Fajan's rule important tool for inorganic chemistry

All d best my all students 

Evaluate your "basic concept of chemistry"

   A sample containing Cu₂S, Cu₂O and 50 millimole of an inert gas is treated with 500 meq. Of acidic KMnO₄. The entire Cu is converted to Cu⁺² and the resulting solution is diluted to 50 ml. A 5 ml. of diluted solution is treated with excess of KI solution, Iodine is liberated and it is filtered. Now it is diluted to 50 ml. 5 ml of this liberated iodine requires 10 ml of 0.2 N Na₂S₂O₃. Find out the mole fraction of Cu₂S and Cu₂O in the given sample. 

physical chemistry for iit jee

Physical Chemistry requires a good understanding of concepts. Try to understand all the formulae used in physical chemistry, instead of just applying them blindly, as a particular formula may not be applicable to the question. (Example, instead of a dilute solution, the given solution may be concentrated, making most formulae invalid, in that case understanding of the basic principles will help.) Try to solve various types of problems for perfection in physical chemistry,.physical chemistry is like mathematics it required lot of practice.

weightage of organic chemistry in iit jee

General Organic Chemistry	9%
Alcohol Phenol and Ether	7%
Aldehyde  Ketone	5%
Carboxylic Acids	2%
Nitrogen Containing compounds	2%
Biomolecules	2%
Hydrocarbons	2%

How to Study Organic Chemistry for IIT JEE?

How to Study Organic Chemistry for IIT JEE?

Although Organic Chemistry in itself is quite vast but at IIT JEE level it deals with General Organic Chemistry i.e. basics organic chemistry.It starts with the concept of Hybridization, IUPAC Nomenclature, Isomerism, Resonance, Hyperconjugation, etc. Stereoisomerism is a critical part which becomes a bottleneck for most of the students. These concepts are used in almost every reaction mechanisms which are a part of topics like Hydrocarbons, Halides, Alcohols & Phenols, Aldehydes & Ketones, Carboxylic acids and derivatives and hence these are basics of organic chemistry. Named Reactions are like drivers direction for students and maximum come from the chapter of Aldehydes & Ketones, which is no less than a climax of a movie.

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Electrochemical cell

Daniel cell

Standard reduction potential

Electrochemical series :- Order of oxidizing and reducing strength

Electrochemical series

Demjanov reaction :- Ring Expansion

Ring expansion mechanism

Nobel prize facts

Why are the individuals and organisations awarded a Nobel Prize called Nobel Laureates?

The word "Laureate" refers to being signified by the laurel wreath. In Greek mythology, the god Apollo is represented wearing a laurel wreath on his head. A laurel wreath is a circular crown made of branches and leaves of the bay laurel (In latin: Laurus nobilis). In Ancient Greece, laurel wreaths were awarded to victors as a sign of honour - both in athletic competitions and in poetic meets.

Facts on the Nobel Prize in Chemistry

On 27 November 1895, Alfred Nobel signed his last will and testament, giving the largest share of his fortune to a series of prizes, the Nobel Prizes. As described in Nobel's will one part was dedicated to “the person who shall have made the most important chemical discovery or improvement”. 

Announcement of the Nobel Prize in Chemistry

The Chemistry Laureates are announced in October every year. Earlier the same day, the members of the Chemistry Committee of the Royal Swedish Academy of Sciences selects the Nobel Laureates through a majority vote. The decision is final and without appeal. The names of the Nobel Laureates in Chemistry are then announced at a press conference.
The Nobel Prize in Chemistry 2013 will be announced on Wednesday 9 October, 11:45 a.m. at the earliest.

The Nobel Prize Amount

Alfred Nobel left most of his estate, more than SEK 31 million (today approximately SEK 1,702 million) to be converted into a fund and invested in "safe securities." The income from the investments was to be "distributed annually in the form of prizes to those who during the preceding year have conferred the greatest benefit on mankind." 

The Nobel Prize amount for 2013 is set at Swedish kronor (SEK) 8.0 million per full Nobel Prize.

some facts of chemistry

1.Natural gas has no odour. The smell is added artificially    so that leaks can be detected.


2.  Water expands by about 10% as it freezes.


3.The lighter was invented before the match (in 1816 by      J.W. Dobereiner).

HAPPY INDEPENDENCE DAY

           
  

TRY TO SOLVE SOME NUMERICALS OF SOLID STATE

Q.1 Gold crystallizes in a face centered cubic unit cell. If the edge length is a. Calculate the closest distance between the gold atom and  an impurity atom if it occupies  a)  a tetrahedral void   b) an octahedral void without any change in volume of he unit cell.

