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NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances

NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances

Edited By Irshad Anwar | Updated on Oct 03, 2023 08:40 PM IST

NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances

NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances: Class 6 Science Chapter 5 Question Answer is part of NCERT Solutions for Class 6 Science. In our daily life, there are many instances when we separate a substance from a mixture of materials, like when tea leaves are separated from the liquid with a strainer, the grain is separated from stalks, cotton separates its seeds from the fibre, milk or curd is churned to separate the butter, etc. You will get questions based on such concepts in separation of substances Class 6 solutions. In this article, you will get NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances.

In NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances, you will get many questions related to these separation methods, which will give you more clarity. On the other hand, students must try to complete the NCERT Class 6 Syllabus for Science as soon as possible so that they can refer to the separation of substances Class 6 NCERT solutions and revise well. You can get NCERT Solutions by clicking on the above link. NCERT solutions for Class 6th science chapter 5, provide answers to each question in the NCERT textbook. Students can refer to this link for NCERT Solutions for Class 6.

NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances

According to the latest CBSE syllabus 2023-24, the chapter "separation of substances" has been renumbered as Chapter 3.

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NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances-Exercises

Question 1:Why do we need to separate different components of a mixture? Give two examples.

Answer: We see that, before we use a substance, we need to separate harmful or useless substances that may be mixed with it. Sometimes, we separate even useful components if we need to use them separately. For examples:

(a)The process that is used to separate the grain from stalks etc. is threshing.

(b) Sieving allows the fine flour particles to pass through the holes of the sieve while the bigger impurities remain on the sieve.

Question 2: What is winnowing? Where is it used?

Answer: This method of separating components of a mixture is called winnowing. Winnowing is used to separate heavier and lighter components of a mixture by wind or by blowing air.

This method is commonly used by farmers to separate lighter husk particles from heavier seeds of grain.

Question 3:How will you separate husk or dirt particles from a given sample of pulses before cooking.

Answer: We will separate husk or dirt particles from a given sample of pulses by the handpicking method.

The handpicking method can be used for separating slightly larger sized impurities like the pieces of dirt, stone, and husk from wheat, rice or pulses.

Question: 4 What is sieving? Where is it used?

Answer: Sieving allows the fine flour particles to pass through the holes of the sieve while the bigger impurities remain on the sieve.

In a flour mill, impurities like husk and stones are removed from wheat before grinding it by sieving.

Question: 5 How will you separate sand and water from their mixture?

Answer: We can separate sand and water from their mixture by the following steps:

  • First by sedimentation, when the heavier component that is sand settles down after water is added to it.
  • Then by decantation where clear water formed an upper layer is removed carefully to another glass.

Question: 6 Is it possible to separate sugar mixed with wheat flour? If yes, how will you do it?

Answer: Yes, it is possible to separate sugar mixed with wheat flour. We will follow the below steps:

  • Add a lot of water to mix wheat flour and sugar both.
  • Then filter this mixture using filter paper.
  • On the filter paper, we have wheat flour.
  • Dry it to get Wheat flour.
  • The filtrate has sugar and water mixed so,
  • Evaporate the water to get sugar.

Question: 7 How would you obtain clear water from a sample of muddy water?

Answer: We can obtain clear water from a sample of muddy water by the following steps:

  • Allow the muddy water to stand for a time.
  • You will see that mud settles down at the bottom.
  • Upper layer water contains some mud particles.
  • Decant it to get that water and filter it to remove traces of mud particles.

Question: 8 Fill up the blanks

(a) The method of separating seeds of paddy from its stalks is called ___________.

(b) When milk, cooled after boiling, is poured onto a piece of cloth the cream (malai) is left behind on it. This process of separating cream from milk is an example of ___________.

(c) Salt is obtained from seawater by the process of ___________.

(d) Impurities settled at the bottom when muddy water was kept overnight in a bucket. The clear water was then poured off from the top. The process of separation used in this example is called ___________.

Answer: (a) The method of separating seeds of paddy from its stalks is called threshing.

(b) When milk, cooled after boiling, is poured onto a piece of cloth the cream (malai) is left behind on it. This process of separating cream from milk is an example of filtration.

(c) Salt is obtained from seawater by the process of evaporation.

(d) Impurities settled at the bottom when muddy water was kept overnight in a bucket. The clear water was then poured off from the top. The process of separation used in this example is called decantation.

Question: 9 True or false?

(a) A mixture of milk and water can be separated by filtration.

(b) A mixture of powdered salt and sugar can be separated by the process of winnowing.

(c) Separation of sugar from tea can be done with filtration.

(d) Grain and husk can be separated with the process of decantation.

Answer: (a) False

(b) False

(c) False

(d) False

Question: 10 Lemonade is prepared by mixing lemon juice and sugar in water. You wish to add ice to cool it. Should you add ice to the lemonade before or after dissolving sugar? In which case would it be possible to dissolve more sugar?

