4. Measurement of Matter exercise question answer 9th science and technlogoy
Class 9 science and technology chapter 4 question answer Mesurment of matter | Mesurment of matter class 9 exercise answers Maharashtra board | Mesurment of matter 9 solution state board.
1. Give examples.
a. Positive radicals
Ans:
Sodium
ion Na+
Calcium
ion Ca2+
Silver
ion Ag+
Aluminium
ion Al3+
b. Basic radicals
Ans:
Mg2+
K+
Fe2+
Cu2+
c. Composite radicals
Ans:
H3O+
NH4+
CO32-
SO42-
d. Metals with variable valency
Ans:
Cu …..Cu+ , Cu2+
Hg ……Hg+ , Hg2+
Fe …….Fe+ , Fe2+
e. Bivalent acidic radicals
Ans:
S2-
O2-
f. Trivalent basic radicals
Ans:
Al3+
Cr3+
Fe3+
9th class science and technology question answer pdf | Class 9 science and technology chapter 4 question answer Maharashtra board | Class 9 science and technology chapter 4 question answer pdf
2. Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.
Mercury, potassium, nitrogen, copper, sulphur,
carbon, chlorine, oxygen
Ans:
|
Elements |
Symbols |
Charge |
|
Mercury |
Hg |
Hg2+ |
|
Potassium |
K |
K+ |
|
Nitrogen |
N |
N3- |
|
Copper |
Cu |
Cu2+ |
|
Sulphur |
S |
S2- |
|
Carbon |
C |
C+ |
|
Chlorine |
Cl |
Cl– |
|
Oxygen |
O |
O2- |
1. Write the steps in deducing the chemical formulae of the following compounds.
Sodium
sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium
hydroxide
Ans:
1)
Sodium sulphate (Na2SO4):
The formula
is obtained as follows:
The
symbol and valencies of sodium and sulphate
Symbol Na SO4
valency 1 2
Cross
multiply the valencies
Na × 2 and SO4 × 1
The
chemical formula of sodium sulphate is Na2SO4
2)
Potassium nitrate (KNO3)
The formula is obtained as follows:
The symbol and valencies of Potassium and nitrate
Symbol K NO3
valency 1 1
Cross multiply the valencies
K × 1 and NO3 × 1
The
chemical formula of Potassium nitrate is K NO3
3) Frriec phosphate (FePO4)
The formula is obtained as follows:
The symbol and valencies of Frriec and phosphate
Symbol Fe PO4
valency 3 3
Cross multiply the valencies
Fe × 3 and PO4 × 3
The
chemical formula of Frriec phosphate is FePO4
4)
Calcium oxide (CaO)
The formula is obtained as follows:
The symbol and valencies of Calcium and oxide
Symbol Ca O
valency 2 2
Cross multiply the valencies
Ca × 2 and O× 2
The
chemical formula of Calcium oxide is CaO
5) Aluminium hydroxide Al(OH)
The formula is obtained as follows:
The symbol and valencies of Aluminium and hydroxide
Symbol Al OH
valency 3 1
Cross multiply the valencies
Al× 1 and OH× 3
The
chemical formula of Aluminium hydroxide is Al(OH)3
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4. Write answers to the following questions and explain your answers.
a. Explain the monovalency of the element sodium.
Ans:
The
electronic configuration of sodium is 2 , 8 , 1. There is only one electron in
the outermost orbit of Na . If sodium atom gives one electron to an atom of
some other elements , a sodium ion is formed , Hence the valency of sodium is
one i.e. monovalent . After the give and take of electrons is over , the
electronic configuration of both the resulting ions have complete octet .
b. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals: sulphate and phosphate
Ans:
The
symbol and valencies of M and sulphate
Symbol M SO4
valency 2 2
Cross multiply the valencies
M × 2 and SO4 × 2
The
chemical formula of Aluminium hydroxide is MSO4
c. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.
Ans:
(1) 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 introduced .
(2) As the hydrogen atom (H) is the lightest it was taken as the reference atom in old days . The nucleus of H atom contains only one proton , therefore its mass was taken as one unit .
(3) 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
(4) 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 x 12 u 12 is equal to 1 u . ( Note : 1 u = 1.66053904 x 1027 kg ) .
(5) 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 proton.)
d. What is meant by Unified Atomic Mass?
Ans:
In the 20th century various
instruments for accurate determination of atomic masses were developed.
Therefore, instead of the relative atomic mass, it is now possible to determine
the atomic masses in kg. Now unified atomic mass unit, called the Dalton has
been accepted as the unit for expressing atomic mass. 1 u = 1.66053904 × 10-27
kg.
e. Explain with examples what is meant by a ‘mole’ of a substance.
Ans:
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 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 a
substance ( n ) Mass of substance in grams Molecular mass of substance
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5. Write the names of the following compounds and deduce their molecular masses.