ANSWER: - a) closest distance if an impurity occupies a tetrahedral void: - √3/4 a

b) Closest distance if an impurity occupies an octahedral void: - a/2

Q.2 a) Spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in  ccp pattern. The normal spinel has one- eight of the tetrahedral voids occupied by one type of metal ion and one half of the octahedral holes occupied by another type of metal ion . such a spinel is formed by Zn⁺², Al⁺³ and O^-2 with Zn⁺² in tetrahedral holes. Give the formula of the spinel.

b) if all species in above question touch each other, determine the fraction of he volume occupied by ions in the unit cell.

ANSWER :- a) ZnAl₂O₄                     b) 0.77

Q.3 a strong current of trivalent gaseous boron passed through a germanium crystal decrease the density of the crystal due to part replacement of germanium by borone and due to interstitial vacancies created by missing Ge atoms. In one such experiment, one gram of germanium istaken and theboron atoms are found to be 150 ppm by weight, when the density of the germanium crystal decreases by 4 % Calculate the percentage of missing vacancies due to germanium, which are filled up by boron atoms

ANSWER :- 2.376%

EUDIOMETRY (GAS ANALYSIS) :- APPLICATION OF POAC ON EUDIOMETRY (gas analysis) NUMERICALS and HOW TO FIND OUT MOLECULAR FORMULA OF ANY HYDROCARBON USING "POAC"

Eudiometry:- Eudiometry means gas analysis. It means in this type of numericals either you have to find out vol. of GAS From the data given in question or you have to find out any other thing like molecular formula of hydrocarbon ,no. of mole , mass of any other reactant and vol. of GAS is given. Specific gases are absorbed in specific reagents as given below

Reactants                                                  Absorbed gases

KOH solution                                             CO₂ & SO₂

Alkaline Pyrogallol                                     O₂

Solution of ammonical cuprous chloride    CO

And if the mixture is cooled, H₂O vapour changes to liquid.

Hence, when temperature is above 100 ⁰C (373 K) then only volume of H₂O is considered.

Eudiometry is mainly based on Avogadro's law,

     A _(g) +       B _(g) →  C_(g)  +   D_(g)

 a volume   b volume   c volume   d volume

 a moles    b moles     c moles    d moles

Q.1 Three moles of Na₂CO₃ is reacted with 6 moles of HCl solution. Find out 

the volume of CO₂ gas produced at STP

Sol.  Na₂CO₃→ NaCl + CO₂

Apply POAC on C atom

No. of moles of C atoms in Na₂CO₃ = no. of moles of C atoms in CO₂

1* no. of moles of Na₂CO₃ = 1* no. of moles of CO₂

So no. of moles of Na₂CO₃ = no. of moles of CO₂ = 3 moles (given)

Vol. of CO₂ = 3 * 22.4 litre = 67.2 litre

  v HOW TO FIND OUT MOLECULAR FORMULA OF ANY HYDROCARBON USING POAC:-

Q. 10 ml. of a gaseous hydrocarbon was burnt completely in 80 ml. of O₂ 

at   NTP. The remaining gas occupied 70 ml. at NTP. This vol. became 50

ml on   treatment with KOH solution. What is the formula of hydrocarbon?

Solution: -  C_x H_y + O₂ → CO₂+ H₂O

 Vol. of ( CO₂ + unreacted O₂) = 70 ml.

Because CO₂  can be absorbed by KOH .  So when this 70 ml. mix. Is treated it 

becames 50 ml it means vol. of CO₂ in the mixture was 20 ml. which is absorbed

 by KOH and remaining 50 ml. was unreacted O₂  ( O₂  can NOT absorbed by 

KOH)

So vol. of CO₂ = 20 ml.

Vol. of O₂ reacted = 80 -50 = 30 ml.

Vol. of hydrocarbon = 10 ml.

Apply POAC on C atoms

X* no. of mole of C_x H_y = 1* no. of mole of CO₂   -----------------------( i  ) 

X* 10 = 20

So X = 20/10= 2  

Apply POAC on H atoms

Y* no. of moles of C_x H_y = 2* no. of mole of H₂O -----------------( ii )

Apply POAC on O atoms

2* no. of moles of O₂ = 1* no. of mole of H₂O+2* no. of mole of CO₂

From eq. (ii)

2* no. of moles of O₂ = y/2* no. of mole of C_x H_y + 2* no. of mole of CO₂

2*30 = Y/2*10 + 2 *20

20 = 10Y/2

Y = 4

Put the value of X and Y in C_x H_y

The molecular formula of hydrocarbon is C₂H₄

_{TRY TO SOLVE FOLLOWING QUESTION}

Q.2   7.5 ml of hydrocarbon gas was exploded with excess of oxygen. On

 cooling it was found to have undergone a contraction of 15 ml. if vapour 

density of hydrocarbon is 14. Determine its molecular formula.

