Saturday, 30 July 2011

CBSE Class IX ( 9th) Science Chapter 4. Structure Of The Atom | Lesson Exercises

Question 1. Compare the properties of electrons, protons and neutrons.
Answer :
ElectronsProtonsNeutron
Electrons are Negatively charged particles
Electrons are present in outer shells with in an atom and  revolve around the nucleus in well-defined orbits or discrete orbits
The mass of an electron is about 1/ 2000 times
the mass of an hydrogen atom.
A electron  is represented as 'e-'
Protons are Positively charged particles.
Protons are present in the nucleus of all atoms
Number of protons determines the atomic number of an element
The mass of a proton is taken as one unit and equals to neutron.
A proton is represented as 'p+'
Neutrons do not carry any charge and are neutral.
Neutrons are present in the nucleus of all atoms, except hydrogen
The mass of a neutron is taken as one unit and equals to that of proton.
A neutron is represented as 'n'

Question 2. What are the limitations of J.J. Thomson's model of the atom?
Answer : Thomson proposed that: an atom consists of a positively charged sphere and the electrons are
embedded in it.The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.Thomson's model of the atom fails to explain Rutherford's  α-particle scattering experiment in which most of the fast moving α-particles passed straight through the gold foil.Only Some of the α-particles were deflected by the foil by small angles.Which clearly established atom has a lot of empty space and positive charge is concentrated in a very small volume within the atom.

Question 3. What are the limitations of Rutherford's model of the atom?
Answer : As per Rutherford's nuclear model of an atom , an atom has a very small sized nucleus with positive charge   inside and has  electrons revolving around this nucleus in well-defined orbits. Nearly all
the mass of an atom resides in the nucleus.
Rutherford's model of the atom failed to explain the the stability of the atom.  As any particle in a
circular orbit would undergo acceleration. During acceleration, revolving electron as charged particles would lose energy and finally fall into the nucleus. This may lead to a collapsed atomic structure, resulting in  very unstable atoms.Contrarily  atoms are mostly stable.


Question 4. Describe Bohr's model of the atom.
Answer . Neils Bohr put forward the following postulates about the model of an atom:
(i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.
(ii) While revolving in discrete orbits the electrons do not radiate energy.
 These orbits or shells are called energy levels.These orbits or shells are represented by the letters K,L,M,N,… or the numbers, n=1,2,3,4,…
.
Question 5. Compare all the proposed models of an atom given in this chapter.
Answer :
THOMSON'S Model of An AtomRUTHERFORD'S Model of An AtomBOHR' S Model of An Atom
(i) An atom consists of a positively charged sphere and the electrons are embedded in it.
(ii) The negative and positive charges are equal in magnitude. So, the atom as a  whole is electrically neutral.
(i) There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an   atom resides in the nucleus.
(ii) The electrons revolve around the nucleus in well-defined orbits.
(iii) The size of the nucleus is very small as compared to the size of the atom.
(i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.
(ii) While revolving in discrete orbits the electrons do not radiate energy.

Question 6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer : The following rules are followed for writing the distribution of electrons in different energy
levels or shells for the first eighteen elements :
(i) The maximum number of electrons present in a shell is given by the formula 2n2, where 'n' is the orbit
number or energy level index, 1,2,3,….Hence the maximum number of electrons in different shells are as
follows:
first orbit or K-shell will be = 2 × 12 = 2,
second orbit or L-shell will be = 2 × 22 = 8,
third orbit or M-shell will be = 2 × 32= 18,
fourth orbit or N-shell will be = 2 × 42= 32,
(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.
(iii) Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner.

Distribution of electrons in various shells for the first eighteen elements
ElementSymbolAtomic NumberNumber of Electron
Distribution of electrons
KLMN
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminium
Silicon
Phosphorus
Sulphur
Chlorine
Argon
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
-
-
1
2
3
4
5
6
7
8
8
8
8
8
8
8
8
8
-
-
-
-
-
-
-
-
-
-
1
2
3
4
5
6
7
8
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-


 
Question 7. Define valency by taking examples of silicon and oxygen.
Answer : The valency may be defined as the combining capacity    of the atoms of  elements,  or their tendency to  form molecules with atoms of the same or different elements. The electrons in an atom are arranged in different shells/orbits. The electrons present in the outermost shell of an atom are known as the valence electrons. The outermost shell of an atom can accommodate a maximum of 8 electrons. All atoms of various elements have tendency to have a fully-filled outermost shell with 8 electrons or octet . This is achieved through reaction with atoms of other elements,  by sharing, gaining or losing electrons. 
  • In case of  Silicon, it has 4 electrons in the outer most shell, thus it is deficient by 4 electrons to have a fully filled outer most shell. Hence Its valency is 4.
  • In case of  Oxygen, it has 6 electrons in the outer most shell, thus it is deficient by 2 electrons to have a fully filled outer most shell. Hence Its valency is 2.

