NCERT solutions for class 12 chemistry Chemistry Chapter 6 General Principles and Processes of Isolation of Elements 2021

Q: What is the weightage of NCERT Class 12 Chemistry chapter 6 in JEE Mains?

One question can be expected for JEE Main from the chapter General Principles and Processes of Isolation of Elements

Q: What is the weightage of NCERT Class 12 Chemistry chapter 6 in the CBSE board exam?

Around 3 marks of questions will be asked in the CBSE exam from General Principles and Processes of Isolation of Elements. Follow NCERT book for a good score. More problems can be solved using NCERT exemplar for Class 12 Chemistry.

Q: What are the important topics of General Principles and Processes of Isolation of Elements?

The important topics of this chapter are Leaching reactions Blast furnace reactions Extraction processes

NCERT solutions for class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

Edited By Apoorva Singh | Updated on Aug 16, 2022 12:18 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements - Students appearing in Class 12 board exams must refer to this NCERT Chemistry Class 12 chapter 6 to score well in the examination. To develop a grip on every topic you must solve topic wise questions and exercise questions of NCERT book Cass 12 Chemistry chapter 6. The NCERT solutions for Class 12 Chemistry chapter 6 General Principles and Processes of Isolation of Elements are prepared by subject experts to help students in clarifying their doubts.

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NCERT solutions for class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

Also, the NCERT solutions for other classes and subjects can assist you in your preparation. By referring to the NCERT, students can understand all the important concepts and practice questions well enough before their examination.

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Find NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements Below:

Solutions to In Text Questions Ex 6.1 to 6.4

Question 6.1 Which of the ores mentioned in Table 6.1 can be concentrated by magnetic separation method?

Table 6.1: Principle Ores of some Important Metals

Metal	Ores	Composition
Aluminium	Bauxite Kaolinite(a form of clay)	AlO _x (OH) _3-2x _{[Where 0<x<1]} [Al ₂ (OH) ₄ Si ₂ O ₅ ]
Iron	Haemtite Magnetite Siderite Iron pyrites	Fe ₂ O ₃ Fe ₃ O ₄ FeCO ₃ FeS ₂
Copper	Copper pyrites Melachite Cuprite Copper glance	CuFeS ₂ CuCO ₃ .Cu(OH) ₂ Cu ₂ O Cu ₂ S
Zinc	Zinc blande or Sphalerite Calamine Zincite	ZnS ZnCO ₃ ZnO

Answer :

Ores which contains some magnetic particles or they can attract by the magnetic field only they can be concentrated by magnetic separation method. Among the ores mentioned in table 6.1, the ores of iron such as haematite $(Fe_{2}O_{3})$ , magnetite $(Fe_{3}O_{4})$ , siderite $(FeCO_{3})$ , and iron pyrites $(FeS_{2})$ can be separated by the process of magnetic separation

Question 6.2 What is the significance of leaching in the extraction of aluminium?

Answer :

In the extraction of aluminium, the significance of leaching is to concentrate the pure alumina ( $Al_{2}O_{3}$ ) from bauxite ore.

$\\Al_{2}O_{3}(s)+2NaOH(aq)+3H_{2}O\rightarrow 2Na[Al(OH)_{4}](aq)\\ SiO_{2}(l)+2NaOH(aq)\rightarrow NaSiO_{3}(aq)+H_{2}O$

Impurities are then filtered and the solution is neutralized by carbon dioxide gas

$2Na[Al(OH)_{4}](aq)+CO_{2}(g)\rightarrow Al_{2}O_{3}.xH_{2}O(s)+2NaHCO_{3}(aq)$

(hydrated alumina)

$Al_{2}O_{3}.xH_{2}O(s)\overset{\Delta }{\rightarrow} Al_{2}O_{3}+xH_{2}O$

Question 6.3 The reaction, $Cr_2 O_3 + 2 AI \rightarrow Al_2 O_3 + 2 Cr$ $(\Delta G^{o} = - 421 kJ)$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

Answer :

$Cr_2 O_3 + 2 AI \rightarrow Al_2 O_3 + 2 Cr$
This reaction is thermodynamically feasible, but it does not take place in room temperature because at room temperature all the reactants are solids. At high temperature, the chromiums start melting and the reaction becomes faster.

