Solution-Class 9-Science-Chapter-4-Measurement of Matter-Maharashtra Board

Positive ions :

(i) Sodium Ions Na⁺

(ii) Silver ions Ag⁺

(iii) Aluminium ions  Al³⁺ ,

(iv) Chromium ion Cr³⁺

(v) Calcium ion Ca²⁺

Basic Radical :

(i) Na⁺

(ii) Fe²⁺

(iii) Ag⁺

(iv) Al³⁺ 

(v) Cr³⁺

(vi) Sn^2+.

Composite Radical :

(i) SO₄ ^2- 

(ii) H₃O⁺

(iii) NH₄⁺ 

(iv) HCO₃^-  

(v) SO₄^2- 

(vi) NO₄^-.

Metal With Variable Valency :

(i) Cu → Cu⁺ Cu²⁺,

(ii) Hg → Hg⁺, Hg²⁺

(iii) Fe →  Fe²⁺ , Fe³⁺.

Bivalent Acidic Radical :

(i) S^2- 

(ii) O^2- 

(iii) Se^2- 

Trivalent Basic Radical :

(i) Al³⁺ 

(ii) Cr³⁺

(iii) Fe³⁺

(iv) Au³⁺. 

Sodium sulphate :

Step 1 : Write the symbols and valencies of the radicals.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound.     

The chemical formula of  Sodium sulphate is Na₂SO₄

Potassium nitrate :

Step 1 : Write the symbols and valencies of the radicals.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound.

The chemical formula of Potassium nitrate is KNO₃

Ferric phosphate :

Step 1 : Write the symbols and valencies of the radicals.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound.

The chemical formula of  Ferric phosphate is FePO₄

Calcium oxide :

Step 1 : Write the symbols and valencies of the radicals.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound.

 The chemical formula of Calcium oxide  CaO

Aluminium hydroxide :

Step 1 : Write the symbols and valencies of the radicals.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound. 

The chemical formula of  Aluminium hydroxide  Al(OH)₃

Sodium is a metal which has atomic number 11. It has 1 electron in its last shell which means it has 1 valence electron. In order to get stability by acquiring nearest noble gas configuration. Sodium prefers to loose 1 electron. Hence, we say that sodium has a valency of one and is thus monovalent.

M is a bivalent metal that is : M²⁺

Step 1 : Write the symbols and valencies of of M and sulphate.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound.

The chemical formula is  MSO₄

M is a bivalent metal that is : M²⁺

Step 1 : Write the symbols and valencies of of M and Phosphate.

Step 2 : Cross-multiply symbols of radicals with their respective valency.

Step 3 : Write down the chemical formula of the compound.

The chemical formula is  M₃(PO4)₂

• Since the atom is very very small it is very difficult to determine its mass very accurately, Hence the concept of the relative mass was
• As the hydrogen atom (H) is the lightest it was taken as the reference atom in old The nucleus of H atom contains only one proton, therefore its mass was taken as one unit.
• The mass of nitrogen atom (N) is 14 times than that of the H atom. Hence the relative mass of the N atom was taken as 14 units. In this manner the relative masses of various elements were determined.
• In 1961 it was decided to take the C-12 atom as the reference atom. In this the mass of C-12 atom is taken as 12 u. The mass of the hydrogen atom relative to that of the C-12 atom is equal to 1/12 x 12 u is equal to 1 u. (Note: 1 u =
• 1.66053904 x 10^-27 kg ).
• On the relative atomic masS of scale, the mass of the proton is very nearly 1 u. Similarly the mass of the neutron is very nearly 1 u. (Note : The mass of the neutron is slightly greater than that of the)

Instead of relative atomic mass, it is now possible to determine atomic masses in kg.  The Unified Atomic Mass Unit or Dalton is a standard unit of mass that quantifies mass on an atomic or molecular scale.

• One unified atomic mass unit is approximately equivalent to 1g/mol.
• It is denoted by a symbol :1u or Da = 1.66053904 x 10^-27 kg

A mole is that quantity of a substance whose mass in grams is equal in magnitude to the molecular mass of that substance in daltons. The SI unit is mole.

Examples

The molecular mass of oxygen is 32 u; 32 g oxygen is 1 mole of oxygen.

The atomic mass of carbon is 12 u; 12 g carbon is 1 mole of carbon.

The molecular mass of H2O is 18 u; 18 g H2O is 1 mole of water.

Number of moles of Mass of substance (n) =  \(\frac{\text{Mass of substance in grams}}{Molecular mass of substance}\)

(i) Na₂SO₄ - Sodium sulphate.

Molecular mass= sum of masses of individual components

Atomic mass of, Na = 23u, S=32u, O=16u

∴ Molecular mass of Na₂SO₄ = 2x(23) + 32 + 4x(16)= 142u

(ii) K₂CO₃ - Potassium carbonate.