Question 8. Explain with examples 
(i) Atomic number, (ii) Mass number, (iii) Isotopes and ( iv) Isobars. Give any two uses of isotopes.

Answer : 
(i) Atomic number : Atomic number of  an element  is defined as  the number of protons present in its atom.  It is denoted by 'Z'.

(ii) Mass number : The mass number is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom.

(iii) Isotopes :Isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers.

( iv) Isobars : Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobars.

Two uses of isotopes : 
(i) An isotope of uranium is used as a fuel in nuclear reactors.
(ii) An isotope of cobalt is used in the treatment of cancer.

Question 9. Na+ has completely filled K and L shells. Explain.
Answer : Na+  represent Sodium with atomic number 11. Which means it has 11 protons. Number of Electrons is equal to number of protons. So Sodium has 11 electrons in the outer shells distributed as  2 electrons in 'k' shell, 8 electron in 'L' shell and 1 electron in 'M' shell.  The distribution of electron in the shells is governed by the formula i.e. 2n2. Where 'n' represents the number of shells. The first shell is represented as 'K', second shell as 'L', third shell as 'M', and fourth shell as 'N' and so on.
There for, as per the formula above,   the 'K' shell  i.e. 1st  shell will contain  2 x 12   =2 electrons and 'L' shell will contain 2 x 22 = 8 electrons.Which means it has completely filled 'K' and 'L' shells.

Question 10. If bromine atom is available in the form of, say, two isotopes
 
79

Br (49.7%)
35
and
81

Br (50.3%).
35
calculate the average atomic mass of bromine atom.
Answer : The average atomic mass of bromine atom  (79 × 49.7 ) / 100 + ( 81 × 50.3)/ 100 =(3926.3 + 4074.3 ) / 100 = 8000.6 / 100= 80.06 u

Question 11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes
  
16

X
8
and 
  
18

X
8
in the sample?

Answer :
Suppose the percentage of 
  
16

X  in given sample = x
8
:.  The percentage of 
18

X  in given sample = 100-x
8
  
 As given average atomic mass of a sample of an element X = 16.2 u

=> (16  × x /100) + [18  × (100-x) /100] = 16.2
 => (16x + 1800-18x) / 100 = 16.2
=>  (16x + 1800-18x) = 16.2 × 100
=>  -2x + 1800 = 1620
=> -2x = 1620-1800
= -2x = -180
=> x = 90

:.  The  percentage of 
16

X  in given sample = x = 90 
8

:.  The  percentage of 
16

X  in given sample = 100-x =100-90 = 10
8
 
Question 12. If Z = 3, what would be the valency of the element? Also, name the element.
Answer :
 
Given Z= 3
we know that Z = Atomic Number = Number of protons = Number of electrons
:.  Number of electrons = 3
Distrubution of electrons is given by the formula 2n2 where n is the number of orbit or shell represented by letters K, L, M, N etc
Hence for first orbit i.e. 'k' the number of electrons = 2 × 1 = 2
for second orbit i.e. 'L' the number of electron = total electrons - numbers of electrons in 'K' = 3-2=1
Valency of element is 1 as it will readily lose its outermost valance electrons than to gather 7 more to have a fully filled outermost orbit.
The element is Lithium

Question 13. Composition of the nuclei of two atomic species X and Y are given as under :

                      X            Y 
Protons         = 6          6
Neutrons      = 6           8 
Give the mass numbers of X and Y. What is the relation between the two species?

Answer :
We know that :
Mass number of an atom  = Number of Protons + number of Neutrons
 Atomic Number of an atom= Number of Protons
:.  Atomic Number of X =  6
:.  Atomic Number of y =  6
:. Mass number of X =  6+6 = 12
:.  Mass number of  y = 6+8 = 14
X and Y have similar  Atomic Number (6) but have different mass numbers, therefor both are isotopes of same element.