Question 6.4 Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What are those conditions ?

Answer :

Yes, temperature below 1623K, Mg can reduce $Al_{2}O_{3}$ to Al and temperature above 1623K Aluminium can reduce $MgO$ to $Mg$

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Ssolation of Elements- Exercise Questions

Question 6.1 Copper can be extracted by hydrometallurgy but not zinc. Explain.

Answer :

Copper can be extracted by hydrometallurgy but not zinc because the reduction potential of zinc is lower than that of copper. So, that copper can be reduced from its solution by zinc.

$Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)$

But to reduce zinc from its solution we need higher reactive metal, which reduction potential is lower than zinc like $Al,Mg,Ca,K$ But these metals readily react with water with the evolution of $H_{2}$ gas. As a result, these metals are cannot use to displace zinc ion from its solution.

$2K(s)+2H_{2}O(l)\rightarrow KOH(aq)+H_{2}(g)$

Question 6.2 What is the role of depressant in froth floatation process?

Answer :

Role of depressant in froth flotation method is to separate two sulphid ores by adjusting one ore to form froth. For example- $NaCN$ is used as a depressant it selectively prevents $ZnS$ to coming to the froth but allows PbS with the froth. NaCN react with $ZnS$ to form $Na_{2}[Zn(CN)_{4}]$

Reaction-

$4NaCN+ ZnS\rightarrow Na_{2}[Zn(CN)_{4}]+ Na_{2}S$

Question 6.3 Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Answer :

Gibbs free energy of formation $(\Delta _{f}G^{0})$ of $Cu_{2}S$ is less than that of $H_{2}S$ and $CS_{2}$ so that hydrogen and carbon are unable to reduce

$Cu_{2}S$ $\rightarrow Cu$ . But the Gibbs free energy of formation $(\Delta _{f}G^{0})$ of $Cu_{2}O$ is higher than that of CO. So, that carbon can able to reduce

$Cu_{2}O$ $\rightarrow Cu$ . Therefore the extraction of copper from its pyrite ore is more difficult than its oxide ore through reduction.

Question 6.4 Explain: (i) Zone refining

Answer :

ZONE REFINING- This method is based on that the impurities are more soluble in the molten state than n the solid state. In this process, a circular mobile heater surrounding the rod of impure metal is fixed at one end. Move the heater from one end to another end so that the impurities present in the rod also moves to the other end of the rod. Repeat the process several times again and again. Impurities get concentrated at separate ends of the rod. This end is cut off.

1594816216537

Question 6.4 Explain: (ii)Column chromatography.

Answer :

Column chromatography-
The principle of chromatography is based on, the different component of the mixture is getting adsorbed to a different extent on an adsorbent. In this, there are two phases one is mobile and the other is immobile phase. In column chromatography, $Al_{2}O_{3}$ column is generally used as a stationary phase. The mobile phase is may be a gas or liquid. The mobile phase is forced to move over the stationary phase. The adsorbed component is removed with the help of a suitable solvent.
1594816234027

Question 6.5 Out of C and CO, which is a better reducing agent at 673 K ?

Answer :

At 673K $(\Delta G^{0})$ for change of $(CO\rightarrow CO_{2})$ is less than $(\Delta G^{0})$ of $(C\rightarrow CO)$ . Therefore CO can easily oxidized to $CO_{2}$ than C to CO.

Hence CO is better reducing agent than C at 673K.

Question 6.6 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Answer :

The following elements present in the anode mud in electrolytic refining of copper are selenium, silver, tellurium, gold, platinum and antimony. They are present because they are very less reactive and are not affected by the purification process. So, therefore they settle down below as anode mud.

Question 6.7 Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Answer :

At 500-800K

$\\3Fe_{2}O_{3}+CO\rightarrow 2Fe_{3}O_{4}+CO_{2}\\ Fe_{3}O_{4}+CO\rightarrow 3Fe+4CO_{2}\\ Fe_{2}O_{3}+CO\rightarrow 2FeO+CO_{2}$

limestone also decomposes to calcium oxide which reverses silicate impurity of the ore as slag $CaCO_{3}\rightarrow CaO+CO_{2}$

At 900-1500K

$\\C+CO_{2}\rightarrow 2CO\\ FeO+CO\rightarrow Fe+CO_{2}$

$CaO+SiO_{2}\rightarrow CaSiO_{3}(slag)$

Question 6.8 Write chemical reactions taking place in the extraction of zinc from zinc blende.

