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Sunday, October 18, 2015

9th(IX) Chemistry Solved Paper_Structure of Atom

Class 9 CBSE Test paper Solved Chapter 3: Structure of atoms -1
1. Q. Write the charge and mass of an electron.
Ans:
Particle       Symbol       Charge                        Mass
Electron         e–                 -1.60×10-19 C          9.1×10-31  kg

2. Q. List three subatomic particles of an atom. Compare them on the basis of relative mass and charge in a tabular form.
Ans: 

Particle
Symbol
Charge
Mass
Electron
e
-1.60×10-19 C
9.1×10-31  kg
Proton
p+  (H+)
1.60×10-19 C
1.672×10-27 kg
Neutron
n0
0.00 C
1.674×10-27 kg

3. Q. Find the number of electrons, protons and neutron possessed by the alpha particles used in the gold leaf experiment.

Ans: The number of electrons, protons and neutron are 0,2,2

4. Q. Draw Bohr’s model for an atom with
(i) valency = 1,       (ii) number of orbits = 3       (iii) mass number =23.
Also identify the element. What conclusion can be drawn about the reactivity of the element?  

Ans: Sodium has 11 protons and 12 neutrons.
So, The mass number is 11 + 12  =  23.
This element is metal and highly reactive

5. Q. (a) You are given an element. Find out     
(i) Number of protons, electrons and neutrons in ‘X’.   (ii) Valency of ‘X’.
(iii) Electronic Configuration of ‘X’.  
(b) If bromine atom is available in the form of, say two isotopes 35  (49.7%) and  (50.3%). Calculate the average atomic mass of bromine atom.  

Ans: (i) Number of protons, electrons and neutrons in ‘X’. are 7, 7 and 7 respectively
     (ii) Valency of ‘X’. is = 3 [NH3]
(iii) Electronic Configuration of ‘X’.  = 2, 5
(b) The average atomic mass of bromine atom =  +  = 80u

6. Q. (a) From Rutherford’s a - particle scattering experiment give the experimental evidence for deriving the conclusion that
(i) Most of the space inside the atom is empty.  
(ii) Whole mass of an atom is concentrated in its centre.
(iii) The nucleus of an atom is positively charged.

(b) An element has mass number 31 and atomic number 15 find :

(i) the number of neutrons in the element, and
(ii) the number of electrons in the outermost shell.

Ans: (a) 
(i) Most of the alpha particles passed through gold foil with getting deflected.
(ii) Very few particles were deflected from their path by 1800 indicating that whole mass of the atom is present in its centre.
(iii) Few particles deflected at small and large angle from their path indicating that centre is positively charged

(b) (i) the number of neutrons in the element = 31-15 = 16
(ii) Electronic configuration 2,8,5  then the number of electrons in the outermost shell. = 5

7. Q. Give reasons:
(a) Mass number of an atom excludes the mass of an electron.
(b) Nucleus of an atom is charged.
(c) Alpha particle scattering experiment was possible by using gold foil only and not by foil of any other metal.

Ans: (a) Mass number is the sum of the number of proton and neutron present in nucleus of atom therefore Mass number of an atom excludes the mass of an electron.
(b) Nucleus of an atom is charged as they contain positively charged proton.
(c) This is because because gold has high malleability and can be hammered easily into thin sheet.      `
8. Q. Laws of conservation of mass is not fully applicable for:
(a)Precipitation reaction            (b) Redox reaction      
(c) Nuclear reaction                  (d) Displacement reaction

Ans: (c) Nuclear reaction  

9. Q. (a) Define the following terms with one example each. (i) Isotope (ii) Isobar
(b) Name the elements whose isotopes are used in:
(i) Nuclear Reactor (ii) treatment of cancer. (iii) Treatment of cancer

Ans:(a) Isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers.
For example, hydrogen atom, it has three atomic species, namely Protium, deuterium and tritium. The atomic number of each one is 1, but the mass number is 1, 2 and 3, respectively.

Other such examples are carbon, C12, C14 and C16 where as chlorine, C35 and Cl37

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

For example calcium, atomic number 20, and argon, atomic number 18 but the mass number of both these elements is 40.

(b) (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.
(iii) An isotope of iodine is used in the treatment of goitre.

10.Q. (a) How many electrons are present in the valence shell of nitrogen, oxygen and argon ?
(b) Nucleus of an atom has 5 protons and 6 neutrons. What is the atomic number, mass number and electronic configuration of the atom?

Ans: (a) 5, 6 8 electrons are present in the valence shell of nitrogen, oxygen and argon respectively.
(b) Nucleus of an atom has 5 protons and 6 neutrons.
The atomic number = no of proton = 5, mass number = p + n = 5 + 6 = 11
This represents the element Boron
Electronic configuration of the atom of Boron = 2,3

11.Q. State the problem of atomic structure which was solved after the discovery of neutron.


Ans: when the neutron was not discovered, many scientists found that the atomic mass of many atoms was found to be double or more than double the mass of total number of protons (as the mass of electron was so small that it was assumed to be negligible), so by the discovery of neutron led to the solution of this problem. For e.g.­ the mass of Carbon­ 12 atom is 12 u but the no. of protons were 6 so it becomes 6u.

Thursday, October 15, 2015

CBSE BOARD PAPER QUESTIONS IX Atoms and molecules (Sample Test Paper 2016)

QUESTIONS FROM BOARD PAPER

Questions: 5 g of calcium combine with 2 g of oxygen to form a compound. Find the molecular formula of the compound. (Atomic mass of Ca = 40 u ; O =16 u)

Solution: Number of moles in 5g of calcium = mass / molar mass = (5 / 40) = 0.125

Number of moles in 2g of oxygen = mass / molar mass = (2 / 16) = 0.125

Now we will calculate the simplest ratio of the element by dividing the number of moles of each element by the smallest value.

