Science And Technology Solutions Solutions for Class 9 Science Chapter 4 Measurement Of Matter are provided here with simple step-by-step explanations. These solutions for Measurement Of Matter are extremely popular among class 9 students for Science Measurement Of Matter Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Science And Technology Solutions Book of class 9 Science Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Science And Technology Solutions Solutions. All Science And Technology Solutions Solutions for class 9 Science are prepared by experts and are 100% accurate.
Page No 57:
Question 1:
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+,Sn2+.
c. Composite Radical : SO42- ,NH4+ ,HCO3- , HSO4- , NO4-.
d. Metal With Variable Valency : Cu(1+,2+), Hg(1+,2+) ,Fe(2+,3+).
e. Bivalent Acidic Radical : S2- ,O2- ,Se2- .
f. Trivalent Basic Radical : Al3+ ,Cr3+,Fe3+,Au3+.
Page No 57:
Question 2:
Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen
Answer:
Element |
Symbols | Free radicals |
1. Mercury | Hg | Mercurous- Hg+ |
2. Potassium | K | Potassium- K+ |
3. Nitrogen | N | Azide- N3- Nitrate- NO3- Nitrite-NO2- |
4. Copper | Cu | Cuprous- Cu+ |
5. Sulphur | S | Sulphide- S2- Sulphate- SO42- Sulphite- SO32- |
6. Carbon | C | Carbide-C- ,triple carbene(:CH2) |
7. Chlorine | Cl | Chloride- Cl- |
8. Oxygen | O | Oxide- O2- |
Page No 57:
Question 3:
Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide
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
Page No 57:
Question 4:
a. Explain how the element sodium is monovalent.
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.
Page No 57:
Question 5:
Na2 SO4 , K2 CO3 , CO2 , MgCl2 , NaOH, AlPO4 , NaHCO3
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
Page No 57:
Question 6:
‘sample m’ mass : 7g
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".
Page No 57:
Question 7:
32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine.
Answer:
a. 32 g of oxygen
mol
1 mole of O2 contains-----6.022 x 1023 molecules
b. 90 g of water
moles
5 moles of H2O contains-----5 x 6.022 x 1023 molecules = 30.11 x 1023 molecules
c. 8.8 g of CO2
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
0.1 moles of Cl2 contains-----0.1 x 6.022 x 1023 molecules = 0.6022 x 1023 molecules
Page No 57:
Question 8:
Sodium chloride, magnesium oxide, calcium carbonate
Answer:
a. We know that,
Molar mass= sum of constituent atomic masses
Molar mass of NaCl= 23 + 35.5 = 58.5 g/mol
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
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
g
We need, 20 g of CaCO3 for obtaining 0.2 moles of CaCO3.
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