Maharashtra Board Class 9 Science Solution Chapter 4 – Measurement of Matter
Balbharati Maharashtra Board Class 9 Science Solution Chapter 4: Measurement of Matter. Marathi or English Medium Students of Class 9 get here Measurement of Matter full Exercise Solution.
Std |
Maharashtra Class 9 |
Subject |
Science Solution |
Chapter |
Measurement of Matter |
1.) Give examples:
Ans:
a) Positive radicals:
The examples of positive radicals are H+, Na+, K+, Li+ etc.
b) Basic radicals:
The examples of basic radicals are Na+, Ba2+, Fe3+ etc.
c) Composite radicals:
The examples of composite radicals are SO42-, NH4+, NO4- etc.
d) Metals with variable valency:
The examples of metals with variable valency are copper, tin, iron and mercury etc.
e) Bivalent acidic radicals:
The examples of Bivalent acidic radicals are Se2-, O2-, S2- etc.
f) Trivalent basic radicals:
The examples of Trivalent basic radicals are Al3+, Au3+, Bi3+, Cr3+, Fe3+ etc.
2.) Write the symbols of the following elements and the radicals obtained from them and indicates the charge on the radicals.
Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen.
Ans:
- Mercury:
Symbol: Hg
Radicals: Hg+
- Potassium:
Symbol: K
Radicals: K+
- Nitrogen:
Symbol: N
Radicals: NO2-, NO3-, N3-
- Copper:
Symbol: Cu
Radical: Cu+
- Sulphur:
Symbol: S
Radicals: SO32-, SO42-, S2-
- Carbon:
Symbol: C
Radicals: C-
- Chlorine:
Symbol: Cl
Radicals: Cl-
- Oxygen:
Symbol: O
Radical: O2-
3.) Write the steps in deducing the chemical formulae of the following compound.
Sodium sulphate, Potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide.
Ans:
Sodium sulphate:
- Step 1: In sodium sulphate the radicals involved with their symbol are Na and SO4.
- Step 2: The valency of Na is 1 and the valency of SO4 is 2.
- Step 3: As the valency of Na is 1 it requires only one SO4 radical.
While the valency of SO4 is 2 hence it requires two Na radicals.
Thus, compound formed is of the form Na2SO4.
Potassium nitrate:
- Step 1: In potassium nitrate the radicals involved with their symbol are K and NO3.
- Step 2: The valency of K is 1 and the valency of NO3 is 1.
- Step 3: As the valency of K is 1 it requires only one NO3 radical.
While the valency of NO3 is 1 hence it requires only one K radicals.
Thus, compound formed is of the formed is KNO3.
Ferric phosphate:
- Step 1: In ferric phosphate the radicals involved with their symbol are Fe and PO4.
- Step 2: The valency of Fe is 3 and the valency of PO4 is 3.
- Step 3: As the valency of Fe is 3 and the valency of PO4 is 3 hence the compound formed is of the form FePO4.
Calcium oxide:
- Step 1: In calcium oxide the radicals involved with their symbol are Ca and O.
- Step 2: The valency of Ca is 2 and the valency of O is 2.
- Step 3: As the valency of Ca is 2 and the valency of O is 2 hence the compound formed is of the form CaO
Aluminium hydroxide:
- Step 1: In aluminium hydroxide the radicals involved with their symbol are Al and OH.
- Step 2: The valency of Al is 3 and the valency of OH is 1.
- Step 3: As the valency of Al is 3 it requires three OH radical.
While the valency of OH is 1 hence it requires only one Al radicals.
Thus, compound formed is of the form Al(OH)3.
4.) Write answers to the following questions and explain your answers.
a) Explain the monovalency the element sodium.
Ans:
We know that, the atomic number of sodium Na is 11.
Hence, its electronic configuration is 2,8,1.
That means it has only one electron in its Valence shell i.e. last shell. Hence, we can say that the valency of Na is 1.
