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NCERT Exemplar Class 12 Chemistry Solutions Chapter 6 Principles and Processes of isolation of elements

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6 Principles and Processes of isolation of elements

Edited By Sumit Saini | Updated on Sep 16, 2022 05:34 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 6 is important from an examination point of view. The professionals who have deep knowledge in this field have prepared NCERT exemplar Class 12 Chemistry chapter 6 solutions to provide the students with correct and precise information. The experts have used simple language to answer the questions to prepare NCERT exemplar Class 12 Chemistry chapter 6 solutions. The students usually study from this study material to get the learning experience that is best for them. NCERT exemplar Class 12 Chemistry solutions chapter 6 pdf download is very useful to access it offline.

This Story also Contains
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: MCQ (Type 1)
  2. Question:1
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: MCQ (Type 2)
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Short Answer Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Matching Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Assertion and Reason Type
  7. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Long Answer Type
  8. Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 6 Principles and Processes of Isolation of Elements
  9. NCERT Exemplar Class 12 Chemistry Solutions Chapter 6 Principles and Processes of Isolation of Elements
  10. NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: MCQ (Type 1)

Question:1

In the extraction of chlorine by electrolysis of brine ____________.
(i) oxidation of \text {Cl}^{-} ion to chlorine gas occurs.
(ii) reduction of \text {Cl}^{-} ion to chlorine gas occurs.
(iii) For the overall reaction, \Delta \text {G}^{\ominus } has a negative value.
(iv) a displacement reaction takes place.
Answer:

The answer is the option (i). The oxidation reaction of \text {Cl}^{-} ion is responsible for the extraction of chlorine from brine
2H_{2}O(l)+2Cl^{-}(aq)\rightarrow H_{2}(g)+Cl_{2}(g)+2OH^{-}(aq)
The \Delta G^{0} for this reaction is +422\; kJ
\Delta G^{0}=-nFE^{0}
E^{0}=\frac{\Delta G^{0}}{nF}=-2.2 V

Question:2

When copper ore is mixed with silica, in a reverberatory furnace copper matte is produced. The copper matte contains ____________.
(i) sulphides of copper (II) and iron (II)
(ii) sulphides of copper (II) and iron (III)
(iii) sulphides of copper (I) and iron (II)
(iv) sulphides of copper (I) and iron (III)
Answer:

The answer is the option (iii). Inside a reverberatory furnace iron silicate on mixing copper ore with silica in a reverberatory furnace, iron silicate comes off as slag producing copper in the form of copper matte containing copper sulphide [Cu_{2}S(I)] and some iron sulphide [FeS(II)].

Question:3

Which of the following reactions is an example of autoreduction ?
(i) \text {Fe}_{3}\text {O}_{4}+\text {4CO}\rightarrow \text {3Fe}+\text {4CO}_{2}
(ii) \text {Cu}_{2}\text {O}+\text {C}\rightarrow \text {2Cu}+\text {CO}
(iii) \text {Cu}^{2+}(aq)+Fe (s)\rightarrow \text {Cu (s)}+Fe^{2+}(aq)
(iv) \text {Cu}_{2}\text {O}+\frac{1}{2}\text {Cu}_{2}\text {S}\rightarrow \text {3Cu }+\frac{1}{2}\text {SO}_{2}
Answer:

\text {Cu}_{2}\text {O}+\frac{1}{2}\text {Cu}_{2}\text {S}\rightarrow \text {3Cu }+\frac{1}{2}\text {SO}_{2}
In this reaction reduction of copper (I) oxide by copper (I) sulphide takes place. Since, copper is reduced by itself, this process is known as auto-reduction and the solidified copper so, obtained is known as blister copper. We get the copper metal on reaction of sulphide ore of copper with its oxide.

Question:4

A number of elements are available in earth’s crust but most abundant elements are ____________.
(i) \text {Al and Fe}
(ii) \text {Al and Cu}
(iii) \text {Fe and Cu}
(iv) \text {Cu and Ag}
Answer:

The answer is the option (i). A number of elements are available in the earth’s crust but the most abundant elements in the earth’s crust include \text {Al and Fe}.

Question:5

Zone refining is based on the principle that ___________.
(i) impurities of low boiling metals can be separated by distillation.
(ii) impurities are more soluble in molten metal than in solid metal.
(iii) different components of a mixture are differently adsorbed on an absorbent.
(iv) vapours of the volatile compound can be decomposed in pure metal.
Answer:

The answer is the option (ii). The basic principle of zone refining method is that impurities are more soluble in molten metal compared to the solid state of the metal.

Question:6

In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu_{2}O with
(i) \; FeS
(ii) \; CO
(iii) \; Cu_{2}S
(iv) \; SO_{2}
Answer:

The answer is the option (iii). In the process of extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu_{2}O with Cu_{2}S
Since, copper is reduced by itself the process is known as autoreduction.
Chemical reaction occuring in this reaction is Cu_2 O+\frac{1}{2} Cu_2 S\rightarrow 3Cu+\frac{1}{2} SO_2
In this process, copper appears as blister copper.