Answer :

Chemical reactions in the extraction of zinc from zinc blende -

After concentration of ore (removal of gauge from zinc blend ( $ZnS$ ) by froth floatation method).

Conversion to the oxide b y the roasting method -

$2ZnS+3O_{2}\rightarrow ZnO+2SO_{2}$

Extraction of zinc from zinc oxide (by reduction)

$ZnO+C\overset{673K}{\rightarrow}Zn+CO$

Electrolytic refining of impure zinc (electrolyte used is acidified zinc sulphate $ZnSO_{4}$ )

Anode- $Zn\rightarrow Zn^{2+}+2e^{-}$

Cathode (made of the pure copper strip)- $Zn^{2+}+2e^{-}\rightarrow Zn$

Question 6.9 State the role of silica in the metallurgy of copper.

Answer :

The role of silica in metallurgy of copper is to remove the $FeO$ (iron oxide) produced during roasting as slag. If the sulphide ore contains iron, it mixed with silica ( $SiO_{2}$ ) as flux before heating. $FeO$ combines with silica and produce iron silicate ( $FeSiO_{3}$ )

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

Question 6.10 Which method of refining may be more suitable if element is obtained in minute quantity?

Answer :

Chromatographic method of refining is particularly suitable if the element is obtained in minute quantity.

Question 6.11 Which method of refining will you suggest for an element in which impurities present have chemical properties close to the properties of that elements?

Answer :

Zone refining is the method for an element in which impurities present have chemical properties close to the properties of those elements.

Question 6.12 Describe a method for refining nickel.

Answer :

Nickel is refined by Mond's Process, in which Nickel is heated in steam in presence of Carbon monoxide ( $CO$ ) to form nickel carbonyl complex . $(Ni(CO)_{4})$ The complex is volatile in nature.

$Ni+4CO\overset{330K-350K}{\rightarrow} Ni(CO)_{4})$

This complex is decomposed at high temperature to get a pure nickel.

$Ni(CO)_{4}\overset{450K-470K}{\rightarrow} Ni+4CO$

Question 6.13 How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.

Answer :

The separation of alumina carried out in the following steps-

The powdered ore is treated with the concentrated $NaOH$ solution at 473K-523K and 35-36 bar pressure.
$\\Al_{2}O_{3}+2NaOH(aq)+3H_{2}O\rightarrow 2Na[Al(OH)_{4}](aq)\\ SiO_{2}+NaOH\rightarrow Na_{2}SiO_{3}+H_{2}O$
The $CO_{2}$ gas is passed through the resulting solution to neutralise the aluminate, which results in ppt. of hydrated alumina.
$2Na[Al(OH)_{2}](aq)+CO_{2}\rightarrow Al_{2}O_{3}+2NaHCO_{3}(aq)$
The obtained hydrated alumina is filtered, dried and heated to get pure alumina.
$Al_{2}O_{3}.xH_{2}O\rightarrow Al_{2}O_{3}+x.H_{2}O$

Question 6.14 Giving examples, differentiate between ‘roasting’ and ‘calcination’.

Answer :

Roasting- The process of converting sulphide ore to oxide. In this process, the ore is heated in the furnace with a regular supply of air, below its melting temperature: example- sulphide ore of zinc, lead, and copper.

$\\2ZnS+3O_{2}\rightarrow 2ZnO+2SO_{2}\\ 2PbS+3O_{2}\rightarrow 2PbO+2SO_{2}\\ 2Cu_{2}S+3O_{2}\rightarrow 2Cu_{2}O+2SO_{2}$

Calcination- Process of converting hydroxide and carbonate ore to oxide in a limited supply of air. The temperature should be below the metals melting point in this process volatile matter which escapes, leaving behind the metal oxide.