In order to get stability, Na atom easily donate 1 electron and complete its octave.
This explains the monovalency of sodium atom.
b) M is Bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with radicals: sulphate and phosphate.
Ans:
Given that,
M is Bivalent metal. That means the valency of metal M is 2.
Compound formed by Bivalent metal M with radical sulphate:
- Step 1: In this the radicals involved with their symbols are M and SO4.
- Step 2: The valency of metal M is 2 and the valency of SO4 is 2.
- Step 3: Thus in order to make stable compound with radical SO4, metal M forms the compound in the form of MSO4.
b) Compound formed by Bivalent metal M with radical PO4:
- Step 1: In this the radicals involved with their symbol are metal M and PO4.
- Step 2: The valency of Bivalent metal M is 2 and the valency of radical PO4 is 3.
- Step 3: As the valency of metal M is 2 hence it requires two PO4 radicals.
While as the valency of radical PO4 is 3 hence it requires three atoms of metal M.
Hence the stable compound formed here is M3(PO4)2.
c) Explain need for the reference for atomic mass. Give some information about two reference atoms.
Ans:
We know that,
- The central part of the atom at which whole mass of atom is supposed to be concentrated is the nucleus.
- Nucleus contains positively charged protons and neutral neutrons.
- Hence, the total charge on nucleus is positive due to which negatively charged electrons are revolving around the nucleus in particular orbits.
- The mass of any atom is the sum of number of protons and neutrons it contains.
- But, it is very difficult to determine the mass of atoms so to make it easy the reference atom is considered. Initially, hydrogen atom is considered as reference for determining mass of other atoms.
- Here, the relative mass of hydrogen atom is taken as 1. Hence, we can say that the mass of oxygen atom is 16 which is 16 times the mass of reference atom hydrogen.
- After that, carbon is considered as reference atom whose relative mass is taken as 12.
- Hence, here we can say that the mass of hydrogen atom is 1/12 times the mass of carbon atom.
- In this way, for easy measurements of mass of atoms reference atoms and their relative mass is considered.
d) What is mean by Unified Atomic mass?
Ans:
- The atomic masses are measured in unified atomic mass unit.
- It is denoted by simply u.
- And it is also defined as the, it is the 1/12th mass of carbon-12.
- It is also named as Dalton.
e) Explain with example what is mean by mole of a substance?
Ans:
- One mole of substance is defined as the amount of substance which contains how many total number of atoms or molecules or ions in it.
- One mole of each substance contains 6.023*1023 molecules in it.
- And the number of molecules in one mole always remains constant.
- And this number is called as Avogadro’s number
5.) Write the name of the following compounds and deduce their masses.
Na2SO4, K2CO3, CO2, MgCl2, NaOH, AlPO4, NaHCO3.
Ans:
1) Na2SO4: Sodium Sulphate
Molecular mass= 2(mass of Na) + mass of S + 4( mass of O)
= 2(23) + 32 + 4(16)
= 46 + 32 + 64
= 142 g
2) K2CO3: Potassium Carbonate
Molecular mass = 2(mass of potassium) + mass of carbon + 3(mass of oxygen)
= 2(39) + 12 + 3(16)
= 78 + 12 + 48
= 138 g
3) CO2: Carbon Dioxide
Molecular mass= mass of C + 2(mass of oxygen)
= 12+2(16)
= 12+ 32
= 44 g
4) MgCl2: Magnesium Chloride
Molecular mass= mass of Mg + 2(mass of chlorine)
= 24 + 2(35.5)
= 24 + 71
= 95 g
5) NaOH: Sodium Hydroxide
Molecular mass= mass of Na + mass of O + mass of H
= 23 + 16 + 1
= 40 g
6) AlPO4: Aluminium Phosphate
Molecular mass= mass of Al + mass of P + 4(mass of O)
= 27 + 31 + 4(16)
= 27 + 31 + 64
= 122 g
7) NaHCO3: Sodium Bicarbonate
Molecular mass= mass of Na + mass of H + mass of C + 3(mass of O)
= 23 + 1 + 12 + 3(16)
= 23 + 1 + 12 + 48
= 84 g
6.) Two samples m and n of slaked lime were obtained from two different reactions. The details about their composition are as follows.