Question:7

Brine is electrolysed by using inert electrodes. The reaction at anode is ________.
(i) Cl^{-}(aq.)\rightarrow \frac{1}{2}Cl_{2}(g)+e^{-};E^{o}cell=1.36 \; V
(ii) 2H_{2}O(l)\rightarrow O_{2}(g)+4H^{+}+4e^{-};E^{o}cell=1.23 \; V
(iii) Na^{+}(aq)+e^{-}\rightarrow Na (s);E^{0}cell=2.71\; V
(iv) H^{+}(aq)+e^{-}\rightarrow \frac{1}{2}H_{2} (g);E^{0}cell=0.00\; V

Answer:

The answer is the option (i). Electrolysis of brine solution is a process of decomposition of NaCl solution using the electric current. Anode is positive; therefore it attracts negative hydroxide and chlorine ions. Chloride ions get oxidized by electron loss to chlorine molecules.

Question:8

In the metallurgy of aluminium ________________.
(i) Al^{3+} is oxidised to Al (s).
(ii) graphite anode is oxidised to carbon monoxide and carbon dioxide.
(iii) the oxidation state of oxygen changes in the reaction at the anode.
(iv) the oxidation state of oxygen changes in the overall reaction involved in the process
Answer:

The answer is the option (ii). The graphite anode is oxidized to carbon monoxide (CO) and carbon dioxide (CO_{2}) in the process of metallurgy of aluminium .
C(s)+O^{2-}(melt)\rightarrow CO(g)+2e^{-}
C(s)+2O^{2-}(melt)\rightarrow CO_{2}(g)+4e^{-}

Question:9

Electrolytic refining is used to purify which of the following metals?
(i) \; Cu \; and\; Zn
(ii) \; Ge \; and \; Si
(iii)\; Zr \; and\; Ti
(iv) \; Zn \; and\; Hg
Answer:

The answer is the option (i). Copper and zinc can be purified by using electrolyte refining method. In this process, impure metal acts as anode and a strip of pure metal is used as a cathode. Impurities from the blister copper or impure zinc deposit are used as anode mud. The electrolytic refining of copper and zinc is done to get high grade metals.

Question:10

Extraction of gold and silver involves leaching the metal with CN^- ion. The metal is recovered by ________________.
(i) displacement of metal by some other metal from the complex ion.
(ii) roasting of the metal complex.
(iii) calcination followed by roasting.
(iv) thermal decomposition of the metal complex.
Answer:

The answer is the option (i). Extraction of gold and silver involves leaching with CN- ion. The metal in cyanide process is recovered by displacement of metal by some other metal from the complex ion. The reaction taking place is as:
4M+8CN^{-}+2H_{2}O+O_{2}\rightarrow 4[M(CN)_{2}]^{-}+4OH^{-}
2[M(CN)_{2}]^{-}+Zn\rightarrow [Zn(CN)_{4}]^{2-}+2M

Question:11

Answer the questions 11-13 on the basis of Fig. 6.1.

Choose the correct option of temperature at which carbon reduces \text {FeO} to iron and produces \text {CO}.
(i) Below temperature at point A.
(ii) Approximately at the temperature corresponding to point A.
(iii) Above temperature at point A but below a temperature at point D.
(iv) Above temperature at point A.
Answer:

The answer is the option (iv) \text {FeO+C}\rightarrow \text {Fe+CO}
It can be seen as a couple of two simpler reactions
\text {FeO}\rightarrow \text {Fe+}\frac{1}{2}\text {O}_{2}(\Delta \text {G}^{0}_{FeO.Fe})
\text {C+}\frac{1}{2}\text {O}_{2}\rightarrow \text {CO}(\Delta \text {G}^{0}_{C.CO})
Total Gibbs energy change becomes \Delta \text {G}^{0}_{C.CO}+\Delta\text {G}_{FeO.Fe}^{0}
In \Delta \text {G}^{0} versus T plot, the plot \text {Fe} to \text {FeO} goes upward and the plot for \text {C} to \text {CO} goes downwards.
At temperature above point A, the \text {C} to \text {CO} lines comes below \text {Fe} to \text {FeO} and \Delta \text {G}^{0}_{C,CO} < \Delta \text {G}^{0}_{Fe, FeO}
So in this range, \text {C} will reduce \text {FeO} to \text {Fe} and itself be oxidised to \text {CO}.

Question:12

Below point ‘A’ \text {FeO} can ______________.
(i) be reduced by carbon monoxide only.
(ii) be reduced by both carbon monoxide and carbon.
(iii) be reduced by carbon only.
(iv) not be reduced by both carbon and carbon monoxide.
Answer:

The answer is the option (i). Below point A, \text {FeO} can be reduced by carbon monoxide only.

Question:13

For the reduction of \text {FeO} at the temperature corresponding to point D, which of the following statements is correct?
(i) \Delta \text {G} value for the overall reduction reaction with carbon monoxide is zero.
(ii) \Delta \text {G} value for the overall reduction reaction with a mixture of 1 mol carbon and 1 mol oxygen is positive.
(iii) \Delta \text {G} value for the overall reduction reaction with a mixture of 2 mol carbon and 1 mol oxygen will be positive.
(iv) \Delta \text {G} value for the overall reduction reaction with carbon monoxide is negative.
Answer:

