Science And Technology Solutions Solutions for Class 9 Science Chapter 4 - Measurement Of Matter

Answer:

Given examples of the following are as follows :

a. Positive Radicals : Na⁺, Fe²⁺,Ag⁺Al³⁺,Cr³⁺,Fe³⁺,Au³⁺,Co²⁺,Ni²⁺,Hg²⁺,Sn²⁺.

b. Basic Radical : Na⁺, Fe²⁺,Ag⁺,Al³⁺,Cr³⁺,Fe³⁺,Au³⁺,Co²⁺,Ni²⁺,Hg²⁺,Sn^2+.

c. Composite Radical : SO₄^2- ,NH₄⁺ ,HCO₃^-  , HSO₄^- , NO₄^-.

d. Metal With Variable Valency : Cu(1+,2+), Hg(1+,2+) ,Fe(2+,3+).

e. Bivalent Acidic Radical : S^2- ,O^2- ,Se^2- .  

f. Trivalent Basic Radical : Al³⁺ ,Cr³⁺,Fe³⁺,Au³⁺. 

Answer:

Answer:

a.Sodium sulphate :
Step 1 : Write the symbols of the radicals.
                           Na                         SO₄
Step 2 : Write the valency below the respective radical.
                           Na                         SO₄
                            1                             2
Step 3 : Cross-multiply symbols of radicals with their respective valency.

                         
Step 4 : Write down the chemical formula of the compound.
        
              Na₂SO₄


b. Potassium nitrate :
 Step 1 : Write the symbols of the radicals
                         K                              NO₃

Step 2 : Write the valency below the respective radical.
                          K                              NO₃
                                1                                1
Step 3 : Cross-multiply symbols of radicals with their respective valency.
                        
Step 4 : Write down the chemical formula of the compound.

              KNO₃
 

c. Ferric phosphate :
Step 1 : Write the symbols of the radicals
                         Fe                             PO4

Step 2 : Write the valency below the respective radical.
                          Fe                             PO4
                                3                               3
Step 3 : Cross-multiply symbols of radicals with their respective valency.
                        
Step 4 : Write down the chemical formula of the compound.

              FePO4


d. Calcium oxide :
Step 1 : Write the symbols of the radicals
                         Ca                             O

Step 2 : Write the valency below the respective radical.
                         Ca                             O
                          2                               2

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


e. Aluminium hydroxide :
Step 1 : Write the symbols of the radicals
                         Al                             OH

Step 2 : Write the valency below the respective radical.
                         Al                             OH
                          3                               1

Step 3 : Cross-multiply symbols of radicals with their respective valency.
                        
Step 4 : Write down the chemical formula of the compound. 
            
              Al(OH)₃

Answer:

a. 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.

b. M is a bivalent metal that is : M²⁺
Step 1 :              M                          SO₄
                    
Step 2 : 
                          M                          SO₄

                          2+                          2-

Step 3 :
                       

Step 4 :              MSO₄


M is a bivalent metal that is : M²⁺

Step 1 :
                         M                         PO4

Step 2 :             M                          PO4
                                2                            3

Step 3 :         
                        
                       
Step 4 :            M₃(PO4)₂

c. Mass of  an atom is concentrated in its nucleus. It is th sum of proton(p) and neutron(n). The number(p+n) is called atomic mass number.
As we know that atom is very small in size, then how do we determine its mass? Therefore, the concept of reference atom for atomic mass is taken into consideration.
Initially hydrogen atom being lightest was chosen as the reference atom.The relative mass of hydrogen atom was accepted as (1). this is how relative atomic masses of various elements were determined in reference of hydrogen. For example: the mass of one nitrogen atom is fourteen(14) times that of a hydrogen atom.

