GSEB Solutions for Class 9 Science and Technology – Properties of Matter

GSEB Solutions for Class 9 Science and Technology – Properties of Matter (English Medium)

GSEB SolutionsMathsScience
Exercise 57:

Solution 1.1:

D. Solid
In solid substances, the inter-molecular forces are very high giving it a specific shape and structure.

Solution 1.2:

B.1:8
Weight of water is 18 amu = 2 (1 amu of Hydrogen) + 1 (16 amu of Oxygen)
Thus, ratio of Hydrogen : Oxygen is 2:16 = 1:8

Solution 1.3:

D. Lavoisier
Lavoisier defined element as the basic form or the basic unit of any substance.

Solution 1.4:

D. Volume increases
When the temperature of a state of matter is increased, heat is absorbed by the substance in the form of energy. The kinetic energy of the molecules making up the substance increases. The molecules experience a lot of vibration and occupy more space. As a result, a substance expands and its volume increases.

Solution 1.5:

Inter-molecular forces are maximum
In gases, inter-molecular forces are minimum as compared to their liquid and solid forms.

Exercise 58:

Solution 2.1:

The force which exists between the particles in solid, liquid or gaseous state of matter is known as inter-molecular force.

Solution 2.2:

A physical change is a change in the physical form or physical property of matter, with the chemical identity remaining the same.
Examples – Dissolution of salt in water, Chopping of wood, Melting of ice

Solution 2.3:

A chemical change is a change of matter where there is reaction between two or more substances and a new substance is formed.
Examples – Burning of wood, Rusting of iron, Reaction of an acid with a base to form salt and water

Solution 2.4:

When the temperature of a state of matter is increased, heat is absorbed by the substance in the form of energy. The kinetic energy of the molecules making up the substance increases. The molecules experience a lot of vibration and occupy more space. As a result, a solid substance expands and becomes liquid and a liquid turns to gas.
Thus, on increasing the temperature of a substance, the following change in the states of matter is noted:
Solid → Liquid → Gas
Similarly, if the temperature of a substance is decreased:
Gas → Liquid → Solid.

Solution 2.5:

The state of matter of a substance is attributed to the distance between the composite particles which make the substance. Thus, when pressure is applied on gas, it compresses to liquid and when pressure is increased on liquids, it becomes solid. The effect of pressure on solids is negligible.
Thus, if pressure is applied on a substance:
Gases → Liquid → Solid.

Solution 2.6:

A mixture in which the components are uniform in the whole solution, its formation in the whole solution is same and appears like a single substance is called a homogeneous mixture. Example – Air is a homogeneous mixture of gases.

Solution 2.7:

A mixture in which each component of the mixture is different and identifiable and the properties of each component are different is called a heterogeneous mixture.

Example – A mixture of slat and iron powder is a heterogeneous mixture.

Solution 2.8:

A homogeneous solution which is stable and formed as a result of complete dissolution of the solute in the solvent is called a true solution. Example – Air, sugar solution etc.

Solution 2.9:

A heterogeneous solution in which the solute disperses in the solvent is called a suspension solution. The particles of the solute in the suspension solution remain floating in the solvent.

Example – solution of limestone in water, solution of barium sulphate in water.

Solution 2.10:

A heterogeneous solution, in which the solute disperses evenly throughout the solvent and remains in the solvent giving it the appearance of a homogeneous solution, is a colloidal solution.

Example – Foam, smog, cloud etc.

Solution 2.11:

The scattering of light by particles of colloid in a suspension is called Tyndall effect.

Solution 2.12:

Dispersed phase are the colloidal particles, generally of the solute which remain distributed evenly in the dispersion medium.

Exercise 59:

Solution 2.13:

A dispersion medium is a continuous medium, generally a solvent, in which the dispersed phase is distributed.

Solution 2.14:

A solute when dissolved in a solvent dissolves completely. If more solute is added, it does not dissolve but remains as a mere substance. Such a solution which contains the solute to its highest capacity at normal temperature is called a saturated solution.

Solution 2.15:

The molar mass of a substance is the sum total of the atomic masses of the atoms present in that molecule or compound.

Solution 2.16:

The relative proportion of atoms of each element present in the compound is expressed by a formula called the empirical formula.

Solution 2.17:

John Dalton proposed the atomic theory and gave the laws of chemical combination.

Solution 2.18:

An element is the basic unit of a substance; formed of only one type of atom.
Example of liquid metal – Mercury
Example of liquid non-metal – Bromine.

Solution 2.19:

A mixture is a substance prepared by mixing two or more substances such that the mixed substances maintain their original properties.
Mixtures are of two types:

1. Homogeneous mixture: Example – Air
2. Heterogeneous mixture: Example – mixture of salt and sulphur.

Solution 2.20:

1. Starch mixed in sugar – homogeneous mixture
2. Water mixed in milk – homogeneous mixture
3. Iron powder mixed in tea – heterogeneous mixture
4. Kerosene mixed in petrol – homogeneous mixture

Solution 3.1:

A physical change is the change in the physical form or physical property of matter, with the chemical identity remaining the same.
A chemical change is a change of matter where there is a reaction between two or more substances and a new substance is formed.
For example:
Chopping of wood is a physical change, whereas burning of wood is a chemical change.
Chopping of wood results only in decreasing the particle size of wood but its properties remain the same. On the other hand, burning of wood results in the formation of ash, carbon dioxide etc. i.e. a new substance is formed.

Solution 3.2:

A mixture in which the components are uniform in the whole solution, its formation in the whole solution is the same and it appears like a single substance, is called a homogeneous mixture.
Example – Air is a homogenous mixture of gases.
Salt solution is a homogeneous mixture of common salt and water.
A mixture in which each component of the mixture is identifiable and different and the properties of each component are different is called a heterogeneous mixture.
Example – Mixture of slat and iron powder is a heterogeneous mixture.
Mixture of salt and sulphur powder is a heterogeneous mixture.