The answer is the option (i). At point D, the curves of oxidation of iron and oxidation of \text {CO} intersect each other.
Thus, \Delta \text {G}^{0} for the reaction \text {FeO+CO}\rightarrow \text {Fe+CO}_{2} is zero.
\text {2FeO+O}_{2}\rightarrow \text {2FeO};\Delta \text {G}^{0}=-280
\text {2CO+O}_{2}\rightarrow \text {2CO};\Delta \text {G}^{0}=-280
Resulting in \text {2FeO+2CO}_{2}\rightarrow \text {2Fe+2CO}_{2}
\Delta \text {G}^{0}= (-280+280)=0

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: MCQ (Type 2)

Question:14

At the temperature corresponding to which of the points in Fig.6.1, \text {FeO} will be reduced to \text {Fe} by coupling the reaction \text {2FeO}\rightarrow \text {2Fe+O}_{2} with all of the following reactions?
(i) \text {C+O}_{2}\rightarrow \text {CO}_{2}
(ii) \text {2C+O}_{2}\rightarrow \text {2CO} and
(iii) \text {2CO+O}_{2}\rightarrow \text {2CO} _{2}
(i) Point A
(ii) Point B
(iii) Point D
(iv) Point E
Answer:

The answer is the option (ii, iv) The reduction reaction ranges can be determined from the intersection points on \Delta \text {G}^{0} versus T plots.
The reduction of \text {FeO} to \text {Fe} takes place at point B and E because at these points \Delta \text {G}^{0} value of all three oxidation reactions is lesser than that of reduction reaction.

Question:15

Which of the following option are correct?
(i) Cast iron is obtained by remelting pig iron with scrap iron and coke using hot air blast.
(ii) In the extraction of silver, silver is extracted as cationic complex.
(iii) Nickel is purified by zone refining.
(iv) \text {Zr and Ti} and are purified by van Arkel method
Answer:

The answer is the option (i, iv) Iron obtained from blast containing about 4% carbon and many impurities such as Mn, P, Si, etc is called pig iron. It is remelted with scrap iron to obtain cast iron which has slightly less carbon content (3%) compared to pig iron and is extremely hard and brittle.
For refining of \text {Zr and Ti} Van Arkel method based on the thermal decomposition of metal components is used. In this method, the impurities i.e. oxygen and nitrogen are removed by heating metal in an evacuated vessel with iodine.
\text {Zr (s)}+\text {2 I}_{2}\overset{\text {280 k}}{\rightarrow}\text {ZrI}_{4}\text {(g)} \overset{\text {2075 k}}{\rightarrow}\text {Zr (s)}\text {(pure)}+\text {2 I}_{2}
\text {Ti}+\text {2I}_{2}\overset{\text {530 k}}{\rightarrow}\text {TI}_{4}\overset{\text {1700 k}}{\rightarrow}\text {Ti}+\text {2I}_{2}
Zone refining is used for semiconductors like Ge, Si. For purification of Nickel, Mond’s process is used.

Question:16

In the extraction of aluminium by Hall-Heroult process, purified \text {Al}_{2}\text {O}_{3} is mixed with \text {CaF}_{2} to
(i) lower the melting point of \text {Al}_{2}\text {O}_{3}
(ii) increase the conductivity of molten mixture.
(iii) reduce \text {Al}^{3+} into Al(s).
(iv) acts as a catalyst
Answer:

The answer is the option (i, ii) Purified \text {Al}_{2}\text {O}_{3} is mixed with \text {CaF}_{2} in Hall-Heroult process to lower the melting point of \text {Al}_{2}\text {O}_{3} and increase the conductivity of the molten mixture. Hence, option a and b are correct.

Question:17

Which of the following statements is correct about the role of substances added in the froth floatation process?
(i) Collectors enhance the non-wettability of the mineral particles.
(ii) Collectors enhance the wettability of gangue particles.
(iii) By using depressants in the process two sulphide ores can be separated.
(iv) Froth stabilisers decrease wettability of gangue.
Answer:

The answer is the option (i, iii) To enhance the non-wettability of mineral particles as well as froth stabilisers, collectors like sodium ethyl xanthate are used.
For example, sodium cyanide acts as a depressant for zinc sulphide in the process of concentration of galena ore. Depressants can be used for the separation of two sulphide ores.
Froth stabilizers are used to stabilize the froth in the process.

Question:18

In the Froth Floatation process, zinc sulphide and lead sulphide can be separated by ______________.
(i) using collectors.
(ii) adjusting the proportion of oil to water.
(iii) using depressant.
(iv) using froth stabilisers.
Answer:

The answer is the option (ii, iii) Froth floatation method is used in extraction of metal from sulphide ore. The two sulphide ores can be separated by using depressant and adjusting the proportion of oil to water in the froth floatation process. Depressant used in separation of zinc sulphide and lead sulphide is NaCN. It is utilized in selective prevention of ZnS from coming to the froth.

Question:19

Common impurities present in bauxite are ____________.
(i) \text {CuO}
(ii) \text {ZnO}
(iii) \text {Fe}_{2}\text {O}_{3}
(iv) \text {SiO}_{2}
Answer:

The answer is the option (iii, iv) some of the major impurities in bauxite are iron oxides (goethite and hematite), silicon dioxide, kaolinite and small amounts of anatase.

Question:20

Which of the following ores are concentrated by froth flotation?
(i) Haematite
(ii) Galena
(iii) Copper pyrites
(iv) Magnetite
Answer:

The answer is the option (ii, iii) Sulphide ores are concentrated by froth floatation process as pine oil selectively wets the sulphide ore. Galena and copper pyrite, being sulphide ores are concentrated by froth floatation process.