Question 14. For the following statements, write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1 2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer :
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.(F)
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.(F)
(c) The mass of an electron is about 1/ 2000 times that of proton.(T)
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.(T)


Put tick () against correct choice and cross (×) against wrong choice in questions 15, 16 and 17
Question 15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of 
(a) Atomic Nucleus
(b) Electron 
(c) Proton 
(d) Neutron
Answer : 
(a) Atomic Nucleus (√)
(b) Electron (×)
(c) Proton (×)
(d) Neutron (×)

Question 16. Isotopes of an element have (a) the same physical properties (b) different chemical properties (c) different number of neutrons (d) different atomic numbers.
Answer :
 (a) the same physical properties (×)
(b) different chemical properties (×)
(c) different number of neutrons (√)
(d) different atomic numbers. (×)

Question 17. Number of valence electrons in Cl ion are: (a) 16 (b) 8 (c) 17 (d) 18  
Answer :
(a) 16 (×)
(b)  8 (√)
(c) 17 (×)
(d) 18 (×)

Question 18. Which one of the following is a correct electronic configuration of sodium?
(a) 2,8          (b) 8,2,1          (c) 2,1,8         (d) 2,8,1.  
 

Answer :(d) 2,8,1.

Question 19. Complete the following table.


Atomic NumberMass NumberNumber of NeutronsNumber of ProtonsNumber of ElectronsName of the Atomic Species
9-10---
1032---Sulphure
-24-12--
-2-1--
-1010-

Answer : 

Atomic NumberMass NumberNumber of NeutronsNumber of ProtonsNumber of ElectronsName of the Atomic Species
9191099Fluorine
1632161616Sulphure
1224121212Magnesium
12111Deuterium
11011Protium

Thursday, 28 July 2011

Solved Exercises : CBSE Class VI (6th) Science Chapters 1 to 16

CBSE Class IX ( 9th) Science | Chapter 3. Atoms And Molecules | Lesson Exercises

Question 1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer :
Mass of boron as given = 0.096 g
Mass of oxygen as given  = 0.144 g
Mass of sample as given   = 0.24 g
  Percentage of boron by weight in the compound = (0.096 ÷ 0.24 )  x 100 = 40%
  Percentage of oxygen by weight in the compound = (0.144 ÷ 0.24)  x 100 = 60%

Question 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide willbe formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer : We know that in burning process carbon utilises oxygen to form Carbon Dioxide .
  Carbon + Oxygen => Carbon Dioxide + Heat Energy

Here as given,  in burning, 3.0 g of carbon utilises 8.00 g of oxygen to form 11.00 g of carbon dioxide.When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from  the total of 50 gm of Oxygen to produce 11 gm of Carbon Dioxide. 42 gm of Oxygen will be left behind without any change.
The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction.
Here in terms of mass 3g of Carbon and 50 gm of Oxygen combines to give 11 gm of Carbon Dioxide and 42 gm of unused Oxygen keeping the total mass of combining

chemicals 53 gm (3+50 gm)  by weight constant before and after the reaction.

2. Law of constant proportions which states that In a chemical substance the elements are always present in definite proportions by mass.Here respective of the excess quantity of Oxygen available for burning of same weight of Carbon, the amount of carbon dioxide will remain same.

Question 3. What are polyatomic ions? Give examples.
Answer : Polyatomic ions are clusters or group of atoms that always have a  fixed  charge attach to them. This charge may be positive or negative .
Ammonium  NH4+
Hydroxide OH-
Nitrate NO3 -
Carbonate CO32-
Sulphate  SO42-

Question 4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Answer :
(a) Magnesium chloride     => MgCl2
(b) Calcium oxide         => CaO       
(c) Copper nitrate        => Cu(NO3)2.
(d) Aluminium chloride    => AlCl3
(e) Calcium carbonate.    => CaCO3


Question 5. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

Answer :

CompoundChemical FormulaElements


(a) Quick lime

(b) Hydrogen bromide


(c) Baking powder


(d) Potassium sulphate.

CaO


HBr


NaHCO3 


K2SO4
 Calcium, Oxygen


Hydrogen, Bromine


Sodium, Hydrogen, Carbon, Oxygen


Potassium, Sulphur, Oxygen

Question 6. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3

Answer :
(a) Molar Mass of Ethyne, C2H2; =     2 × 12 + 2 × 1 =  26 g
(b) Molar Mass of Sulphur molecule, S8 =   8 × 32  =  256 g
(c) Molar Mass of Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)  =   4 × 31  =  124 g
(d) Molar Mass of Hydrochloric acid, HCl  =   1 + 35.5 =  36.5 g
(e) Molar Mass of Nitric acid, HNO3 =   1 + 14 + 3 × 16  =  63 g

Question 7. What is the mass of.
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium= 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?

Answer :The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The atomic mass of an element gives us the mass of one atom of that element in atomic mass units (u). To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
Molar mass of atoms is also known as gram atomic mass. For example, atomic mass of hydrogen=1u. So, gram atomic mass of
hydrogen = 1 g.

(a) The mass of 1 mole of Nitrogen atoms = 14 g.

