Ncert Solutions Class 9th Physics Chapter 2
NCERT Solutions Class 9 Maths Chapter 2 Is Matter Around Us Pure? – Here are all the NCERT solutions for Class 9 Maths Chapter 2. This solution contains questions, answers, images, explanations of the complete Chapter 2 titled The Fun They Had of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 2 Is Matter Around Us Pure?. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 2 Is Matter Around Us Pure? in one place.
NCERT Solutions Class 9 Science Chapter 2 Is Matter Around Us Pure?
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Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Science for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 2 Is Matter Around Us Pure? , Science, Class 9.
| Class | 9 |
| Subject | Science |
| Book | Science |
| Chapter Number | 2 |
| Chapter Name | Is Matter Around Us Pure? |
NCERT Solutions Class 9 Science chapter 2 Is Matter Around Us Pure?
Class 9, Science chapter 2, Is Matter Around Us Pure? solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: What is meant by a substance? Ans : A pure substance is the one that consists of a single type of particles, i.e., all constituent particles of the substance have the same chemical nature. Pure substances can be Classified as elements or compounds.
Q.2: List the points of differences between homogeneous and heterogeneous mixtures. Ans : A homogeneous mixture is a mixture having a uniform composition throughout the mixture. For example: salt in water, sugar in water, copper sulphate in water A heterogeneous mixture is a mixture having a non-uniform composition throughout the mixture. For example: sodium chloride and iron filings, salt and sulphur, oil and water.
Q.3: Differentiate between homogeneous and heterogeneous mixtures with examples. Ans : A homogeneous mixture is a mixture having a uniform composition throughout the mixture. For example, mixtures of salt in water, sugar in water, copper sulphate in water, iodine in alcohol, alloy, and air have uniform compositions throughout the mixtures. On the other hand, a heterogeneous mixture is a mixture having a non-uniform composition throughout the mixture. For example, composition of mixtures of sodium chloride and iron filings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and water are not uniform throughout the mixtures.
Q.4: How are sol, solution and suspension different from each other? Ans : Sol is a heterogeneous mixture. In this mixture, the solute particles are so small that they cannot be seen with the naked eye. Also, they seem to be spread uniformly throughout the mixture. The Tyndall effect is observed in this mixture. For example: milk of magnesia, mud Solution is a homogeneous mixture. In this mixture, the solute particles dissolve and spread uniformly throughout the mixture. The Tyndall effect is not observed in this mixture. For example: salt in water, sugar in water, iodine in alcohol, alloy Suspensions are heterogeneous mixtures. In this mixture, the solute particles are visible to the naked eye, and remain suspended throughout the bulk of the medium. The Tyndall effect is observed in this mixture. For example: chalk powder and water, wheat flour and water.
Q.5: . To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature. Ans : Mass of solute (sodium chloride) = 36 g (Given) Mass of solvent (water) = 100 g (Given) Then, mass of solution = Mass of solute + Mass of solvent = (36 + 100 )g = 136 g Therefore, concentration (mass by mass percentage) of the solution \( \begin{array}{l}{=\frac{\text { Mass of solute }}{\text { Mass of solvent }} \times 100 \%} \\ {=\frac{36}{136} \times 100 \%} \\ {=26.47 \%}\end{array} \)
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Ncert Solutions Class 9th Physics Chapter 2
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