Sample m mass: 7g
Mass of constituent oxygen: 2g
Mass of constituent calcium = 5g
Sample n mass: 1.4 g
Mass of constituent oxygen : 0.4g
Mass of constituent calcium: 1g
Which law of chemical combination does this prove? Explain.
Ans:
Given that,
Sample m mass: 7g
Mass of constituent oxygen: 2g
Mass of constituent calcium = 5g
Sample n mass: 1.4 g
Mass of constituent oxygen : 0.4g
Mass of constituent calcium: 1g
In case of sample m, the ratio of proportion of calcium and oxygen is 5:2 = Ca:O
And in similar way, in case of sample n the ratio of proportion of calcium and oxygen is 1:0.4= 10:4= 5:2
Thus after simplification of ratio of proportion of mass we get the same ratio in both the samples m and n. And hence this proves the law of constant proportion.
7.) Deduce the number of molecules of the following compounds in the given quantities.
32 g oxygen, 90 g water, 8.8 g carbon dioxide, 7.1 g chlorine.
Ans:
1) 32 g of oxygen:
Molecular mass of O2= 2(16)= 32g
Thus, number of moles in 32g of O2 = mass of O2 in g/ molecular mass of O2
= 32/32= 1 mole
And we know that, 1 mole of any substance contains 6.023*1023 molecules.
Hence, 1 mole of O2 contains 6.023*1023 molecules.
2) 90 g of water:
Molecular mass of H2O= 2(1)+16= 18g
Thus, number of moles in in 90g of water= mass of water/ molecular mass of water
= 90/18= 5moles
90g of water contains 5 moles and 1 mole contains 6.023*1023 molecules.
Thus, in 5 moles 5(6.023*1023)= 30.11*1023 molecules are present.
3) 8.8 g of carbon dioxide:
Molecular mass of CO2= 12 +2(16)= 44g
Thus, number of moles in 8.8g of CO2= mass of CO2/molecular mass of CO2
= 8.8/44= 0.2moles
Thus, 0.2 moles contains 0.2(6.023*1023)= 1.2046*1023 molecules.
4) 7.1 g chlorine:
Molecular mass of chlorine = 2(35.5)= 71 g
Thus, number of moles in 7.1 g of chlorine= mass of chlorine/molecular mass of chlorine
= 7.1/71= 0.1 mole
Hence, 0.1 mole of chlorine contains 0.1(6.023*1023)= 0.6023*1023 molecules.
8.) If 0.2 mole of the following substances are required how many grams of those substances should be taken?
Sodium chloride, magnesium oxide, calcium carbonate.
Ans:
1) Sodium chloride:
Molecular mass of NaCl= 23+35.5= 58.5
We know that,
Number of moles= mass of substance/molecular mass of substance
0.2 moles of NaCl= mass of NaCl/ molecular mass of NaCl
0.2= mass of NaCl/58.5
Hence, mass of NaCl= 11.7g
Thus, 11.7 g of NaCl contains 0.2moles.
2) Magnesium oxide:
Molecular mass of MgO= 24+16= 40g
We know that,
Number of moles= mass of substance/molecular mass of substance
0.2 moles of MgO= mass of MgO/40
Thus, mass of MgO= 0.2*40=8g
Hence, 8g of MgO contains 0.2 moles.
3) Calcium Carbonate:
Molecular mass of CaCO3= 40 + 12 +3(16)
= 100g
We know that,
Number of moles of substance= mass of substance/molecular mass of substance
0.2 moles of CaCO3= mass of CaCO3/100
Thus, mass of CaCO3= 0.2*100=20g
Thus, 20 g of CaCO3 contains 0.2 moles.