(b) The mass of 4 moles of Aluminium atoms is (4 × 27) g = 108 g

(c) The mass of 10 moles of Sodium Sulphite (Na2SO3) =10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Question 8. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.

Answer: To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
1 mole substance = Total Atomic mass of a substance (u)       Or
                             = Total gram atomic mass of substance (g)  Or
                             = Molar Mass
                             = 1 Mole of substance                                Or 
                             =  6.022 × 1023 in number of particles of that substance
Here required mole of substance  =  Given mass of substance in 'g' ÷ Molar Mass

(a) 12 g of oxygen gas =  Given mass of oxygen in 'g' ÷ Molar Mass of Oxygen
                                 = 12 ÷ 32 
                                 = 0.375 mole of oxygen
 
(b) 20 g of water = Given mass of water in 'g' ÷ Molar Mass of Water
                         = 20 ÷ 18
                         = 1.11 mole of water

(c) 22 g of carbon dioxide.= Given mass of Carbon Dioxide in 'g' ÷ Molar Mass of Carbon Dioxide
                                      = 22 ÷ 44
                                      = 0.5 mole of carbon dioxide


Question 9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer : Mass of atoms (m) in given moles (n) = Molar Mass of substance (M)× Given Moles of substance (n)

(a) ∴  Mass of 0.2 mole of oxygen atoms = Molar Mass of oxygen × Given Moles  =  16g × 0.2   = 3.2 g

(b) 
Mass of 0.5 mole of water molecules =Molar Mass of water × Given Moles = 18 g × 0.5  = 9 g

Question 10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Answer :
1 mole of Sulphur (S8 = 8 × 32 g = 256 g
also as
1 mole = 6.022 × 1023 in number of particles of that substance
So 256 g of  sulphur contains = 6.022 × 1023  molecules of sulphur
Therefor 16 g of sulphur will contains = 3.76 ×
1022  molecules (approx)

Question 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Answer :
1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16  = 102 g

We know, 102 g of
Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)
We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number
Then, 0.051 g of Aluminium Oxide (Al2O3) contains
= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3) 
= 3.011 × 1020 molecules of Aluminium Oxide (Al2O3)

The number of Aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of Aluminium ions (
Al3+) present in 3.011 × 1020 molecules (0.051 g ) of Aluminium Oxide (Al2O3)
= 2 × 3.011 × 1020

= 6.022 ×
1020
 

Sunday, 24 July 2011

CBSE Class IX (9th) Science | Chapter 2. IS Matter Around Us Pure ? | Lesson Exercises

Question 1. Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.

Answer :Separation techniques for separation of given substance mixtures :-
Mixtures of SubstanceSeparation techniques
(a) Sodium chloride from its solution in water.By Evaoparation
(b) Ammonium chloride from a mixture containing
sodium chloride and ammonium chloride
By Sublimation
(c) Small pieces of metal in the engine oil of a car.By Filtration
(d) Different pigments from an extract of flower petals.Using Chromatography
(e) Butter from curd.By Centrifugation
(f) Oil from water.Using separating funnel
(g) Tea leaves from tea. By Filtration
(h) Iron pins from sand.By a Magnet
(i) Wheat grains from husk.By Winnowing
(j) Fine mud particles suspended in water.By Centrifugation


Question 2. Write the steps you would use for making tea.
Use the words : solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

 Answer : For making tea, first of all we will take desired cups of  water as solvent in a tea pan. Then it is allowed to boil on stove. Tea leaves, as salute is added to it and  brewed. While brewing , the colour of water as solvent gets changed to tea colour as soluble part of tea as salute gets dissolve in heated water  as solvent. There after,   milk is added to it as solute and is further allowed to boil for some times. Insoluble tea leaves as residue  are removed by passing the mixture through a tea strainer. Sugar as solute according to need is added to filtrate so obtained and stirred with spoon. Sugar gets dissolve in filtrate. Resulting solution is in the form of  Tea, ready for use.

Question 3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below
(results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
Substance DissolvedTemperature in K
283   293    313    333    353

Solubility

Potassium nitrate
Sodium chloride
Potassium chloride
Ammonium chloride
21
36
35
24
32
36
35
37
62
36
40
41
106
37
46
55
167
37
54
66


(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?


