Solutions-Class 8-Science-Chapter-6-Composition of Matter-Maharashtra Board

Composition of Matter

Maharashtra Board Class 8- General Science - Chapter-6

Solution

Question 1:

Choose the appropriate option and rewrite the following statements.

A. The intermolecular force is _______ in the paricles of solid.
(i) Minimum    (ii) Moderate    (iii) maximum    (iv) indefinite.

B. Solids retain their voume even when external pressure is applied. This property is called__________
(i)  plasticity    (ii) Incompressibility    (iii) fluidity    (iv) elasticity

C. Matter is classified into the types mixture, compound and element by applying the criterion______________
(i)  states of matter    ii Phases of matters    iii chemical compositions of matter    iv all of these

D. Matter that contain two or more constituent substances is called__________
(i)  mixture    (ii) compound    (iii) element    (iv) metalloid

E. Milk is an example of type of matter called __________
(i)  solution    (ii) homogeneous mixture    iii heterogeneous mixture    (iv) suspension

F. Water, mercury and bromine are similar o each other, because three are
(i)  liquids    (ii)  compounds     (iii) nonmetals    (iv) elements.

G. valency of carbon is 4 and that of oxygen is 2. From this, we understand that there are _______ chemical bond/bonds between the carbon atom and one oxygen atom in the compound-carbon dioxide.
(i)  1    (ii) 2    (iii) 3     (iv) 4 

Answer

A. The intermolecular force is  maximumin the paricles of solid.

B. Solids retain their voume even when external pressure is applied. This property is called elasticity.

C. Matter is classified into the types mixture, compound and element by applying the criterion Chemical composition of matter.

D. Matter that contain two or more constituent substances is calledmixture.

E. Milk is an example of type of matter called heterogeneous mixture.

F. Water, mercury and bromine are similar to each other, because three are liquids.

G. Valency of carbon is 4 and that of oxygen is 2. From this, we understand that there are chemical bond/bonds between the carbon atom and one oxygen atom in the compound carbon dioxide.

Question 2:

Identify the odd term out and explain

1. Gold, silver, copper, brass

2. Hydrogen, hydrogen peroxide, carbon dioxide, water vapour.

3. Milk, lemon juice, carbon, steel.

4. water, mercury, bromine, petrol.

5. sugar, slat, baking soda, blue vitrol.

6. Hydrogen, sodium, potassium, carbon.

Answer

  1. Brass is odd one out because it is an alloy and gold, silver, copper are element.
  2. Hydrogen is odd one out because it is an element and others are compound.
  3. Carbon is odd one out because it is an element and others are mixture of various elements.
  4. Petrol is odd one out because others are inorganic compounds.
  5. Sugar is odd one out because others are inorganic compounds
  6. Carbon is odd one out because it has 4 valence electrons and others have 1 valence electron.

Question 3:

Answer the following questions.

1. Plants synthesize glucose in sunlight with the help of chlorophyll from carbon dioxide and water and give away oxygen. identify the four compounds in this process and name their types.

Answer

Photosynthesis is a chemical process through which plants, some bacteria and algae, produce glucose and oxygen from carbon dioxide and water, using only light as a source of energy, which is absorbed by chlorophyll.

6CO2 + 6H2O → C6H12O6 + 6O2

Four compounds in this process are as follows:

1.Carbon dioxide = Inorganic compound

2.Water = Inorganic compound

3.Glucose = organic compound

4.Chlorophyll = complex compound

2. In one sample of brass, the following ingredients were found : copper (70%) and zinc (30%).  Identify the solvent, solute and solution from these.

Answer

Brass is an alloy it contains 70% copper, and 30% zinc.

Copper is solvent with largest proportion

Zinc is solute in smaller proportion

The solution is Brass

3. Sea water tastes salty due to the dissolved salt. the salinity (the proportion of salts in water) of some water bodies Lonar lake - 7.9 %, Pacific Ocean 3.5%, Mediterranean sea- 3.8%, Dead sea- 33.7%. Explain two characteristics of mixture from the above information.

Answer

Characteristics of mixtures from above information are:

1.Constituent substances of a mixture (proportion of salt in water) do not combine chemically.

2.The proportion of constituent substances present by weight in a mixture can be variable.

3.The constituent of mixture can be separated by physical process.

Question 4:

Give two examples each

A. Liquid element

B. Gaseous element

C. Solid element

D. Homogeneous mixture

E. Colloid

F. Organic compound

G. Complex compound

H. Inorganic compound

I. Metalloid

J. Element with valency 1

K. Element with valency 2

Answer

A. Liquid element = mercury (Hg), bromine(Br2)

B. Gaseous element = oxygen(O2), nitrogen(N), hydrogen(H)

C. Solid element = sodium(Na), carbon (c), aluminium(Al)

D.Homogeneous mixture = sugar in water, corn oil, Sea water.

E. Colloid = milk, blood, jelly.

F. Organic compound = glucose, urea, carbohydrates.

G. Complex compound = chlorophyll, hemoglobin, cyanocobalamine.

H. Inorganic compound = limestone, rust, common salt, Soda.

I. Metalloid = silicon, germanium, arsenic.

J .Element with valency 1 = sodium(Na), potassium(K), chlorine(Cl).