Solution ( HINT ) :- molecular wt.= 2* vapour density

   Answer:- C₂H₄

Q.3  7.5 ml of hydrocarbon gas was exploded with 36 ml.of oxygen. The 

vol. of gases on cooling was found to be 28.5 ml., 15 ml. of which was 

absorbed by KOH and the rest was absorbed in a solution of alkaline 

pyrogallol. If all vol. are measured under the same conditions, determine 

the molecular formula of hydrocarbon.

       Answer:- C₂H₄

     

   

EVERY SUCCESS STORY IS ALSO A STORY OF GREAT FAILURE.

A man failed in business at the age of 21; was defeated in a legislative race at the age 22; failed again in business at age 24; his wife died when he was age 26; had nervous breakdown at age 27; lost a congressional race at age 32; lost a senatorial race at age 45; failed in an effort to become vice- president at age 47; lost a senatorial race again at age 49; and was elected president of united state at age 52. This man was Abraham Lincoin.

If we study the history, we will find that all stories of success are also stories of great failures. But people don’t see the failures. They only see the end result and think that person got lucky: “he must have been at the right place at the right time.”

"A SMOOTH SEA NEVER MADE A SKILLFUL MARINER. EVERYTHING IS DIFFICULT BEFORE IT BECOME EASY. WE CAN NOT RUNAWAY FROM OUR PROBLEMS. ONLY LOSERS QUIT AND GIVE UP."

One day a partially deaf four year old child came to home with a note in his pocket from his teacher , “ your tommy is too stupid too learn, get him out of school.” His mother read the note and answered, “ my tommy is not too stupid to learn,I will teach him myself.” And that tommy grew up and become the great scientist Thomas Edison. Thomas Edison had only three months of formal schooling.

LEARN TO LIKE THE THINGS THAT NEED TO BE DONE:-

"START BY DOING WHAT IS NECESSARY, THEN WHAT IS POSSIBLE; SUDDENLY YOU ARE DOING THE IMPOSSIBLE."

“LEARNING IS LOT LIKE EATING. IT IS NOT HOW MUCH YOU EAT THAT MATTER, WHAT MATTER HOW MUCH YOU DIGEST.”

GET "FREE OF COST" QUESTIONS ABOUT POAC AND MOLE CONCEPT

IF YOU WANT TO GET SOME MORE QUESTIONS ON POAC, MOLE CONCEPT FOR PRACTICE, YOU CAN SEND A REQUEST FOR THE SAME ON OUR  MAIL ADDRESS 
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 SOON WE WILL PUBLISH ANOTHER IMPORTANT CONCEPTS  OF CHEMISTRY. AND YOU CAN GET THESE ALL IMPORTANT CONCEPT" FREE OF COST" BY SENDING A REQUEST ON OUR MAIL ADDRESS.

SOLVED NUMERICALS ON MOLE CONCEPT AND POAC :- BOOST UP YOUR CONCEPT ABOUT "APPLICATION OF POAC AND MOLE CONCEPT"

Q.1 what amount of CaO will be produced by 1 gm. of Ca?

This numerical we can solve by two methods

 1.     Mole concept  2. POAC method

Advantage of POAC over Mole concept is that we need not to write down the stoichiometry of reaction . means in POAC method NEED NOT to  BALANCE THE CHEMICAL REACTIONS

By mole concept

First wite down the BALANCE CHEMICAL REACTION

2Ca + O₂ → 2CaO

It means when 2mole of ca ( 2 *40=80 gm) react with 1 mole of oxygen it produce 2 

mole of CaO (2*56= 112 gm).

