Std 8 General Science · Chapter 6

6. Composition of Matter

Complete solutions to the Exercises. Maharashtra State Board (Balbharati), Grade Eight General Science.

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Exercises

1Choose the appropriate option and rewrite the following statements.

(a) The intermolecular force is maximum in the particles of solid.

(b) Solids retain their volume even when external pressure is applied. This property is called incompressibility.

(c) Matter is classified into the types mixture, compound and element, by applying the criterion chemical composition of matter.

(d) Matter that contains two or more constituent substances is called mixture.

(e) Milk is an example of type of matter called heterogeneous mixture.

(f) Water, mercury and bromine are similar to each other, because the three are liquids.

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

2Identify the odd term out and explain.

(a) Gold, silver, copper, brass Brass

Brass is a mixture (alloy). The others are elements.

(b) Hydrogen, hydrogen peroxide, carbon dioxide, water vapour Hydrogen

Hydrogen is an element. The others are compounds.

(c) Milk, lemon juice, carbon, steel Carbon

Carbon is an element. The others are mixtures.

(d) Water, mercury, bromine, petrol Petrol

Petrol is a mixture. The others are pure substances.

(e) Sugar, salt, baking soda, blue vitriol Sugar

Sugar is an organic compound. The others are inorganic compounds.

(f) Hydrogen, sodium, potassium, carbon Carbon

The valency of carbon is 4. The valency of the others is 1.

3Answer the following questions.

(a) 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.

  • Carbon dioxide (CO₂) – Inorganic compound.
  • Water (H₂O) – Inorganic compound.
  • Glucose (C₆H₁₂O₆) – Organic compound.
  • Chlorophyll – Complex compound.

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

  • Solvent: Copper (it is more in proportion).
  • Solute: Zinc (it is less in proportion).
  • Solution: Brass.

(c) Sea water tastes salty due to the dissolved salt. The salinity of some water bodies: Lonar lake 7.9 %, Pacific Ocean 3.5 %, Mediterranean sea 3.8 %, Dead sea 33.7 %. Explain two characteristics of mixtures from the above information.

  1. The proportion of constituents in a mixture is not fixed. The salinity of each water body is different.
  2. The constituents of a mixture keep their own properties. So sea water tastes salty.

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4Give two examples each.

(a) Liquid element: Mercury, bromine.

(b) Gaseous element: Oxygen, nitrogen.

(c) Solid element: Iron, copper.

(d) Homogeneous mixture: Salt solution, air.

(e) Colloid: Milk, smoke.

(f) Organic compound: Sugar, urea.

(g) Complex compound: Chlorophyll, haemoglobin.

(h) Inorganic compound: Common salt, baking soda.

(i) Metalloid: Silicon, germanium.

(j) Element with valency 1: Hydrogen, sodium.

(k) Element with valency 2: Oxygen, magnesium.

5Write the names and symbols of the constituent elements and identify their valencies from the molecular formulae given below.
  • KCl: Potassium (K) – 1, Chlorine (Cl) – 1
  • HBr: Hydrogen (H) – 1, Bromine (Br) – 1
  • MgBr₂: Magnesium (Mg) – 2, Bromine (Br) – 1
  • K₂O: Potassium (K) – 1, Oxygen (O) – 2
  • NaH: Sodium (Na) – 1, Hydrogen (H) – 1
  • CaCl₂: Calcium (Ca) – 2, Chlorine (Cl) – 1
  • CCl₄: Carbon (C) – 4, Chlorine (Cl) – 1
  • HI: Hydrogen (H) – 1, Iodine (I) – 1
  • H₂S: Hydrogen (H) – 1, Sulphur (S) – 2
  • Na₂S: Sodium (Na) – 1, Sulphur (S) – 2
  • FeS: Iron (Fe) – 2, Sulphur (S) – 2
  • BaCl₂: Barium (Ba) – 2, Chlorine (Cl) – 1
6Chemical composition of some matter is given in the following table. Identify the main type of matter from their composition.
Name of matterChemical compositionMain type of matter
Sea waterH₂O + NaCl + MgCl₂ + …Mixture
Distilled waterH₂OCompound
Hydrogen gas filled in a balloonH₂Element
The gas in LPG cylinderC₄H₁₀ + C₃H₈Mixture
Baking sodaNaHCO₃Compound
Pure goldAuElement
The gas in oxygen cylinderO₂Element
BronzeCu + SnMixture
DiamondCElement
Heated white powder of blue vitriolCuSO₄Compound
Lime stoneCaCO₃Compound
Dilute hydrochloric acidHCl + H₂OMixture
7Write scientific reason.

(a) Hydrogen is combustible, oxygen helps combustion, but water helps to extinguish fire.

Water is a compound of hydrogen and oxygen. The properties of a compound are different from its constituent elements. Hence, water does not burn and helps to extinguish fire.

(b) Constituent substances of a colloid cannot be separated by ordinary filtration.

The particles of a colloid are very small. They pass through the pores of filter paper. Hence, they cannot be separated by filtration.

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

Lemon sherbat is a mixture of sugar, lemon juice, salt and water. In a mixture, each constituent keeps its own property. So it tastes sweet, sour and salty. Water makes it a liquid. So it can be poured in a glass.

(d) A solid matter has the properties of definite shape and volume.

In solids, the intermolecular force is very strong. The particles are very close and cannot change their places. Hence, solids have definite shape and volume.

SolidLiquidGasparticles close, fixed placesparticles close but free to slideparticles far apart, move freely

Arrangement of particles in the solid, liquid and gaseous states.

8Deduce the molecular formulae of the compound obtained from the following pairs of elements by the cross multiplication method.

Write the symbols. Write the valency below each symbol. Then cross multiply the valencies.

CO42C2O4divide by 2 (smallest whole numbers)CO2SymbolsValency

Cross multiplication method — carbon (valency 4) and oxygen (valency 2) give CO₂.

(a) C (valency 4) & Cl (valency 1)

C1Cl4 → CCl₄

(b) N (valency 3) & H (valency 1)

N1H3 → NH₃

(c) C (valency 4) & O (valency 2)

C2O4 → divide by 2 → CO₂

(d) Ca (valency 2) & O (valency 2)

Ca2O2 → divide by 2 → CaO