K. Element with valency 2 = magnesium(Mg), calcium(Ca)

Question 5:

Write the names and symbols of the constituent eleements and identify their valencies from the molecular formulae given below.
KCl,   HBr,   MgBr2,   K2O,  NaH,   CaCl2,  CCl4,  HI,   H2S, Na2S,   FeS,   BaCl2

Answer

Compounds Name of compounds Symbol of constituent elements Valency of constituent elements
KCl Potassium chloride K, Cl K = 1, Cl = 1
HBr Hydrogen bromide K, Br K = 1, Br = 1
MgBr2 Magnesium bromide Mg, Br Mg = 2, Br = 1
K2O Potassium oxide K, O K = 1, O = 2
NaH Sodium hydride Na, H Na = 1, H = 1
CaCl2 Calcium chloride Ca, Cl Ca = 2, Cl = 1
CCl4 Carbon tetrachloride C, Cl C = 4, Cl = 1
HI Hydrogen iodide H, I H = 1, I = 1
H2S Hydrogen sulphide H, S H = 1, S = 2
Na2S Sodium sulphide NaS Na = 1S = 2
FeS Iron (II) Sulfide Fe, S F = 2, S = 2
BaCl2 Barium chloride Ba, Cl B = 2, Cl = 1

Question 6:

Chemical composition of some matter is given in the following table. Identify the main type of matter from their.

  Name of matter  Chemical composition  Main type of matter 
Sea water   H2O + NaCl + MgCl2  
Distilled water   H2O  
 Hydrogen gas filled  in a ballon H2  
The gas in LPG cylinder  C4H10 + C3H8  
Baking soda   NaHCO3  
Pure gold     Au  
The gas in oxygen cylinder    O2  
Bronze   Cu + Sn  
Diamond   C  
Heated white powder of blue vitroi   CuSO4  
Lime stone   CaCO3  
Dilute hydrochloric acid  HCL+ H2O  

Answer

Name of matter   Chemical composition  Main type of matter
Sea water H2O + NaCl + MgCl2 mixture
Distilled water H2O compound
 Hydrogen gas filled  in a ballon H2 element
The gas in LPG cylinder C4H10 + C3H8 mixture
Baking soda NaHCO3 compound
Pure gold Au element
The gas in oxygen cylinder O2 element
Bronze Cu + Sn mixture
Diamond C element
Heated white powder of blue vitroi CuSO4 compound
Lime stone CaCO3 compound
Dilute hydrochloric acid HCL+ H2O mixture

Question 7:

Write scientific reason.

1. Hydrogen is combustible, oxygen helps combustion, but water helps to extinguish fire.

Answer

  • Water is compound of hydrogen and oxygen.
  • Hydrogen is combustible and oxygen helps combustion.
  • Though water is made up of two atoms of hydrogen and one atom of oxygen by forming an ionic compound, it does not possess the characteristics of them because a compound does not have the properties of its constituent elements.
  • Therefore, water has its own properties, which helps to extinguish fire.

2. Constituent substances of a colloid cannot be separated by ordinary filtration.

Answer

Colloidal solution is heterogeneous. Constituent substances of a colloid cannot be separated by ordinary filtration because the size of the particles in a colloids (or colloidal solution) is bigger than those in a true solution but smaller than those in suspension. It is in between 1μ to 100μ in diameter. The pore size of ordinary filter paper is more than 100μ due to which colloidal particles are passed through the pores of a filter paper. Therefore we prefer to use filter paper so that, filtration of colloidal particles take place easily.

3. Lemon sherbat has sweet, sour and salty taste and it can be poured in a glass.

Answer

  • Lemon sherbat is a mixture. It is made up of lemon juice, sugar, salt and
  • Formation of lemon sherbat does not involve any chemical reaction.
  • The constituents of sherbat retain their individual properties.
  • Hence, lemon sherbet is sweet, sour and salty to taste and it can be poured in a glass.

4. A solid matter has the properties of definite shape and volume.

Answer

A solid matter has the properties of definite shape and volume because of the following reasons:

  • The forces among the constituent particles (atom/molecules) are called intermolecular forces.
  • In solids these forces are strong enough to keep the particles together in fixed positions, as a result solids have a definite shape and volume element.

Question 8:

Deduce the molecular formulae of the compound obtained from the following pairs of elements by the cross multiplication method.

1. C (Valency 4) & Cl (Valency 1)

Answer

Step 1 : Write the symbols of the elements.                       

C     Cl

                     

 

Step 2 : Write the valency below the respective elements.       

C     Cl
4 1

                     

 

 

Step 3 : Cross-multiply symbols of elements with their respective valency.
 
Step 4 : Write down the chemical formula of the compound.

The molecular formula is CCl4

2. N (Valency 3) & H (Valency 1)

Answer

Step 1 : Write the symbols of the elements.                       

H

                     

 

Step 2 : Write the valency below the respective elements.       

N   H
3 1

                     

 

 

Step 3 : Cross-multiply symbols of elements with their respective valency.

Step 4 : Write down the chemical formula of the compound.

The molecular formula is NH3

3. C (Valency 4) & O (Valency 2)

Answer

Step 1 : Write the symbols of the elements.        

O

                     

 

Step 2 : Write the valency below the respective elements.       

C O
4 2

                     

 

 

Step 3 : Cross-multiply symbols of elements with their respective valency.

The molecular formula is CO2

4. Ca (Valency 2) & O  (Valency 2)

Answer

Step 1 : Write the symbols of the elements.                       

Ca O

                     

 

Step 2 : Write the valency below the respective elements.       

Ca O
2 2

                     

 

 

Step 3 : Cross-multiply symbols of elements with their respective valency.

Step 4 : Write down the chemical formula of the compound.
The molecular formula is CaO

 

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