   Question Answers

# 0.27 g of a long chain fatty acid was dissolved in $100c{m^3}$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 = $0.9gc{m^{ - 3}}$, $\pi = 3$]A. ${10^{ - 8}}m$B. ${10^{ - 6}}m$C. ${10^{ - 4}}m$D. ${10^{ - 2}}m$  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 $c{m^3}$). Divide the mass by the volume in order to get an object's Density.
$density = \dfrac{{mass}}{{volume}}$

Formula used: $density = \dfrac{{mass}}{{volume}}$, $volume = area \times height$.
Given that, mass = 0.27g, density = $0.9gc{m^{ - 3}}$, 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).
$\Rightarrow 1c{m^3} = 1ml \\ \Rightarrow 100c{m^3} = 100ml \\$
Then 10ml of hexane contains = $\dfrac{{0.27 \times 10}}{{100}} = 0.027g$ .
We know that, volume = $\dfrac{{mass}}{{density}}$
$\Rightarrow$ Volume of fatty acid over glass plate = $\dfrac{{mass}}{{density}}$
$\Rightarrow$ Volume of fatty acid over glass plate = $\dfrac{{0.027g}}{{0.9g/c{m^3}}}$
$\Rightarrow$ Volume of fatty acid over glass plate = $0.03c{m^3}$
Now, $Volume = area \times height$
$\Rightarrow 0.03c{m^3} = \pi {r^2} \times height \\ \Rightarrow 0.03c{m^3} = 3 \times {(10)^2} \times height \\ \Rightarrow 0.03c{m^3} = 300c{m^2} \times height \\ \Rightarrow \dfrac{{0.03c{m^3}}}{{300c{m^2}}} = height \\ \Rightarrow {10^{ - 4}}cm = height \\$
Hence, height = ${10^{ - 4}}cm$
height = $\dfrac{{{{10}^{ - 4}}}}{{100}}m = {10^{ - 6}}m$
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.
Uses of Oxalic Acid  Acid Anhydride  Acetylsalicylic Acid or Aspirin  Fatty Liver Symptoms  Preparation of Standard Solution of Oxalic Acid  Carboxylic Acid Properties  An Overview of Food Chain  Electron Transport Chain  Value of g  Grazing Food Chain  Important Questions for CBSE Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry  Important Questions for CBSE Class 11 Chemistry  Important Questions for CBSE Class 11 Chemistry Chapter 14 - Environmental Chemistry  Important Questions for CBSE Class 12 Chemistry Chapter 5 - Surface Chemistry  Important Questions for CBSE Class 11 Chemistry Chapter 5 - States of Matter  Important Questions for CBSE Class 11 Chemistry Chapter 2 - Structure of Atom  Important Questions for CBSE Class 11 Chemistry Chapter 3 - Classification of Elements and Periodicity in Properties  Important Questions for CBSE Class 11 Chemistry Chapter 13 - Hydrocarbons  Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics  Important Questions for CBSE Class 11 Chemistry Chapter 12 - Organic Chemistry - Some Basic Principles and Techniques  Chemistry Question Paper for CBSE Class 12  CBSE Class 12 Chemistry Question Paper 2020  Chemistry Question Paper for CBSE Class 12 - 2013  Chemistry Question Paper for CBSE Class 12 - 2015  CBSE Class 12 Chemistry Question Paper 2019 - Free PDF  CBSE Class 12 Chemistry Question Paper 2017 - Free PDF  CBSE Class 12 Chemistry Question Paper 2018 - Free PDF  Previous Year Question Paper for CBSE Class 12 Chemistry - 2014  Chemistry Question Paper for CBSE Class 12 - 2016 Set 1 E  Chemistry Question Paper for CBSE Class 12 - 2016 Set 1 S  NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry in Hindi  Surface Chemistry NCERT Solutions - Class 12 Chemistry  NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry In Hindi  NCERT Solutions Class 11 Chemistry  NCERT Exemplar for Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry (Book Solutions)  NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry in Hindi  NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom In Hindi  NCERT Solutions for Class 11 Chemistry in Hindi  NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter In Hindi  Equilibrium NCERT Solutions - Class 11 Chemistry  