Answer :  Mass of potassium nitrate at 313 K per 100 grams of water in given saturated solution of potassium nitrate = 62 Grams
 Required mass of potassium nitrate at 313 K per 50 grams of water for a saturated solution of potassium nitrate = 62 / 100 ×50 = 31 grams

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
Answer :  A saturated solution of potassium chloride in water at 353 K has a specific solubility.If it is allowed to cool down at room temperature around 293 K, with the decrease in temperature, solubility  of potassium chloride in the solution will also become low. It will dissolve less in water and undissolved potassium chloride result in  formation of  a layer of potassium chloride crystals at the bottom.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

 Answer : The  solubility of a saturated solution for a given temperature is expressed as  Mass of solute in gm per 100 gm of solvent
  1. The solubility of Potassium nitrate in a saturated solution with water at 293 K = 32 gm of Potassium nitrate per 100 gm of water
  2. The solubility of Sodium chloride in a saturated solution with water at 293 K = 36 gm of Sodium chloride per 100 gm of water
  3. The solubility of Potassium chloride in a saturated solution with water at 293 K = 37 gm Potassium chloride per 100 gm of water
  4. The solubility of Ammonium chloride in a saturated solution with water at 293 K = 35 gm Ammonium chloride per 100 gm of water
 
Ammonium chloride has  the highest solubility at the given temperature of 293 K


(d) What is the effect of change of temperature on the solubility of a salt?
Answer : Solubility of salt gets  gets affected with change of temperature. It increases with rise in temperature and decreases with fall in temperature.

Question 4. Explain the following giving examples.
(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension

Answer : 
(a) Saturated solution : A solution at any particular temperature, having maximum amount of solute which  can be dissolve in it, is called a saturated solution. In other words, at a given temperature no more solute can be dissolved in a saturated solution.
(b) Pure substance : A pure substance contains only one kind of pure matter and its composition is the same throughout. Mixtures are constituted by more than one kind of pure substance.A pure substance cannot be separated into other kinds of matter by any physical process.Whatever the source of a substance may be, it will always have the same characteristic properties. Examples Sugar, sodium chloride.
(c) colloid : A colloid or a colloidal solution is a heterogeneous mixture in which the particles of a colloid are uniformly spread throughout the solution. Due to the relatively smaller size of particles,  the mixture appears to be almost homogeneous,  for example, milk.
(d) Suspension : A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium. Particles of a suspension are visible to the naked eye.
The particles of a suspension scatter a beam of light passing through it and make its path visible.The solute particles settle down when a suspension is left undisturbed, that is, a suspension is unstable. They can
be separated from the mixture by the process of filtration. When the particles settle down, the suspension breaks and it does not scatter light any more.

Question 5. Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Answer :  

Homogeneous Mixture Heterogeneous Mixture

  1. Soda water
  2. Air
  3. Vinegar

  1. Wood
  2. Soil
  3. Filtered tea



Question 6. How would you confirm that a colourless liquid given to you is pure water?
 Answer :  For a given atmospheric pressure, every liquid has a distinct boiling point. Water in its pure form  boils at 100° C or 273 K at 1 atmospheric pressure. If given colourless liquid starts boiling below or above 100° C or 273 K, it is either not pure water or some other liquid. So boiling point of a liquid, is an indicator of its purity  at a particular atmospheric pressure.

Question 7. Which of the following materials fall in the category of a "pure substance" ?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air.

 Answer : A pure substance contains only one kind of pure matter and its composition is the same throughout. Mixtures are constituted by more than one kind of pure substance.A pure substance cannot be separated into other kinds of matter by any physical process.Whatever the source of a substance may be, it will always have the same characteristic properties. Therefor as per the description of pure substance above, the following materials fall in the category of a “pure substance”:
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury

Question 8. Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water.



Answer : As we know a solution is a homogeneous mixture of two or more substances. The component  of the solution which is present in larger amount and dissolves the other is called solvent and other component with lesser quantity which gets dissolved in solvent is called the solute.
As per the description above, the following mixtures are solutions:

(b) Sea water
(c) Air
(e) Soda water

Question 9. Which of the following will show "Tyndall effect"?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.

 Answer : Scattering of a beam of light passing through a suspension or a colloidal solution is called the Tyndall effect. Both a suspension and  a colloidal solution is a heterogeneous mixture in which the particles which may or may not be seen with naked eyes,  can scatter a beam of light passing through it.
As per the description above, following will show "Tyndall effect"


 (b) Milk
(d) Starch solution.

Question 10. Classify the following into elements, compounds and mixtures.

(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood

 Answer : Classification of materials as elements, compounds and mixtures.

ElementsCompoundsMixture
(a) Sodium
(d) Silver
 (f) Tin
(g) Silicon
(e) Calcium carbonate


(k) Methane
(l) Carbon dioxide
(b) Soil

(c) Sugar solution
(h) Coal
(i) Air
(j) Soap
(m) Blood

Question 11. Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle

Answer : 
(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food
(g) Burning of a candle

Saturday, 23 July 2011

CBSE Class IX (9th) Science | Chapter 1. MATTER IN OUR SURROUNDINGS | Lesson Exercises

Question 1. Convert the following temperatures to the celsius scale.
(a) 293 K (b) 470 K.