Question:21

Which of the following reactions occur during calcination?
(i) \text {CaCO}_{3}\rightarrow \text {CaO}+ \text {CO}_{2}
(ii) \text {2FeS}_{2}+\frac{11}{2}\text {O}_{2}\rightarrow \text {Fe}_{2}\text {O}_{3}+ \text {4SO}_{2}
(iii) \text {Al}_{2}\text {O}_{3}.x\text {H}_{2}\text {O}\rightarrow \text {Al}_{2}\text {O}_{3}+ x \text {H}_{2}\text {O}
(iv) \text {ZnS}+\frac{3}{\text {2}}O_{2}\rightarrow \text {ZnO}+\text {SO}_{2}
Answer:

The answer is the option (i, iii). In the process of calcination, heating of the ore below its melting in limited supply of air takes place. Ores that contain oxygen like oxide, hydroxides or carbonates are calcined. The reaction that takes place is:
\text {CaCO}_{3}\overset{\Delta }{\rightarrow}\text {CaO}+\text {CO}_{2}
\text {Al}_{2}\text {O}_{3}.\text {xH}_{2}\text {O}\overset{\Delta }{\rightarrow}\text {Al}_{2}\text {O}_{3}+ \text {xH}_{2}\text {O}

Question:22

For the metallurgical process of which of the ores calcined ore can be reduced by carbon?
(i) haematite
(ii) calamine
(iii) iron pyrites
(iv) sphalerite
Answer:

The answer is the option (i, ii). Oxides reduction can be done by using carbon. Haematite (Iron ore - \text {Fe}_{2}\text {O}_{3} ) and calamine \text {ZnCO}_{3}\rightarrow \text {ZnO}+\text {CO}_{2} contain oxygen, so they can be reduced by carbon.
Sulphide ores cannot be reduced by carbon. Iron pyrite and sphalerite are sulphide ores of iron and zinc respectively.

Question:23

The main reactions occurring in the blast furnace during the extraction of iron from haematite are ________.
(i) \text {Fe}_{2}\text {O}_{3}+\text {3CO}\rightarrow \text {2Fe}+\text {3CO}_{2}
(ii) \text {FeO}+\text {SiO}_{2}\rightarrow \text {FeSiO}_{3}
(iii) \text {Fe}_{2}\text {O}_{3}+\text {3C}\rightarrow \text {2Fe}+\text {3CO}
(iv) \text {CaO}+\text {SiO}_{2}\rightarrow \text {CaSiO}_{3}
Answer:

The answer is the option (i, iv). The 2 main reactions taking place in blast furnace during extraction of iron are:
(i) Reduction of \text {Fe}_{2}\text {O}_{3}+\text {CO}\rightarrow \text {2FeO}+\text {CO}_{2}
\text {2FeO}+\text {2CO}\rightarrow \text {2Fe}+\text {2CO}_{2}
\text {Fe}_{2}\text {O}_{3}+\text {3CO}\rightarrow \text {2Fe}+\text {3CO}_{2}
(ii) Slag formation \text {CaO}+\text {SiO}_{2}\rightarrow \text {CaSiO}_{3} \; \text {Fusible slag}

Question:24

In which of the following method of purification, metal is converted to its volatile compound which is decomposed to give pure metal?
(i) heating with a stream of carbon monoxide.
(ii) heating with iodine.
(iii) liquation.
(iv) distillation.
Answer:

The answer is the option (i,ii)
(i) Heating of metal with stream of \text {CO}
\text {Ni}+\text {4CO}\rightarrow \text {Ni}\text {(CO)}_{4}\overset{450-470\text {k}}{\rightarrow}\text {Ni+4CO}\; \; \text {Mond's Process}
(ii) Heating with Iodine
\text {Zr+2I}_{2}\overset{870\text {k}}{\rightarrow}\text {ZrI}_{4}\overset{2075\; \text {K}(Tungsten filament)}{\rightarrow}\text {Zr+2I}_{2}\text {(van arkel method)}

Question:25

Which of the following statements are correct?
(i) A depressant prevents a certain type of particle to come to the froth.
(ii) Copper matte contains \text {Cu}_{2}\text {S} and \text {ZnS}
(iii) The solidified copper obtained from the reverberatory furnace has blistered appearance due to evolution of \text {SO}_{2} during the extraction.
(iv) Zinc can be extracted by self-reduction.
Answer:

The answer is the option (i, iii). Frothing of a particular sulphide can be prevented by depressants. The solidified copper that we get in reverberatory furnace has blistered appearance due to evolution of \text {SO}_{2} during extraction. Hence, it is called blister copper.

