm3c{m^3}). Divide the mass by the volume in order to get an object's Density.
density=massvolumedensity = \dfrac{{mass}}{{volume}}

Complete answer:
Formula used: density=massvolumedensity = \dfrac{{mass}}{{volume}}, volume=area×heightvolume = area \times height.
Given that, mass = 0.27g, density = 0.9gcm−30.9gc{m^{ - 3}}, distance from edge to centre of watch glass = 10cm.
We know that, Density is a measure of mass per unit volume. The average density of an object is equal 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).
⇒1cm3=1ml ⇒100cm3=100ml  \Rightarrow 1c{m^3} = 1ml \\\ \Rightarrow 100c{m^3} = 100ml \\\
Then 10ml of hexane contains = 0.27×10100=0.027g\dfrac{{0.27 \times 10}}{{100}} = 0.027g .
We know that, volume = massdensity\dfrac{{mass}}{{density}}
⇒\Rightarrow Volume of fatty acid over glass plate = massdensity\dfrac{{mass}}{{density}}
⇒\Rightarrow Volume of fatty acid over glass plate = 0.027g0.9g/cm3\dfrac{{0.027g}}{{0.9g/c{m^3}}}
⇒\Rightarrow Volume of fatty acid over glass plate = 0.03cm30.03c{m^3}
Now, Volume=area×heightVolume = area \times height
⇒0.03cm3=πr2×height ⇒0.03cm3=3×(10)2×height ⇒0.03cm3=300cm2×height ⇒0.03cm3300cm2=height ⇒10−4cm=height  \Rightarrow 0.03c{m^3} = \pi {r^2} \times height \\\ \Rightarrow 0.03c{m^3} = 3 \times {(10)^2} \times height \\\ \Rightarrow 0.03c{m^3} = 300c{m^2} \times height \\\ \Rightarrow \dfrac{{0.03c{m^3}}}{{300c{m^2}}} = height \\\ \Rightarrow {10^{ - 4}}cm = height \\\
Hence, height = 10−4cm{10^{ - 4}}cm
height = 10−4100m=10−6m\dfrac{{{{10}^{ - 4}}}}{{100}}m = {10^{ - 6}}m
Therefore, the correct answer is option (B).

Note: Whenever we are asked such types of questions, we will use some basic formulae like density, volume etc. first we have to identify the given parameters and using them we will determine the other required parameters. Then we will find out the volume of the given solution and using the volume and other given parameters we can easily find out the height by using the formula of volume. Through this we will get the required answer.