Na2 SO4
, K2CO3 , CO2 , MgCl2 , NaOH, AlPO4 , NaHCO3
Ans:
Na2 SO4
( Sodium sulphate)
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
Na2
SO4 Sodium sulphate |
Na Sodium |
23 |
2 |
23 × 2 |
46 |
|
S Sulphur |
32 |
1 |
32 × 1 |
32 |
|
|
O oxygen |
16 |
4 |
16 × 4 |
64 |
Molecular
mass = sum of constituent atomic masses
Molecular
mass of Na2 SO4 = (Atomic mass of Na) ×
2 + (Atomic mass of S) × 1 + (Atomic mass of O) × 4
= (23 ×
2) + (32 × 1) + (16 × 4)
= 46 +
32 + 64
= 142u
Molecular
mass of Na2 SO4 = 142u
2.)
K2CO3 ( Potassium carbonate)
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
K2CO3
Potassium carbonate |
K Potassium |
39 |
2 |
39× 2 |
78 |
|
C carbon |
12 |
1 |
12 × 1 |
12 |
|
|
O oxygen |
16 |
3 |
16 × 3 |
48 |
Molecular
mass = sum of constituent atomic masses
Molecular
mass of K2CO3
=
(Atomic mass of K) × 2 + (Atomic mass of C) × 1 + (Atomic mass of O) × 3
= (39× 2) + (12 × 1) + (16 × 3)
= 78 +
12 + 48
= 138u
Molecular
mass of K2CO3 = 138u
3)
CO2 ( Carbon dioxide) :
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
CO2 Carbon dioxide |
C carbon |
12 |
1 |
12 × 1 |
12 |
|
O oxygen |
16 |
3 |
16 × 2 |
32 |
|
|
|
|
|
|
|
Molecular
mass = sum of constituent atomic masses
Molecular
mass of CO2
= (Atomic mass of C) × 1 + (Atomic mass of O) ×
2
= (12 ×
1) + (16 × 2)
= 12 +
32
= 44u
Molecular
mass of CO2 = 44u
4)
MgCl2 (Magnesium chloride)
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
MgCl2
Magnesium chloride |
Mg magnesium |
24 |
1 |
24 × 1 |
24 |
|
Cl chlorine |
35.5 |
2 |
35.5 × 2 |
71 |
|
|
|
|
|
|
|
Molecular
mass = sum of constituent atomic masses
Molecular
mass of MgCl2
= (Atomic mass of Mg ) × 1 + (Atomic mass of
Cl) × 2
= (24 ×
1) + (35.5 × 2)
= 24+
71
= 95u
Molecular
mass of MgCl2 = 95u
5) NaOH
(Sodium Hydroxide)
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
NaOH
Sodium Hydroxide |
Na Sodium |
23 |
1 |
23*1 |
23 |
|
O oxygen |
16 |
1 |
16 × 1 |
16 |
|
|
H Hydrogen |
1 |
1 |
1 × 1 |
1 |
Molecular
mass = sum of constituent atomic masses
Molecular
mass of NaOH
= (Atomic mass of Na) × 1 + (Atomic mass of O)
× 1 + (Atomic mass of H) × 1
= (23 ×
1) + (16 × 1) + (1 × 1)
= 23 +
16 + 1
= 40u
Molecular
mass of NaOH= 40u
6)
AlPO4 (Aluminium Phosphate) NaHCO3
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
AlPO4
Aluminium Phosphate |
Al Aluminium |
27 |
1 |
27 × 1 |
27 |
|
O oxygen |
16 |
4 |
16 × 4 |
64 |
|
|
P phosphate |
31 |
1 |
3 × 1 |
31 |
Molecular
mass = sum of constituent atomic masses
Molecular
mass of AlPO4
= (Atomic mass of Na) × 1 + (Atomic mass of O)
× 1 + (Atomic mass of H) × 1
= (27 ×
1) + (31 × 1) + (16 × 4)
= 27 +
31 + 64
= 122u
Molecular
mass of AlPO4 = 122u
7)
NaHCO3 ( Sodium bicarbonate)
|
Molecule |
Constituent elements |
Atomic masses (u) |
Number of atoms in the molecule |
Atomic mass × number of atoms |
Mass of the constituents |
|
NaHCO3
Sodium bicarbonate |
Na Sodium |
23 |
1 |
23*1 |
23 |
|
H Hydrogen |
1 |
1 |
1 × 1 |
1 |
|
|
C carbon |
12 |
1 |
12× 1 |
12 |
|
|
O oxygen |
16 |
4 |
16 × 4 |
64 |
Molecular mass = sum of constituent atomic masses
Molecular
mass of NaHCO
= (Atomic mass of Na) × 1 + (Atomic mass of H)
× 1 + (Atomic mass of C) × 1 + (Atomic mass of O) ×3
= (23 ×
1) + (1 × 1) + (12 × 1) + (16 × 3)
= 23 +
1 + 12 + 48
= 84u
Molecular
mass of NaHCO3 = 122u
Class 9 science and technology chapter 4 question answer pdf | Class 9 science and technology chapter 4 question answer Mesurment of matter
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 : 1.4g
Mass of
constituent oxygen : 0.4g
Mass of
constituent calcium : 1.0g
Which
law of chemical combination does this prove? Explain.