Answer : As we know 0° C =273.16 K. For convenience, we take 0° C = 273 K after rounding off the decimal.
 To change a temperature on the Kelvin scale to the Celsius scale we have to subtract 273 from the given temperature

(a) 293 K =  293-273  = 20° C
(b) 470 K = 470-273   = 197° C


Question2. Convert the following temperatures to the Kelvin scale.
(a) 25°C (b) 373°C.

Answer : Kelvin is the SI unit of temperature, 0° C =273.16 K. For convenience, we take 0° C = 273 K
after rounding off the decimal. To  convert a temperature on the Celsius scale to the Kelvin scale we have to add 273 to the given temperature.

(a) 25°C = 25 + 273 = 298 K

(b) 373°C = 373 + 273 = 646 K

Question3. Give reason for the following observations.
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.

 Answer : (a) Naphthalene balls disappear with time without leaving any solid. This is due to the property of sublimation. Some solids do not exist in their liquid form or changes from solid to liquid; instead they directly evaporate from their solid form to vapour forms. Naphthalene is a sublime substance with very high vapors pressure,  that show the property of sublimation. So naphthalene balls disappear after some time due to the property of sublimation.

(b) We can get the smell of perfume sitting several metres away. This is due to high rate of  diffusion of perfume particles. Liquid generally vaporize at normal temperature. Perfume particles on vaporizing, intermix on their own with each other and air molecules around at a very fast rate. They do so by getting into the spaces between the particles. This intermixing of particles of two different types of matter on their own is called diffusion.Therefor the smell of perfume reach us sitting metres away.

Question4. Arrange the following substances in increasing order of forces of attraction between the particles. water, sugar, oxygen.

Answer : Substances in increasing order of forces of attraction between the particles.  
  • Oxygen 
  • Water
  • Sugar

Question5. What is the physical state of water at.
(a) 25°C (b) 0°C (c) 100°C ?

Answer : Physical states of water at given temperatures :
(a) 25°C - Liquid as familiar water
(b) 0°C  - Solid  as ice
(c) 100°C- Gaseous as water vaporour
Question6. Give two reasons to justify.
(a) water at room temperature is a liquid.
(b) an iron almirah is a solid at room temperature.

Answer :  (a) water at room temperature is a liquid because its melting temperature is with in range of room temperature. As we know force of molecular attraction of a substance becomes high at low temperature and weak  at high temperature. Due to this  water at 0°C becomes solid due to increased molecular particles attraction.At room temperature arround 20 °C, due to increase in tempreture,  kinetic energy of the water particles increases. Due to the increase in kinetic energy, the particles start vibrating with greater speed. The energy supplied by heat overcomes the forces of attraction between the water particles. The
particles leave their fixed positions and start moving more freely. A stage is reached when the solid ice melts and is converted to water a liquid.

(b) an iron almirah is a solid at room temperature.This is because the melting point of iron is  1536 °C or 2797 °F or 1809 K where as room temperature is around 20°C. At this temperature, kinetic energy of the iron molecules is not enough to  overcomes the forces of attraction between the water particles. There for iron almirah is solid at room temperature.

Question7. Why is ice at 273 K more effective in cooling than water at the same temperature?
Answer : The heat energy  absorbed by ice to change to liquid state, without showing any rise in temperature is known as the latent heat. We already know, the  heat can change the physical state of a matter by overcoming the forces of attraction between the particles. For water freezing point and melting point is 0°C. So water has extra amount of hidden energy as compare to ice.

Question8. What produces more severe burns, boiling water or steam?
Answer : Steam produces more severe burns as compare to boiling water as it has extra high heat energy.

Question9. Name A,B,C,D,E and F in the following diagram showing change in its state

Answer :

Tuesday, 19 July 2011

CBSE Class VIII (8th) Science | Chapter 7. CONSERVATION OF PLANTS AND ANIMALS | Lesson Exercises

Question 1. Fill in the blanks:
(a) A place where animals are protected in their natural habitat is called  wildlife sanctuaries .
(b) Species found only in a particular area is known as Endemic species .
(c) Migratory birds fly to far away places because of  climatic changes.