Question:26

In the extraction of chlorine from brine _____________.
(i) \Delta \text {G}^{0} for the overall reaction is negative.
(ii) \Delta \text {G}^{0} for the overall reaction is positive.
(iii) \text {E}^{0} for the overall reaction has negative value.
(iv) \text {E}^{0} for the overall reaction has positive value.
Answer:

The answeris the option (ii,iii). 3Cl^{-}+2H_{2}O\rightarrow 2OH^{-}+H_{2}+Cl_{2}
\Delta G^{0} for the reaction is +422\; kJ
Using \Delta G^{0}=-nFE^{0}, we get E^{0}=-2.2\; V

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Short Answer Type

Question:27

Why is an external emf of more than 2.2V required for the extraction of Cl_{2} from brine?
Answer:

In the process of extraction of Cl_{2} from brine solution :
2Cl^{-}(aq)+2H_{2}O(l)\rightarrow 2OH^{-}(aq)+H_{2}(g)+Cl_{2}(g) the value of \Delta G^{\Theta } is +ve \; (422\; KJ).
For this the value of E^{\Theta } is \Delta G^{\Theta }=-nFE^{\Theta } or E^{\Theta }=-\frac{\Delta G^{\Theta }}{nF}=-\frac{422\times 10^{3}J}{2\times 96500}=-2.20\; V
Hence, An external emf more than 2.20 V is required for extraction of Cl_{2} from brine solution for electrolysis process.

Question:28

At temperatures above 1073K coke can be used to reduce FeO\; to \; Fe. How can you justify this reduction with the Ellingham diagram?
Answer:

As per Ellingham diagram at temperature greater than 1073 K, \Delta G^{0} for the formation FeO is less negative than \Delta G^{0} for the formation of carbon monoxide from carbon.
Thus, above 1073 K, carbon can reduce FeO, i.e., \Delta_{r} G^{0} for the reaction
FeO+C\overset{>1073}{\rightarrow}Fe+CO\; \text {is negative}

Question:29

Wrought iron is the purest form of iron. Write a reaction used for the preparation of wrought iron from cast iron. How can the impurities of sulphur, silicon and phosphorus be removed from cast iron?
Answer:

Wrought iron is an alloy with a very low carbon content (less than 0.08%).
The reaction is:
Fe_{2}O_{3}+3C\rightarrow 2Fe+3CO

In extraction of cast iron from haematite ore, on addition of limestone as flux, the impurities of sulphur, silicon and phosphorus pass into slag on changing to their oxides.

Question:30

How is copper extracted from low-grade copper ores?
Answer:

Copper is extracted from low grade ores by the process of hydrometallurgy. Acid or bacteria is used in its leaching process. In the process, ore is treated with a suitable regent that dissolves ore but not the impurities. Scrap iron, zinc or H_{2} is used to treat the solution containing copper ions (Cu^{+2}) and the reactions are as:
Cu^{2+} (aq)+H_{2} (g) \rightarrow Cu(s)+2H^{+} (aq)
Cu^{2+} (aq)+Fe (s) \rightarrow Cu(s)+Fe^{2+} (aq)
In this way, copper, is obtained.

Question:31

Write two basic requirements for refining of metal by Mond process and by Van Arkel Method
Answer:

Mond’s process and van arkel method are used for refining of nickel and zinc, respectively. The basic requirements for both the processes are:
(i) A volatile compound should be formed by the metal on reaction with an available reagent.
(ii) The volatile compound formed should be easily decomposable, enabling the metal to be easily recovered.

Question:32

Although carbon and hydrogen are better-reducing agents they are not used to reduce metallic oxides at high temperatures. Why?
Answer:

Despite being better reducing agents, carbon and hydrogen are not used as carbon and hydrogen form carbides and hydrides respectively on reaction with metals at high temperature.

Question:33

How do we separate two sulphide ores by Froth Floatation Method? Explain with an example.
Answer:

The separation of two sulphide ores is done by adjusting proportion of oil to water or by using depressants. NaCN is used for an ore containing ZnS and PbS. Sodium cyanide forms complex with ZnS and prevents it from coming with froth but lead sulphide remains with froth.
4NaCN+ZnS\rightarrow Na_{2}Zn(CN)_{4}+Na_{2}S

Question:34

The purest form of iron is prepared by oxidising impurities from cast iron in a reverberatory furnace. Which iron ore is used to line the furnace? Explain by giving a reaction.
Answer:

Haematite (Fe_{2}O_{3}) is used to line the furnace. It supplies the oxygen and oxidises impurities present in the cast iron like carbon, silicon, manganese and phosphorus to CO, SiO_{2}, MnO and P_{2}O_{5}.
Fe_{2}O_{3}+3C\overset{Heat}{\rightarrow}2Fe+3CO
3Si+2Fe_{2} O_3\rightarrow 4Fe+3SiO_2
3S+2Fe_2 O_3\rightarrow 3SO_2+4Fe
CO and SO_{2} escape and MnO \; and\; SiO_{2} combine to form to form manganous silicate (MnSiO_{3}) as slag MnO+SiO_{2}\overset{Heat}{\rightarrow}MnSiO_{3} (Manganous silicate-slag)

Question:35

The mixture of compounds A and B is passed through a column of Al_{2}O_{3} by using alcohol as eluant. Compound A is eluted in preference to compound B. Which of the compounds A or B, is more readily adsorbed on the column?
Answer:

Compound B is more readily adsorbed on the column as compound A is eluted in preference to compound B

Question:36

Why is sulphide ore of copper heated in a furnace after mixing with silica?
Answer:

Iron oxide that is present as impurity in sulphide ore of copper forms an iron silicate slag and copper is produced in the form of copper matte. This is why, sulphide ore of copper is heated in a furnace after mixing with silica.
FeO+SiO_{2}\rightarrow FeSiO_{3}

Question:37

Why are sulphide ores converted to oxide before reduction?
Answer:

Sulphides are not reduced easily unlike the oxides. This is why, sulphides are first oxidised and then subjected to reduction.