$\\Fe_{2}O_{3}.xH_{2}O \overset{\Delta }{\rightarrow}Fe_{2}O_{3}(s)+xH_{2}O(g)\\ ZnCO_{3} \overset{\Delta }{\rightarrow} ZnO(s)+CO_{2}(g)\\ CaCO_{3}.MgCO_{3} \overset{\Delta }{\rightarrow} CaO(s)+MgO(s)+2CO_{2}(g)$

Question 6.15 How is ‘cast iron’ different from ‘pig iron”?

Answer :

The iron that obtains from the blast furnace is pig iron . it contains 4% carbon and many impurities like $(s,P,Si,Mn)$ .

Cast iron is obtained from pig iron by melting pig iron with scrap iron and coke using hot air blast. It contains slightly less carbon content around 3% and it is hard an brittle.

Question 6.16 Differentiate between “minerals” and “ores”.

Answer :

Minerals are the naturally occurring substance which contains some amount of metals. They are found in the earth crust and obtained by mining.

Ores are the specific minerals in which metals are economically extracted. It has a definite composition. Each and every ore is a mineral but not every mineral is ore.

Question 6.17 Why copper matte is put in silica lined converter?

Answer :

The copper matte composed of $Cu_{2}S$ and $FeS$ . It is put in a silica-lined converter to remove the remaining ferrous oxide ( $FeO$ ) and ferrous sulphide ( $FeS$ ) present in copper matte as slag ( $FeSiO_{3}$ )

Some silica is also added and hot air blast is blown to convert the remaining $(FeS,FeO \&\ Cu_{2}S/Cu_{2}O)$ to the metallic copper.

$\\2FeS+3O_{2}\rightarrow 2FeO+2SO_{2}\\ FeO+SiO_{2}\rightarrow FeSiO_{3}\\ 2Cu_{2}S+3O_{2}\rightarrow 2Cu_{2}O+2SO_{2}\\ 2CU_{2}O+Cu_{2}S\rightarrow 6Cu+SO_{2}$

Question 6.18 What is the role of cryolite in the metallurgy of aluminium?

Answer :

The main purpose of adding cryolite $(Na_{3}AlF_{6})$ in metallurgy of aluminium is-

To increase the electrical conductivity of the alumina $Al_{2}O_{3}$
To decrease the melting point of the mixture

Question 6.19 How is leaching carried out in case of low grade copper ores?

Answer :

In case of low-grade ores, copper is extracted by hydrometallurgy. Leaching is done with the help of acid or bacteria in the presence of air. The solution containing $Cu^{2+}$ is treated with scrap iron or dihydrogen $(H_{2})$ .

$Cu^{2+}(aq)+H_{2}(g)\rightarrow Cu(s)+2H^{+}(aq)$

Question 6.20 Why is zinc not extracted from zinc oxide through reduction using CO?

Answer :

According to the Ellingham diagram, the standard Gibbs free energy of formation of $ZnO$ from $Zn$ is lower than the formation of $CO_{2}$ from CO. So, therefore, CO cannot reduce $ZnO$ to zinc ( $Zn$ ).

Question 6.21 The value of
$\Delta _ f G ^ \theta$ for formation of $Cr _ 2 O _3$ is $- 540 KJ mol ^{-1}$ and that of $Al_2 O_3 \: \: is \: \: -827 KJ mol^{-1}$ . Is the reduction of $Cr _ 2 O _3$ possible with Al?

Answer :

$2Al+3/2O_{2}\rightarrow Al_{2}O_{3}$ ......................... (-827 KJ/mol)
$2Cr+3/2O_{2}\rightarrow Cr_{2}O_{3}$ ......................... (-540 KJ/mol)
_________________________________ (On substracting )
$2Al+Cr_{2}O_{3}\rightarrow Al_{2}O_{3}+2Cr$ ...............(-287 KJ/mol)

In the above reaction, we see that $\Delta_f G^0$ of formation of Cr is negative (-287 KJ/mol) Hence the reduction of $Cr _ 2 O _3$ with aluminium is possible.

Question 6 . 22. Out of C and CO, which is a better reducing agent for ZnO ?

Answer :

1594816261439
The Gibbs free energy formation of $CO_{2}$ from $CO$ is always higher than the Gibbs free energy of formation of $ZnO$ . Thus CO cannot reduce $ZnO$ .
On the other hand, Gibbs free energy of formation of $CO_{2}$ from C is less than gibbs free energy of formation of $ZnO$ . Hence C can easily reduce $ZnO$ to zinc.