Finally, the carbon was selected as reference atom. The relative mass of carbon atom was accepted as 12. The relative atomic mass of hydrogen compared to the carbon atom becomes 12 x 1/12 =1.

d. 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.                                                                                                                                                                            

e. One mole is defined as the amount of substance containing as many elementary entities (atoms, molecules, ions, electrons, radicals, etc.) as there are atoms in 12 grams of carbon-12 (6.023 x 10²³⁾.
The mass of one mole of a substance equals to its relative molecular mass expressed in grams. Mole defines the quantity of a substance. One mole of any substance will always contain 6.022 × 10²³ particles, no matter what that substance is. Therefore, we can say:

• 1 mole of potassium atoms (K) contains 6.022 × 10²³ potassium atoms.
• 1 mole of potassium ions (K⁺) contains 6.022 × 10²³  potassium ions.
• 1 mole of hydrogen atoms (H) contains 6.022 × 10²³ hydrogen atoms.
• 1 mole of hydrogen molecules (H₂) contains 6.022 × 10²³ hydrogen molecules.

Answer:

a. Na₂SO₄ - Sodium sulphate
Molecular mass= sum of masses of individual components
2 (23) + 32 + 4 (16)= 142g

b. K₂CO₃ - Potassium carbonate
2 (39) + 12 + 3 (16) = 138g

c. CO₂ - Carbon dioxide
12 + 2 (16) = 44g

d. MgCl₂ - Magnesium chloride
24 + 2 (35.5) = 95g

e. NaOH - Sodium hydroxide
23 + 16 + 1 = 40g

f. AlPO₄ - Aluminium phosphate
27 + 31 + 4 (16) = 122g

g. NaHCO₃ - Sodium bicarbonate
23 + 1 + 12 + 3 (16) = 84g

Answer:

As it is mentioned already:
Sample m Mass :7g
Mass of constituent oxygen :2g
Mass of constituent calcium :5g
Sample n Mass :1.4g
Mass of constituent oxygen :0.4g
Mass of constituent calcium:1.0g

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".

for e.g: In sample m, the ratio of proportion of elements (calcium:oxygen) is 5:2
Ca:O =5:2
for e.g: In sample n, the ratio of proportion of elements (calcium:oxygen) is 1.0:0.4
Ca:O =1.0:0.4
         =10:4
         =5:2
On simplifying the ratio proportion by mass, we get the same values which verifies "The Law Of Constant Proportion".

Answer:

a. 32 g of oxygen
$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} {\mathrm{O}}_2=\frac{\mathrm{Mass} \mathrm{of} {\mathrm{O}}_2 \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} {\mathrm{O}}_2}$
                                                    $=\frac{32}{32}=1$mol                    
1 mole of O₂ contains-----6.022 x 10²³ molecules


b. 90 g of water
$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} water=\frac{\mathrm{Mass} \mathrm{of} water \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} water}$

                                                       $=\frac{90}{18}=5$moles
5 moles of H₂O contains-----5 x 6.022 x 10²³ molecules = 30.11 x  10²³ molecules

c. 8.8 g of CO_{2
    $$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} {\mathrm{CO}}_2=\frac{\mathrm{Mass} \mathrm{of} {\mathrm{CO}}_2 \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} {\mathrm{CO}}_2} \\ = \frac{8.8}{44}= 0.2mol \end{gathered}$$}
0.2 moles of CO₂ contains-----0.2 x 6.022 x 10²³ molecules = 1.2044 x  10²³ molecules


d. 7.1 g of chlorine

  $$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{in} \mathrm{chlorine}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{chlorine} \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{chlorine}} \\ =\frac{7.1}{71}= 0.1\mathrm{mol} \end{gathered}$$
   0.1 moles of Cl₂ contains-----0.1 x 6.022 x 10²³ molecules = 0.6022 x  10²³ molecules

Answer:

a. We know that,

$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{a} \mathrm{substance}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{substance} \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{substance}}$

Molar mass= sum of constituent atomic masses

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

$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{a} \mathrm{substance}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{substance} \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{substance}}$

$0.2=\frac{\mathrm{x}}{58.5}= 11.7$g

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


b. Molar mass of MgO= 24 + 16 = 40 g/mol

$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{a} \mathrm{substance}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{substance} \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{substance}}$

$0.2=\frac{\mathrm{x}}{40}= 8$g

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


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

$\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{a} \mathrm{substance}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{substance} \mathrm{in} \mathrm{grams}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{substance}}$

$0.2=\frac{\mathrm{x}}{100}= 20$g

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