Question 2. Differentiate between the following:
(a) Wildlife sanctuary and biosphere reserve
(b) Zoo and wildlife sanctuary
(c) Endangered and extinct species
(d) Flora and fauna

Answer :
 (a) Wildlife sanctuary and biosphere reserve :
Wildlife sanctuary Biosphere reserve
Wildlife sanctuary are are areas where animals are protected from any disturbance to them and their habitat wildlife sanctuaries provide protection and suitable living conditions to wild animals. Here  killing (poaching) or capturing of animals is strictly prohibited. Biosphere reserve are large areas of protected land for conservation of wild life, plant and animal resources and traditional life of the tribals living in the area. Biosphere reserves like a complete ecosystem  take care of biodiversity comprising plants, animals and microorganisms generally found in an area.

(b) Zoo and wildlife sanctuary
ZooWildlife sanctuary
Zoo is man made places where animals
kept in artificial setting instead of their
natural habitat  and provided protection .
Zoo can be in the middle of a urbanised city
Wildlife sanctuary provide protection and suitable living
conditions to wild animals just like in theirs natural habitat

(c) Endangered and extinct species
Endangered SpeciesExtinct species
Animals Species whose numbers are diminishing to a level that they might face extinction are known as the endangered speciesExtinct species are those animal species which became extinct a long time ago due to sudden climatic changes . The dinosaurs are the extinct species, which were very large sized animals and were in plenty but vanished from this earth due to their inability to adjust to changes around.
.
(d) Flora and fauna
Flora fauna
The plants found in a particular area are termed
flora of that area. They are naturally occurring and have a indigenous plant life cycle
The animals found in a particular area are termed fauna of that area

Question 3. Discuss the effects of deforestation on the following:
(a) Wild animals
(b) Environment
(c) Villages (Rural areas)
(d) Cities (Urban areas)
(e) Earth
(f) The next generation

Answer :
(a) Wild animals : Deforestation  deprives wild animals  of their natural habitat as result of which they may migrate to other areas or may face extinction.
(b) Environment : Deforestation is  disturbing the whole set of our eco-system , leading to major enviromental changes like Glabal warming.
(c) Villages (Rural areas) : Deforestation results in less rain fall, which means less water for human and crop irrigation. Domestic cattle will have no fodder. Floods  and soil erosion caused by deforestion will further adversly effect the rural economy and they may resort to migration to cities in great numbers.
(d) Cities (Urban areas) : Deforestation is making water scacity a major problem in cities. In the absence of a eco-system for recycling and over grown  urban areas and industries, cities are already facing air pollution and scarcity of resources obtained from forests. Rural migration caused by the depleted resources there, is adding further burden. In future cities may face water scacity, inadequate supplies of food grains, vegitables and other products like milk
(e) Earth : Deforestation is changing earth temerature due to Global Warming. This may endanger the the complete eco-system comprising of land, living beings, water, air and other resources.
(f) The next generation : Deforestation will offer great challenges to the next generation in the form of scarcity of water due to less rain fall, air pollution,  lesser food, lesser organic raw materials, increased temperature,   natural calamities or disasters floods, droughts and storms due to climatic changes.


Question 4. What will happen if:
(a) we go on cutting trees.
(b) the habitat of an animal is disturbed.
(c) the top layer of soil is exposed.

Answer :
(a) we go on cutting trees : Trees are the most important part of our eco-system directly responsible for recycling of polluted air we breath, rainfall augumenting our water reserves which are used for drinking as well as crop production, supply of organic raw materials in the form of food, wood and medicine, providing food and shelter to animals and birds. Trees also prevent soil erosion and floods. Cutting of trees in a longer run will result in catastrophic consequences for the existence of mankind along with whole spectrum of diversity envolvng living being and plants on this planet earth.

(b) the habitat of an animal is disturbed: Each species of a animal is part of a eco-system  chain with diverse elements in the form of land, air, water, flora and fauna.. Each part  essentially  play its role as a team member to support the co-existence of natural diversty in a wholistic and balanced term. If the habitat of an animal is disturbed, it will resort to migrate to some other place for food and shelter. This will disturb the ecosystem of that particular area resulting in long term consequences.

(c) the top layer of soil is exposed : If  the top layer of soil is exposed , rain water will directly hit the ground and impact of this will cause further loosening of humus rich top soil layer. First, this will result in soil erosion in the form of mud flush and reduced soil fertility, secondly it will deprive ground water reserves from recharging as water will flow down without any seapage or percolation. 