Question 6.23 The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Answer :

gpisc
From the graph we can observe that, a mtal oxide can reduce the oxide of other metals,if the $\Delta_f G$ of the first oxide is more than the second oxide.
example- $Mg$ can reduce $ZnO$ to Zn but Zn cannot reduce $MgO$ because $\Delta_f G$ of $MgO$ from Mg is more negative.

Question 6.24 Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?

Answer :

In Down's process, we prepare sodium metals in which chlorine is obtained as a by-product. It involves electrolysis of a fused mixture of $NaCl$ and $CaCl_{2}$ . Sodium is deposited at cathode and chlorine at anode as a by-product.

$NaCl(l)\rightarrow Na^{+}(melt)+Cl^{-}(melt)$

At Cathode - $Na^{+}+e^{-}\rightarrow Na$

At anode- $Cl^{-}\rightarrow \frac{1}{2}Cl_{2}+e^{-}$

If an aqueous solution of NaCl is subjected to electrolysis, dihydrogen gas $(H_{2})$ is evolved at the cathode while chlorine $(Cl_{2})$ is obtained at the anode.

Question 6.26 Outline the principles of refining of metals by the following methods : (i) Zone refining

Answer :

1594816274728

Question 6.26 Outline the principles of refining of metals by the following methods : (ii)Electrolytic refining

Answer :

Electrolytic refining- It is a process of refining impure metals by electricity . In this method, impure metal is act as anode and the same pure metal strip is act as cathode. Same metal salt which is soluble is taken as the electrolyte. When electricity is passed the metal ions deposited at the cathode (pure metal) and the impure metal is starting dissolving in the solution in the form of ions. Impurities present in metal (impure metal) gets collected at bottom of the anode (as anode mud )

Question 6.26 Outline the principles of refining of metals by the following methods :(iii)Vapour phase refining

Answer :

Vapour phase refining- In this process metal is converted into its volatile compound which is collected and decomposed to give pure metals.

Two major requirement-

The metal should form a volatile compound with a suitable reagent.
the volatile compound should be easily decomposable so that to recover easily

Question 6.27 Predict conditions under which Al might be expected to reduce MgO.

Answer :

As per the Ellingham diagram, above 1350, the standard Gibbs free energy of formation $(\Delta G^{0}_{f})$ of $Al_{2}O_{3}$ is lower than the $MgO$ from $Mg$ . So, that's why above $1350^{0}C$ , $Al$ can reduce MgO.

Topics of NCERT Syllabus Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements-

6.1 Occurrence of Metals

6.2 Concentration of Ores

6.3 Extraction of Crude Metal from Concentrated Ore

6.4 Thermodynamic Principles of Metallurgy

6.5 Electrochemical Principles of Metallurgy

6.6 Oxidation Reduction1636.7Refining

6.8 Uses of Aluminium, Copper, Zinc, and Iron

NCERT Solutions 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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Benefits of NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

The step-by-step solutions given in the NCERT class 12 Chemistry solutions chapter 6 will aid you in understanding the chapter easily.
The revision will be a lot much easier because with the help of solutions you will always remember the concepts and get very good marks in your class.
Homework will not be a problem now, all you need to do is check the detailed CBSE Class 12 Chemistry Chapter 6 NCERT solutions and you are good to go.
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Frequently Asked Questions (FAQs)

1. What is the weightage of NCERT Class 12 Chemistry chapter 6 in JEE Mains?

One question can be expected for JEE Main from the chapter General Principles and Processes of Isolation of Elements

2. What is the weightage of NCERT Class 12 Chemistry chapter 6 in the CBSE board exam?

Around 3 marks of questions will be asked in the CBSE exam from General Principles and Processes of Isolation of Elements. Follow NCERT book for a good score. More problems can be solved using NCERT exemplar for Class 12 Chemistry.

3. What are the important topics of General Principles and Processes of Isolation of Elements?

The important topics of this chapter are

Leaching reactions
Blast furnace reactions
Extraction processes

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quartely question paper for mathematics cbse

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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.

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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

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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!

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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.

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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.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT solutions for class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

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