80 gm Ca produce = 112 gm Cao

So 1 gm Ca will produce = 112/ 80 = 1.4 gm CaO (unitary method)

By POAC (ADAVANTAGE OVER mole concept NO NEED TO BALANCE EQUATION)

Ca + O₂ → CaO

 Apply POAC on Ca

Total no. of moles of Ca atoms in reactant side = Total no. of moles of Ca atoms in product side

1* no. of moles of Ca = 1* no. of moles of CaO

 No. of moles of Ca /  no. of moles of CaO = 1

So mole of CaO produced in reaction would be equal to mole of Ca  used in chemical reaction

because we are using 1 gm Ca means ( no. of mole =mass/ atomic wt ; here atomic wt

 of Ca = 40 ; so  no. moles of Ca =1/40 =0.025 moles)  0.025  moles of Ca ,the total moles of Cao would be same means 0.025 moles of Cao means (0.025 * 56 = 

1.4) 1.4 gm Cao . Here m.wt of CaO =56 gm

POAC method is a good method to solve these type of numerical because you can  apply POAC concept directly without balancing of equation. So u can save your time  in competitive exams. Even in starting students may feel to apply POAC something difficult but actually it is better than mole concept and time saving method. So students  have to be practice to solve this type of numerical by POAC method. Here some numericals are solved by POAC method to clarify your concept about POAC .observe these numericals to improve your understanding about POAC.

Q.2 A sample of  KClO₃ on decomposition yielded 448 ml.of oxygen gas at N.T.P. calculate

i. weight of oxygen produced

ii. weight of KClO₃ originally taken, and

iii. wt. of KCl produced

solution:-

i. Calculation of  Wt. of  O₂

  VOL. of O₂ = 448 ml = 0.448 litre

  Mole of  O₂ = 0.448/22.4 = 0.02 moles means

  Mass = no. of mole *  m.wt = 0.02 * 32 = 0.064 gm

ii.Calculation of  Wt of KClO₃

KClO₃  → KCl + O₂
Apply POAC on O atom

3* no. of moles of KClO₃ = 2* no. of moles of O₂

No. of moles of KClO₃ / No. of moles of O₂ = 2/3

No. of moles of KClO₃ = 2/3 * no. of moles O₂

No. of moles of KClO₃ = 2/3 * 0.02 = 0.0133 moles

Mass of  KClO₃ = 0.0133 * 122.5 = 1.633 gm  

Here 122.5 is the m.wt of  KClO₃

iii.Calculation of wt. of  KCl produced 

Apply POAC on K atoms

1* no. of mole of KClO₃ = 1* no. of mole of KCl

So no. of mole of  KCl = NO. of mole of KClO₃ = 0.0133 moles

Mass of KCl = 0.0133 * 74.5 = 0.99 gm

Here 74.5 is the m.wt. of  KCl 

Q.3 A mixture of  KBr and NaBr weighing 0.560 gm was treated with aq. Ag+ ion and all bromide ion was recovered as 0.970 gm of pure AgBr. What was the friction by weight of KBr in the sample?

Solution :- ( KBr   +   NaBr)   + Ag⁺ → AgBr

              (x gm      0.560 –x gm)                                0.970 gm

Because no. of moles of Br atoms are conserved in reaction so

Apply POAC on Br atom

Moles of Br in KBr + moles of Br in NaBr = moles of Br in AgBr 

1* no.of mole of KBr + 1*no. of mole of NaBr =   1* no. of mole of AgBr

Mass of KBr/ m.wt of KBr + Mass of NaBr / M.wt of NaBr  = Mass of AgBr/ M.wt  of AgBr

x/ 119 + 0.560- x / 103 = 0.97/188

x = 0.1332 gm

So KBr in the mixture is 0.1332 gm

Fraction of KBr  in mixture = 0.1332/ 0.560 = 0.2378

Q.4 27.6 gm of K₂CO₃  was treated by a series of reagents so as to convert all of its carbon to K₂Zn₃[ Fe(CN)₆]₂. Calculate the weight of the product.

Solution: - K₂ CO₃   → → → → K₂Zn₃[ Fe(CN)₆]₂

No. of moles of C atoms are conserved in the reaction

Apply POAC on C atoms

Moles of C atoms in K₂ CO₃   = Moles of C atoms in K₂Zn₃[ Fe(CN)₆]₂

1* no. of moles of  K₂ CO₃   =  12* no. of moles of  K₂Zn₃[ Fe(CN)₆]₂

Now you can calculate mass of K₂Zn₃[ Fe(CN)₆]₂

Hint:-

1* mass of K₂ CO₃  / m.wt  of K₂ CO₃  = 12* mass of  K₂Zn₃[ Fe(CN)₆]₂ /m.wt of K₂Zn₃[ Fe(CN)₆]

m.wt of  K₂Zn₃[ Fe(CN)₆]₂ = 698

m.wt of  K₂ CO₃   = 138

answer  mass of K₂Zn₃[ Fe(CN)₆]₂  =   11.6 gm