Ans:
Slaked
lime have two samples ‘m’ and ‘n’ from 7g of sample m, 5g of Ca and 2 g of
oxygen were obtained, that means in sample, the elements Ca and O are in the
proportion 5:2 by weight.
Similarly,
from 1.4g of sample ‘n’, 1 g of Ca and 0.4g of oxygen were obtained that means
in sample n, the elements Ca and O are in the proportion a : 0.4 by weight i.e.
5:2. This shows that the proportion by weight of Ca and O in different samples
(m and n) remains constat.
Sample
m Ca : O 5 : 2
Sample n Ca : O 1 : 0.4 i.e. 5: 2
This
proves the law of constant proportion. The proportion by weight of the
constituent elements (Ca and O) in different samples of a compound (m and n) is
fixed.
7. Deduce the number of molecules of the following compounds in the given quantities.
32g oxygen, 90g water, 8.8g carbon di oxide, 7.1g chlorine.
Ans:
32g
oxygen
Molecular mass of oxygen
= 16 + 16 = 32
O + O
1 mole of oxygen (O) = 32 g of oxygen
1 mole of oxygen contains
6.022 × 10 raise to 23molecules
32 g of oxygen contains
6.022 × 10 raise to 23molecules
2) 90g water
Ans:
Molecular mass of water (H2O)
= ( 2 × 1 ) + 16 = 18
2H O
1 mole of H2O = 18 g H2O
The number of moles of H2O =
Weight of water / Molecular mass of water
90/18
5 mole of H2O
1 mole of water contains
6.022 × 10 raise to 23
5 mole of water contains
= 5 × 6.022 × 10 raise to 23
= 30.022 ×10 raise to 23
90 g water contains = 3.002 × 10 raise to 24
3) 8.8g carbon di oxide
Molecular mass of carbon dioxide (CO2)
= 12 + 16 + 16 = 44
C O O
1 mole of CO2 = 44g CO2
The number of moles of CO2
= Weight of CO2 / Molecular mass of CO2
= 8.8 / 44 = 0.2 mole of CO2
1 mole of CO2 contains
= 6.022 ×10 raise to 23 Molecules
0.2 Mole CO2 contains
= 0.2 × 6.022 × 10 raise to 23
= 1.2044 × 10 raise to 23 Molecules
8.8 g
CO2 contains 1.2044 × 10 raise to 23 Molecules
4) 7.1g chlorine
Ans:
Molecular mass of chlorine (Cl2)
= 35.5 + 35.5 = 71
= Cl Cl
1 mole of chlorine = 71 g chlorine
The number of moles of Cl2
= weight of chlorine / Molecular mass of chlorine
= 7.1 / 71 = 0.1
= 0.1 mole Cl2
1 mole of chlorine contains
= 6.022 × 10 raise to 23 milecules
= 0.1 mole chlorine contains
= 0.1 × 6.022 × 10 raise to 23
= 6.022 × 10 raise to 22 molecules
7.1 g
chlorine contains 6.022 × 10 raise to 22 molecules
8. If 0.2 mol of the following substances are required how many grams of those substances should be taken?
1)Sodium chloride, 2)magnesium oxide, 3)calcium carbonate
Ans:
1)Sodium chloride
Molecular mass of sodium
= 23+35.5 = 58.5
Chloride (NaCl) Na + Cl
1 mole of NaCl = 58.5 g of NaCl
Mass of NaCl in grams
=Number of moles of NaCl × Molecular mass
= 0.2 × 58.5
= 11.7 g
0.2
mole of NaCl = 11.7g of NaCl
2)Magnesium oxide (MgO)
Ans:
Molecular mass of magnesium oxide (MgO)
= 24 + 16
= Mg + O
= 40
1 mole of MgO = 40.0 g of MgO
Mass of MgO in grams =
Number of moles of MgO × Molecular mass
= 0.2 × 40
= 8.0 g
0.2
mole of MgO = 8.0 g of MgO
3) Calcium carbonate (CaCO3)
Ans:
Molecular mass of calcium carbonate (CaCO3)
= 40 + 12 + 48
Ca + C + O
= 100
1 mole of CaCO3 × Molecular mass
= 0.2 × 100
= 20 g
0.2
Mole of CaCO3 = 20 g of CaCO3
***************