Solution 3.3:

Concentration of solution is the measure of the amount of solute present in a known amount of solvent. Concentration of solution is expressed in two ways:

• Weight based method: In weight based method, weight of known amount of solute dissolved in known weight of solvent is expressed. It is expressed in percentage and is written as W/W%.

In second method, weight of known amount of solute dissolved in known volume of solvent is expressed. It is also expressed in percentage and is written as W/V%

• Volume based method: In volume based method, volume of known amount of solute dissolved in known volume of solvent is expressed. It is expressed in percentage and is written as V/V%

Solution 3.4:

The mass of an individual atom of an element is known as its atomic mass.
The molecular mass of a substance is the sum total of the atomic masses of the atoms present in that molecule or compound.

Solution 3.5:

A mole of a substance is defined as the mass of substance containing the same number of fundamental units as there are in 12 grams of carbon (i.e. 6.022 x 10²³).
Mole is itself an SI unit of measurement of chemical compounds.
The molar mass of a substance is the sum total of the atomic masses of the atoms present in that molecule or compound.

Solution 3.6:

The scattering of light by particles of colloid in a suspension is called Tyndall effect.
Suspension solutions and colloidal solutions show Tyndall effect.
Suspension and colloidal solutions are both heterogeneous solutions where the solute does not get completely dissolved in the solvent. The particles of the solute remain freely suspended in the solution and thus scatter light showing Tyndall effect.

Solution 3.7:

Assumptions of Dalton’s atomic theory are:

1. Element is made up of extremely small particles which cannot be seen with naked eyes. This particle called atom is the smallest indivisible particle of an element.
2. All atoms of a given element are the same. Their weights and properties are all alike. The atoms of one element differ from the atoms of other elements and each element has its independent identity.
3. An atom cannot be created or destroyed. Under normal conditions, an atom of one element cannot be changed into an atom of another element (except during nuclear reaction).
4. During a chemical combination between two or more elements, atoms combine in a definite proportion. This proportion can be expressed by simple whole numbers. The resulting particles are called molecules and are bigger than combining atoms.
5. Atom is extremely small and indivisible. It maintains its existence during any physical or chemical change. Example – when Na and Cl combine, NaCl will be formed and not KCl or any molecule with another element.

Solution 3.8:

Molecular mass of K₂Cr₂O₇ = 2 (Atomic mass of K) + 2 (Atomic mass of Cr) + 7 (Atomic mass of O)
= 2 (39) + 2 (52) +7 (16)
= 78 + 104 + 112
=294 amu
Molecular mass of Cr₂(SO₄)₃ = 2 (Atomic mass of Cr) + 3 [(Atomic mass of S) + 4 (Atomic mass of O)]
= 2 (52) + 3 [(32) + 4 (16)]
=2 (52) + 3 [32+ 64]
= 104 + 3 [96]
= 104 + 288
= 392 amu

Solution 3.9:

According to Avogadro’s number, the number of molecules of nitrogen gas = mole x Avogadro number
= 1.5 x 6.022 x 10²³ molecules
= 9.033 x 10²³ molecules of nitrogen gas.
However, nitrogen gas has two atoms of nitrogen per molecule.
∴Number of atoms of nitrogen = 9.033 x 10²³ x 2
= 18.066 x 10²³ atoms of nitrogen are present in 42 grams of nitrogen gas.

Solution 3.10:

One molecule of PCl₅ contains one atom of phosphorous and 5 atoms of chlorine.

Solution 3.11:

Atomic mass of gold (Au) = 197 amu = 1 mole of gold (by def)
∴1 mole of gold will be collected in 197 seconds or in 3.28 minutes.

Solution 3.12:

Molecular mass of NH₄Cl = (Atomic mass of N) + 4 (Atomic mass of H) + (Atomic mass of Cl)
= (14) + 4(1) + (35.5)
=53.5 amu

Solution 3.13:

• In a hydrocarbon, C and H are present.
• Percentage of elements is 85.7 and 14.3 respectively.

• Ratio of simple whole number is 1:2
• Empirical formula is CH₂
• Mass calculated on the basis of empirical formula = 12 + 2(1) = 14
• The multiplication number (n) can be determined from the ratio of molecular mass and empirical formula mass.

As the empirical formula mass and molecular mass are in the ratio 1:4, the empirical formula has to be multiplied by 3.
Thus, (CH₂)_n = (CH₂)₄ = C₄H₈
Thus, the molecular formula of the hydrocarbon is C₄H₈.

Solution 3.14:

Molecular mass of NaHCO₃ = (Atomic mass of Na) + (Atomic mass of H) + (Atomic mass of C) + 3 (Atomic mass of O)
= 23 + 1 + 12 + 3(16)
= 84

According to Avogadro’s number, number of molecules of NaHCO₃ = mole x Avogadro number
= 0.5 x 6.022 x 10²³
= 3.011 x 10²³ molecules
42 grams sodium hydrogen carbonate (NaHCO₃) contains 3.011 x 10²³ molecules.

Solution 3.15:

According to Avogadro’s number, the number of atoms of a compound = no. of moles x Avogadro’s number

Solution 3.16:

Molecular mass of water= 2 (Atomic mass of H) + (Atomic mass of O)
= 2(1) + (16)
= 18
According to Avogadro’s number, number of molecules of water = mole x Avogadro’s number
= 5 x 6.022 x 10²³
= 30.11 x 10²³ molecules
= 3.011 x 10²⁴ molecules
90 ml of water contains 3.011 x 10²⁴ molecules of water.