Answer: We should add ice to the lemonade after dissolving the sugar as more sugar will be going to dissolve as solubility is more before adding ice. Because if we add ice before dissolving the sugar, the temperature of the water gets down and the property of dissolving the sugar will decrease.

NCERT Solutions for Class 6 Science Chapter 5 - Separation of Substances

The solution for Class 6 Chapter 5 Science consists of a total of ten questions, including the subjective type of question, true/false, and fill-in-the-blank types of questions. These solutions for NCERT Class 6 Science Chapter 5 are prepared by experienced subject matter experts and are available in PDF format, so students can download and use them according to their comfort.

NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances

Subtopics of separation of substances Class 6 are given below:

Section Name

Topic Name

5

Separation of substances

5.1

Methods of separation

Subtopics of Separation of Substances Class 6 NCERT Science Chapter 5 :

Topics for Chapter 5 Class 6 Science are given below:

  • Methods of Separation
  • Handpicking
  • Threshing
  • Winnowing
  • Sieving
  • Sedimentation, Decantation and Filtration
  • Evaporation
  • Use of more than one method of separation
  • Can water dissolve any amount of a substance?

There are Many Methods for Substance Separation from the Mixture of the Materials Which are Listed Below:

  • Handpicking:- Husk and stones could be separated from grains by handpicking
  • Winnowing:- Husk is separated from heavier seeds of grain by winnowing.
  • Sieving:- The difference in the size of particles in a mixture can be separated by sieving.
  • Sedimentation:- This method is used to separate impurities when the heavier component in a mixture settles after water is added to it.
  • Decantation:- In a mixture of sand and water, the heavier sand particles settle down at the bottom and the water can be separated by decantation.
  • Filtration:- This method can be used to separate components of a mixture of an insoluble solid and a liquid.
  • Evaporation:- It can be used to separate a solid dissolved in a liquid.

NCERT Solutions for Class 6 Science: Chapter-wise

NCERT Solutions for Class 6: Subject-wise

NCERT Solutions for Class 6 Maths
NCERT Solutions for Class 6 Science

Benefits of NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances

  • Separation of substances NCERT solutions are available in very simple language that can be understood by the average student.
  • Class 6 Science Chapter 5 NCERT Solutions will provide you with in-depth knowledge of your subject.
  • Class 6 Separation of Substances NCERT Solutions will build your fundamental concepts of science, which are required to study science in the upper class.
  • You will develop a logical approach and methodology towards science and other subjects, as well as the separation of substances Class 6 solutions.
  • You can score good marks in the exam with the help of CBSE NCERT solutions for Class 6 Science Chapter 5 Separation of Substances.
  • Your homework will be easier as you will get all Class 6 science chapter 5 question answer, including practice questions given after every topic.
  • The Class 6 Science Chapter 5 NCERT Solutions PDF is easy to download and use offline
  • Separation of substances Class 6 questions and answers are prepared by subject experts as per the latest CBSE syllabus.

Also Check NCERT Books and NCERT Syllabus here

We hope you will ace your 6th grade exam with the help of NCERT Solutions for Class 6 Science Chapter 5 Separation of Substances.

Frequently Asked Questions (FAQs)

1. Does CBSE provides the solutions of NCERT Class 6 ?

No, CBSE doesn’t provide NCERT solutions for any class or subject

2. What are the topics are covered in NCERT solution Class 6 Science Chapter 5

The following are covered in NCERT solution Class 6 Science chapter 5 

  • Methods of Separation  
  • Handpicking  
  • Threshing  
  • Winnowing  
  • Sieving  
  • Sedimentation, Decantation and Filtration  
  • Evaporation  
  • Use of more than one method of separation  
  • Can water dissolve any amount of a substance? 
3. What are the benefits of referring to the NCERT Solutions for Class 6 Science Chapter 5?

NCERT Solutions for Class 6 Science Chapter 5 provide accurate and comprehensive solutions in simplified language, covering the entire syllabus. They aid in exam preparation and enhance problem-solving skills. Furthermore, these solutions are available in PDF format, allowing students to download and utilise them according to their convenience.

4. What are the important topic of Class 6 Science Chapter 5 Separation of Substances

Chapter 5 of NCERT Solutions for Class 6 Science covers important concepts such as methods of separation, including hand-picking, winnowing, threshing, sieving, sedimentation, decantation, and filtration. The chapter also explains the process of evaporation and the use of multiple methods of separation. Additionally, it addresses the question of whether water can dissolve any amount of a substance. To understand these concepts better, students can refer to the NCERT Solutions provided by the Careers360 faculty. These solutions are available in PDF format and can be downloaded by students at their convenience.

5. What is separation of substances Class 6?

Separation of Substances in Class 6 Science refers to dividing a mixture into individual components based on physical properties. It involves learning about various methods such as sedimentation, decantation, filtration, and more.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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