Question:38

Which method is used for refining Zr \; and\; Ti ? Explain with equation.
Answer:

Van Arkel method is used in refining of Zr \; and\; Ti . The crude metal, on heating with iodine forms volatile unstable compound. Pure metal is then obtained by decomposing the compound.
The equations are : \text {Zr (s) (impure)+2I}_{2}\overset{870\; K}{\rightarrow}\text {ZrI}_{4}
\text {ZrI}_{4}\overset{1800\; K}{\rightarrow}\text {Zr (pure)+2I}_{2}
\text {Ti (impure)+2I}_{2}\overset{523\; K}{\rightarrow}\text {TiI}_{4} \text {(g)}
\text {TiI}_{4}\overset{1700\; K}{\rightarrow}\text {Ti (pure)+2I}_{2}

Question:39

What should be the considerations during the extraction of metals by electrochemical method?
Answer:

During the extraction of metals by electrochemical method we should keep in mind the following factors
(i) Reactivity of the metal.
(ii) Suitability of the electrode.

Question:40

What is the role of flux in metallurgical processes?
Answer:

Flux is a chemical cleaning agent.
The ores contain some gangue even after concentration. Certain substances are mixed with concentrated ore for the purpose of removing the gangue. They combine with the gangue to form a fusible material that is insoluble in molten metal. These substances are called fluxes. Some examples of materials used as flux: Limestone, silica, dolomite, lime, borax etc.
Fluxes are classified as acidic or basic based on the type of impurities they remove.
SiO_{2}\; (Acidic\; acid)+CaO\; (basic\; impurity)\rightarrow CaSiO_{3}(Slag)
MgO\; (basic\; flux)+SiO_{2}\; (Acidic\; impurity)\rightarrow MgSiO_{3}(Slag)

Question:41

How are metals used as semiconductors refined? What is the principle of the method used?
Answer:

Semiconductor like silicon and germanium is produced by zone refining method and the principle used is that the impurities are more soluble in melt than the corresponding solid state of metals.

Question:42

Write down the reactions taking place in Blast furnace-related to the metallurgy of iron in the temperature range 500-800K.

Answer:

The reactions that take place in blast furnace during the metallurgy of iron at the temperature range 500-800 K are:
3Fe_{2}O_{3}+CO\rightarrow 2Fe_{3}O_{4}+CO_{2}
Fe_{3}O_{4}+4CO\rightarrow 3Fe+4CO_{2}
Fe_{2}O_{3}+CO\rightarrow 2FeO+CO_{2}

Question:43

Give two requirements for vapour phase refining.
Answer:

The two requirements for vapour phase refining are:
(i) A volatile compound should be formed by the metal on reaction with an available reagent.
(ii) The volatile compound formed should be easily decomposable, enabling the metal to be easily recovered.

Question:44

Write the chemical reactions involved in the extraction of gold by the cyanide process. Also, give the role of zinc in the extraction.
Answer:

In extraction of gold by cyanide process, gold particles present in the ore are treated with a dilute solution of NaCN in the presence of atmospheric oxygen. Gold particles dissolve by forming complex cyanide that is soluble in nature.
4Au+8NaCN+2H_{2}O+O_{2}\rightarrow 4Na[Au(CN)_{2}]+4NaOH
On adding electropositive metal like zinc Au is recovered from solution.
2Na [Au(CN)_{2}]+Zn\rightarrow Na_{2}[Zn(CN)_{4}]+2Au
The process of extraction of gold by cyanide oxidation as well as reduction. oxidation of gold takes place during the formation of cyanide complex whereas reduction takes place during its recovery. Zn acts as a reducing agent.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Matching Type

Question:45

Match the items of Column I with items of Column II and assign the correct code:

Column I

Column II

(A) Pendulum

(1) Chrome steel

(B) Malachite

(2) Nickel steel

(C) Calamine

(3) Na_{3}AlF_{6}

(D) Cryolite

(4) CuCO_{3}.Cu(OH)_{2}



(5) ZnCO_{3}

(i) A (1) B (2) C (3) D (4)
(ii) A (2) B (4) C (5) D (3)
(iii) A (2) B (3) C (4) D (5)
(iv) A (4) B (5) C (3) D (2)

Answer:

Option (ii) is the answer.

Question:46

Match the items of Column I with the items of Column II and assign the correct code :

Column I

Column II

(A) Coloured bands

(1) Zone refining

(B) Impure metal to volatile complex

(2) Fractional distillation

(C) Purification of Ge \; and \; Si

(3) Mond Process

(D) Purification of mercury

(4) Chromatography



(5) Liquation

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (1) D (2)
(iii) A (3) B (4) C (2) D (1)
(iv) A (5) B (4) C (3) D (2)

Answer:

Option (ii) is the answer.

Question:47

Match items of Column I with the items of Column II and assign the correct code :

Column I

Column II

(A) Cyanide process

(1) Ultrapure Ge

(B) Froth Floatation Process

(2) Dressing of ZnS

(C) Electrolytic reduction

(3) Extraction of Al

(D) Zone refining

(4) Extraction of Au



(5) Purification of Ni

(i) A (4) B (2) C (3) D (1)
(ii) A (2) B (3) C (1) D (5)
(iii) A (1) B (2) C (3) D (4)
(iv) A (3) B (4) C (5) D (1)

Answer:

Option (i) is the answer.