Question 5. Answer in brief:
(a) Why should we conserve biodiversity?
Answer : We should conserve biodiversity, because biodiversity represent whole eco-system comprising land, water, air, fauna and flora. Each component of this biodiversity chain is inter dependent on each other for a sustainability.
(b) Protected forests are also not completely safe for wild animals. Why?
Answer : Even protected forests are not safe for wild animals because people living in the neighbourhood encroach upon them and destroy them.
(c) Some tribals depend on the jungle. How?
 Answer : Some tribals depend on the jungle as  they live  in the deep forest like primitive people had lived in anciant time.They are still away from the basic amenities of modern life   Forest provide them food items like seed grains, fruit, raw vegetables and other products like honey, meat of hunted wild animals for themselves, fodder for theirs animals, wood fuel for heating and cooking even clothing using natural fibres, leaves and animal skins.
(d) What are the causes and consequences of deforestation?
Answer : The main causes of deforestation are exploitation of forest resources by the human being   to meet the ever growing requirements of an increasing population, rapid urbanization, industrialization. The consequences of deforestation will result in putting in danger, the whole existence  of human being along with biodiversity on the planet earth.
(e) What is Red Data Book?
Answer : Red Data Book is the source book which keeps a record of all the endangered animals and plants
(f) What do you understand by the term migration?
Answer : Migration is the phenomenon of movement of a species from its own habitat to some other habitat for a particular time period every year for a specific purpose like breeding.

Question 6. In order to meet the ever-increasing demand in factories and for shelter, trees are being continually cut. Is it justified to cut trees for such projects?
Discuss and prepare a brief report.
Answer : It is not justified to cut the trees at present rate for such project. Trees as a part of  forest are the most important part of our eco-system.Word wide, reduced forest cover have already started showing its devastating effects in terms of climatic changes. We must reduce our dependency on our forest resources to meet the ever-increasing demand in factories and for shelter. We should look for some other alternatives for the same. We have to start reforestation at much larger scale taking it as a top priority.

Question 7. How can you contribute to the maintenance of green wealth of your locality?
Make a list of actions to be taken by you.

Answer :
  • We can plant trees in open spaces, park, along the side of road and motivate others.
  • We can make use of roof water or water from kitchen for water harvesting to recharge the ground water table.
  • We can recycle our organic waste from kitchen for making  compost and use it in our kitchen garden or plants.
  • We can reduce use of paper .We can save it, reuse used paper and recycle it.

Question 8. Explain how deforestation leads to reduced rainfall.
Answer : Deforestation increases the temperature and pollution level on the earth. It increases the level of carbon dioxide in the atmosphere. Ground water level also gets lowered. Deforestation disturbs the whole eco-system. As we know, plants need carbon dioxide for photosynthesis. Fewer trees would mean that less carbon dioxide will be used up resulting in its increased amount in the atmosphere. This will lead to global warming as carbon dioxide traps the heat rays reflected by the earth. The increase in temperature on the earth disturbs the water cycle and may reduce rainfall.

Question 9. Find out the information about the national parks in your state. Identify and show their location on the outline map of India.
 Answer : Our state Punjab has just one national park.
Harike Wetland also known as "Hari-ke-Pattan", with the Harike Lake in the deeper part of it, is the largest wetland in northern India in the Ferozepur district of the Punjab state in India. The wetland and the lake were formed by constructing the head works across the Sutlej river, in 1953. The headworks is located downstream of the confluence of the Beas and Sutlej rivers. The rich biodiversity of the wetland which plays a vital role in maintaining the precious hydrological balance in the catchment with its vast concentration of migratory fauna of waterfowls including a number of globally threatened species (stated to be next only to the Keoladeo National Park near Bharatpur) has been responsible for the recognition accorded to this wetland in 1990, by the Ramsar Convention, as one of the Ramasar sites in India, for conservation, development and preservation of the ecosystem
(Source : en.wikipedia.org/wiki/Harike_Wetland)
 
Question 10. Why should paper be saved? Prepare a list of ways by which you can save paper.
Answer : It takes 17 full grown trees to make one tonne of paper. Therefore, we should save paper. Paper can be recycled five to seven times for use. If each student saves at least one sheet of paper in a day, we can save many trees in a year. We should save, reuse used paper and recycle it. By this we not only save trees but also save energy and water needed for manufacturing paper. Moreover, the amount of harmful chemicals used in paper making will also be reduced.
We can also save paper by using Information technology for storage of printed data, books in  digital form.


Question 11. Complete the word puzzle:
Down
1. Species on the verge of extinction.
2. A book carrying information about endangered species.
5. Consequence of deforestation.
Across
1. Species which have vanished.
3. Species found only in a particular habitat.
4. Variety of plants, animals and microorganisms found in an area.

Answer : 

EXTINCT-------
N-------------
D-------------
A-------------
N-R-----------
G-E-----------
ENDEMIC-------
R-B-----------
E-O-----------
D-O-----------
--K-----------
--BIODVERSITY-
-----E--------
-----S--------
-----E--------
-----R--------
-----T--------
-----S--------