Since number of moles of each element is 0.125, therefore calcium and oxygen are present in a ratio of 1 : 1. Thus the empirical formula of the compound is CaO.

For calculating the molecular formula, we need the molecular mass of the compound. However, because a compound with the formula CaO is known,

Therefore the molecular formula of the compound is CaO.

Questions: When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is produced.
What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.00 g of oxygen ?
Which law of chemical combination will govern your answer ? State the law.

Solution: The mass of the carbon dioxide formed will be only 11 grams. The remaining oxygen is not used up.

This indicates the law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.

Class 9 Atoms and molecules solved CBSE Test Papers New addition

Class 9 Atoms and molecules solved CBSE Test Paper-1

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Class 9 Atoms and molecules solved CBSE Test Paper-2

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Class 9 Atoms and molecules solved CBSE Test Paper-3

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Class 9 Atoms and molecules solved CBSE Test Paper-4

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Friday, August 8, 2014

Metal and Nonmetal class 10th Solved Questions



Question: Why is froth floatation process selected for concentration of the sulphide ore?

Ans. : Sulphide ore particles are welted by oil (pire oil) and gangue particles by water]

Question: Why cannot Al be reduced by carbon?

Ans. Aluminium is stronger reducing agent than carbon and therefore, cannot be reduced by it.

Question: Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Ans. The anode mud in the electrolytic refining of copper contains antimony, selenium tellurium, silver, gold and platinum. These are present as impurities in blister copper. They are less reactive and are not affected by CuSO4-H2SO4 solution and hence settle down near anode as anode mud.

Question: Which is the cheapest and most abundant reducing agent which is used in, the extraction of metals?

Ans. Carbon in the form of coke.

Question: What is the significance of leaching in the extraction of aluminium ?

Ans. The principal ore of aluminium is bauxite (Al203). It contains Si02, iron oxides, titanium oxide as impurities. The significance of leaching in the extraction of aluminium from bauxite is to remove the impurities from the ore.

Question: What is the role of graphite rod in the electrometallurgy of aluminium?

Ans. The graphite rod is useful in the electrometallurgy of aluminium for reduction of alumina to aluminium. ;  
    
 2Al203+ 3C → 4Al + 3C02

Question: Write chemical reactions taking place in the extraction of zinc from zinc blende.

Ans. (i) The concentrated zinc blende ore (ZnS) is roasted in the presence of excess air about
1200  to convert it to zinc oxide.

2ZnS                  +         3O2    →    2ZnO            +         2S02

Zinc blende                                        Zinc oxide
(ii) Zinc oxide is reduced to zinc by heating with crushed coke at 1673 K

ZnO + C →Zn + CO

(iii) The impure copper is refined by electro refining method. In this method, the impure zinc is made anode and a plate of pure zinc is made cathode in an electrolytic bath containing zinc sulphate and a small amount of  dilute H2SO4. On passing current, the following reaction occur:

At anode: Zn →  Zn2+ + 2e-

At cathode: Zn2+ + 2e- →Zn

The zinc gets deposited on cathode and is collected.   

Question: What are the different methods of the purification of metals

Answer: 
PURIFICATION OR REFINING OF METALS

The metal obtained by any of the above methods is impure and is known as crude metal. The process of purifying the crude metal is called refining. The method of refining depends upon the nature of the metal and the impurities which are present in the metal. Some of the methods generally applied for refining metals are discussed below.

Liquation: 

This process is used for refining the metals having low melting points. Such as tin, lead, bismuth. etc. this is based on the principle that the metal to be refined is easily fusible (melt easily) but the impurities do not fuse easily. In this process, the impure metal is placed on the sloping hearth of the furnace and is gently heated. The hearth is maintained at a temperature slightly above the melting point of the metal. The metal melts and flows down to the bottom of sloping hearth and the impurities are left behind. The pure metal is collected to the bottom of the sloping hearth. 

Distillation:
This method is used for the purification of volatile metals (which form vapours readily). Such as mercury and zinc. In this method the impure metal is heated strongly in a vessel (called retort). The pure metal distils over and its vapours  are condensed separately in a receiver to get pure metal. The non- volatile impurities are left behind in the retort.

Oxidation method (Oxidative refining):

This method is used for the refining of metal in those cases in the impurities have greater tendency to get oxidised than metal itself.
For example: Impure iron (Pig or cast iron) is refined oxidative refining method. Pig iron contains carbon, sulphur, phosphorous, silicon and manganese as impurities. When a blast of air is blown over molten pig iron these impurities are oxidised to their oxides (CO2, SO2, P2O5 etc.) and get removed. The pure iron is left behind. Similarly, Silver is refined by this method.

Electrolytic refining:

This is the most widely method for the refining of impure metals. Many metals such as copper, zinc, tin, nickel, silver, gold etc. are refined electrolytically. It is based upon the phenomenon of electrolysis, in this process, the impure metal is made of the anode and a thin strip of pure metal is made of the cathode. A solution of the metal salt is used as an electrolyte. On passing the electric current through the electrolyte. The pure metal from the anode dissolved into electrolyte. An equivalent amount of pure metal gets deposited on the cathode. The soluble impurities go into the solution, whereas, the impurities settle down at the bottom of the anode and are known as anode mud.

For example: In the electrolytic refining of copper (The apparatus is set up as shown in figure as below) crude copper is made the anode, a thin sheet of pure copper is made the cathode. The electrolyte is a solution of copper sulphate containing a small amount of dilute H2SO4 acid. On passing the electric current copper dissolve from the anode into the electrolyte. An equivalent amount of copper is deposited at the cathode in the from pure at metal. The following reactions occurs at the electrodes.

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