Question:48

Match the items of Column I with the items of Column II and assign the correct code:

Column I

Column II

(A) Sapphire

(1) \text {Al}_{2}\text {O}_{3}

(B) Sphalerite

(2) \text {NaCN}

(C) Depressant

(3) \text {Co}

(D) Corundum

(4) \text {ZnS}



(5) \text {Fe}_{2}\text {O}_{3}

(i) A (3) B (4) C (2) D (1)
(ii) A (5) B (4) C (3) D (2)
(iii) A (2) B (3) C (4) D (5)
(iv) A (1) B (2) C (3) D (4)

Answer:

Option (i) is the answer.


Question:49

Match the items of Column I with items of Column II and assign the correct code :

Column I

Column II

(A) Blistered Cu

(1) Aluminium

(B) Blast furnace

(2) 2Cu_{2}O+Cu_{2}S\rightarrow 6Cu+SO_{2}

(C) Reverberatory furnace

(3) Iron

(D) Hall-Heroult proess

(4) FeO+SiO_{2}\rightarrow FeSiO_{3}



(5) 2Cu_{2}S+3O_{2}\rightarrow 2Cu_{2}O+2SO_{2}

(i) A (2) B (3) C (4) D (1)
(ii) A (1) B (2) C (3) D (5)
(iii) A (5) B (4) C (3) D (2)
(iv) A (4) B (5) C (3) D (2)
Answer:

Option (i) is the answer.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Assertion and Reason Type

Question:50

In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Nickel can be purified by Mond process.
Reason:Ni(CO)_{4} is a volatile compound which decomposes at 460K to give pure Ni.
(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true but the reason is not the correct explanation of assertion.
(iii) The assertion is true but the reason is false.
(iv) The assertion is false but the reason is true.
(v) Assertion and reason both are wrong.
Answer:

The answer is the option (i). For Mond’s process, the pre-requisites include:
(i) A volatile compound should be formed by the metal on reaction with an available reagent.
(ii) The volatile compound formed should be easily decomposable, enabling the metal to be easily recovered.
Ni(CO)_{4} is a volatile compound which decomposes to give pure nickel on heating at 460 K while CO (Carbon monoxide) is removed as gas.

Question:51

In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Zirconium can be purified by Van Arkel method.
Reason: ZrI_{4} is volatile and decomposes at 1800K.
(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true but the reason is not the correct explanation of assertion.
(iii) The assertion is true but the reason is false.
(iv) The assertion is false but the reason is true.
(v) Assertion and reason both are wrong.
Answer:

The answer is the option (i). Zirconium is also purified by Van Arkel method as it forms ZrI_{4} on treating with iodine and is easily decomposable to get pure zirconium.

Question:52

In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Sulphide ores are concentrated by Froth Flotation method.
Reason: Cresols stabilise the froth in the Froth Flotation Method.
(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true but the reason is not the correct explanation of assertion.
(iii) The assertion is true but the reason is false.
(iv) The assertion is false but the reason is true.
(v) Assertion and reason both are wrong.
Answer:

The answer is the option (ii). Sulphide ores on being wetted by oil, becomes lighter and rises to the surface along with froths. On the other hand, impurities wetted by water settle down as they get heavier.

Question:53

In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Zone refining method is very useful for producing semiconductors.
Reason: Semiconductors are of high purity.
(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true but the reason is not the correct explanation of assertion.
(iii) The assertion is true but the reason is false.
(iv) The assertion is false but the reason is true.
(v) Assertion and reason both are wrong.
Answer:

The answer is the option (ii).The principle of the zone refining method is that the the impurities of semiconductors are more soluble in the molten zone compared to the solid state and the ultrapure semiconductor crystallizes.

Question:54

In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal.
Reason: Copper is extracted by hydrometallurgy.
(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true but the reason is not the correct explanation of assertion.
(iii) The assertion is true but the reason is false.
(iv) The assertion is false but the reason is true.
(v) Assertion and reason both are wrong.
Answer:

The answer is the option (ii). Copper is extracted by hydrometallurgical process. Suitable solvent like water is used for dissolving salts of metal. They are then reduced by more electropositive element in this process.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 6: Long Answer Type

Question:55

Explain the following:

(i) CO_{2} is a better reducing agent below 710K whereas CO is a better reducing agent above 710K.
(ii) Generally sulphide ores are converted into oxides before reduction.
(iii) Silica is added to the sulphide ore of copper in the reverberatory furnace.
(iv) Carbon and hydrogen are not used as reducing agents at high temperatures.
(v) Vapour phase refining method is used for the purification of Ti.
Answer:

(i) As per the Ellingham diagram, below 710 K
\Delta G_{C,CO_{2}}^{0}<\Delta G_{C,CO}^{0}
Hence, CO_{2} is better reducing agent compared to CO.
At temperature above 710 K, \Delta G_{C,CO_{2}}^{0}<\Delta G_{C,CO}^{0}
Hence CO is better reducing agent.
(ii) Sulphide ores cannot be reduced easily but oxide ores can be esily reduced compared to them, hence sulphide ores are generally converted into oxides before reduction
(iii) In addition to copper sulphide, copper pyrites contain iron sulphide. In reverberatory furnace copper ore is roasted to give oxides.
FeO is removed by adding silica from the matte containing copper sulphide and some iron sulphide.
2FeS+3O_{2}\rightarrow 2FeO+2SO_{2}
FeO+SiO_{2}\rightarrow FeSiO_{3}
(iv) Carbon and hydrogen react with metals at high temperature to form carbides and hydrides respectively hence they are not used as reducing agents.
(v) Ti reacts with iodine to form TiI_{4} which is volatile and decomposes to give Ti at high temperature to give extra pure titanium. That is why vapour phase refining method is used for purification of Ti
Ti + 2I_{2}\overset{530 K}{\rightarrow}TiI_{4}\overset{1800 K}{\rightarrow}Ti+2I_{2}

Also, check NCERT solutions for class 12, other subjects

Introduction of NCERT exemplar Class 12 Chemistry solutions chapter 6 Principles and Processes of isolation of elements

The GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS is very important as it deals with general principles and process of isolation of elements, which deals with metallurgy and oxidation. The extraction and isolation of metals from ores involve major steps that are included in this chapter.

As per NCERT exemplar Class 12 Chemistry solutions chapter 6, the chapter deals with learning about the different metals, their principles, oxidation-reduction, and refining. There are different topics in this chapter that will help you to understand the concept of metallurgy. The chapter will help the students in understanding the working of different elements related to metallurgy and how it is used.

Major Subtopics in NCERT Exemplar Class 12 Chemistry Solutions Chapter 6 Principles and Processes of Isolation of Elements

  • Occurrence of Metals
  • Concentration of Ores
  • Extraction of Crude Metal from Concentrated Ore
  • Thermodynamic Principles of Metallurgy
  • Electrochemical Principles of Metallurgy
  • Oxidation-Reduction
  • Refining
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NCERT Exemplar Class 12 Chemistry Solutions Chapter 6 Principles and Processes of Isolation of Elements

In Class 12 Chemistry NCERT exemplar solutions chapter 6 the students will get to know about various topics that they should know when it comes to general principles and processes of isolation of elements. The main topics covered in this chapter are definition of minerals, ore, gangue, metallurgy and different important sub topics. Every topic covered in this NCERT exemplar for Class 12 Chemistry chapter 6 solutions is very well equipped with all the answers and also correctly explained.

The entire scientific and technological process used for isolation of the metal from its ores is known as metallurgy. The most abundant metal found in the Earth’s crust is Aluminium. This chapter will help you to understand and gather all the knowledge related to metallurgy and the isolation of elements. Overall, NCERT exemplar Class 12 Chemistry solutions chapter 6 along with just giving information is also fun and interesting for all the students who wish to learn more about isolation of elements. This chapter is helpful for them who wish to study the subject very deeply.

NCERT Exemplar Class 12 Chemistry Solutions

Important Topics in NCERT Exemplar Class 12 Chemistry solutions chapter 6 Principles and Processes of isolation of elements

- The importance and use of metallurgy elements and the isolation of elements. There are different factors that are related to each other. Everything related to metallurgy is mentioned in the NCERT exemplar Class 12 Chemistry solutions chapter 6.

- NCERT Exemplar Class 12 Chemistry chapter 6 solutions include different aspects of the chapters that will help and control the use of the elements and they can be different based on its use. The chapter will help you to understand all the concepts with a broad perspective and also help you clear your doubts.

- The chapter also provides important information about the different topics that are included in this chapter. It encompasses the relevant and fairly detailed different diagrams that will help the students in getting a better and clear understanding for the students.

NCERT Exemplar Class 12 Solutions

Check NCERT Solutions for Class 12 Chemistry

Chapter 1

The Solid State

Chapter 2

Solutions

Chapter 3

Electrochemistry

Chapter 4

Chemical Kinetics

Chapter 5

Surface chemistry

Chapter 6

General Principles and Processes of isolation of elements

Chapter 7

The P-block elements

Chapter 8

The d and f block elements

Chapter 9

Coordination compounds

Chapter 10

Haloalkanes and Haloarenes

Chapter 11

Alcohols, Phenols, and Ethers

Chapter 12

Aldehydes, Ketones and Carboxylic Acids

Chapter 13

Amines

Chapter 14

Biomolecules

Chapter 15

Polymers

Chapter 16

Chemistry in Everyday life

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Frequently Asked Questions (FAQs)

1. 1. How are these questions solved by the teachers?

-        The teachers use different techniques and also the information that is given to them with their practical knowledge to solve the questions in a very detailed way.

2. 2. How can I download these NCERT Exemplar Class 12 Chemistry Solutions Chapter 6?

-        You can download the solutions in the PDF format from the page of solutions given by a single link on the given link.

3. 3. Who can benefit from these NCERT Exemplar Class 12 Chemistry Solutions Chapter 6?

-        The students who are preparing for their board exams or are preparing for some entrance examination or both, those students can benefit from these NCERT solutions.

4. 4. What can one learn from this NCERT Exemplar Class 12 Chemistry Chapter 6 Solutions?

-        The students will get to learn the various principles and also the processes of isolation of elements from the NCERT solutions.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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