Atomic mass of, K=39u, C=12u, O=16u

∴ Molecular mass of K₂CO₃ = 2x(39) + 12 + 3x(16) = 138u

(iii) CO₂ - Carbon dioxide.

Atomic mass of, C=12u, O=16u

∴ Molecular mass of CO₂ = 12 + 2x(16) = 44u

(iv) MgCl₂ - Magnesium chloride.

Atomic mass of, Mg=24u, Cl=35.5u

∴ Molecular mass of MgCl₂ = 24 + 2x(35.5) = 95u

(v) NaOH - Sodium hydroxide.

Atomic mass of, Na=23u, O=16u, H=1u

∴ Molecular mass of NaOH = 23 + 16 + 1 = 40u

(vi) AlPO₄ - Aluminium phosphate

Atomic mass of, Al= 27u, P=31u, O=16u

∴ Molecular mass of AlPO₄ = 27 + 31 + 4x(16) = 122u

(vii) NaHCO₃ - Sodium bicarbonate

Atomic mass of, Na=23u, H= 1u, C=12u, O=16u

∴ Molecular mass of NaHCO₃ = 23 + 1 + 12 + 3x(16) = 84u

Slaked lime have two samples ‘m’ and ‘n’

Sample m Mass :7g
Mass of constituent oxygen :2g
Mass of constituent calcium :5g

In sample m, the ratio of proportion of elements (calcium:oxygen) is 5:2
Ca:O =5:2 by wight

Sample n Mass :1.4g
Mass of constituent oxygen :0.4g
Mass of constituent calcium:1.0g

In sample n, the ratio of proportion of elements (calcium:oxygen) is 1.0:0.4
Ca:O =1.0:0.4 by weight  i.e.
         =10:4
         =5:2

Out of all the laws of chemical combination, this is proved by "Law Of Constant Proportion".

Law Of Constant Proportion states that "The proportion by weight of the constituent elements in various samples of compound is fixed in ratio".

On simplifying the ratio proportion by mass, we get the same values which verifies "The Law Of Constant Proportion".

(i) 32 g of oxygen :
Molecular mass of O₂ = 2x16=32

Number of moles of O₂ = Mass of O₂ in grams/Molecular mass of O₂ = 32/32 = 1mol                    
∴ 1 mole of O₂ contains-----6.022 x 10²³ molecules
∴ 32g of oxygen contains 6.022 x 10²³ molecules

(ii) 90 g of water :
Molecular mass of water H₂O= 2x(1)+16 =18

Number of moles in water = Mass of water in grams/Molecular mass of water = 90/18 = 5mole

∴ 5 moles of H₂O contains-----5 x 6.022 x 10²³ molecules = 30.11 x  10²³ molecules

 ∴ 90g of water contains 30.11 x  10²³ molecules

(iii) 8.8 g of CO₂ :

Molecular mass of Carbon dioxide CO₂ =12+2x(16) =44
Number of moles in CO₂ = Mass of CO₂ in grams / Molecular mass of CO₂ = 8.8/44 = 0.2mol

  ∴0.2 moles of CO₂ contains-----0.2 x 6.022 x 10²³ molecules = 1.2044 x 10²³ molecules 

∴8.8 g of CO₂ contains 1.2044 x  10²³ molecules

(iv) 7.1 g of chlorine
Molecular mass of Chlorine Cl₂ = 2x(35.5)=71
  No. of moles in chlorine=Mass of chlorine in grams/Molecular mass of chlorine  =7.1/71 = 0.1mol

∴0.1 moles of Cl₂ contains-----0.1 x 6.022 x 10²³ molecules = 0.6022 x 10²³ molecules

∴ 7.1 g of chlorine contains 0.6022 x 10²³ molecules

(i) We know that,

Molar mass= sum of constituent atomic masses

Molar mass of NaCl= 23 + 35.5 = 58.5 g/mol

Number of moles of a substance= \(\frac{\text{Molar mass of substance in grams}}{\text{Molecular mass of the substance}}\)

0.2=\(\frac{x}{58.5}\)

x = 0.2×58.5 = 11.7g

We need, 11.7 g of NaCl for obtaining 0.2 moles of NaCl.

(ii) Molar mass of MgO= 24 + 16 = 40 g/mol

Number of moles of a substance= \(\frac{\text{Molar mass of substance in grams}}{\text{Molecular mass of the substance}}\)

0.2=\(\frac{x}{40}\)

x = 0.2×40 = 8g

We need, 8 g of MgO for obtaining 0.2 moles of MgO.

(iii)  Molar mass of CaCO₃= 40 + 12  + 3 (16) = 100 g/mol

Number of moles of a substance= \(\frac{\text{Molar mass of substance in grams}}{\text{Molecular mass of the substance}}\)

0.2=\(\frac{x}{100}\)

x =0.2×100= 20 g

We need, 20 g of CaCO₃ for obtaining 0.2 moles of CaCO₃.