Cathode is negative and anode is positive inthe electrolytic cell.

Cathode (-ve) and Anode (+ve) sign for electrolytic cell.

Calculating molar conductivity at infinite dilution for weak electrolytes

Degree of dissociation ¢cm÷¢m^infinity(¢=lambda)

Dissociation constant k_c=c(alpha)²÷1-(alpha)

Solubility of soluble NaCl

Answers to all queries related to CBSE 12th compartment result 2017 are are authentically  available at  cbse.12thresultsnic.in/compartment-result.html  However ,looking to your incusitiveness the following  is reproduceed -

• The CBSE 12th Supply Exam Results willl be available on official website in third week of August Month ,( may be 20th of the month ). All candidates are eligible to see their results through @ cbseresullts.nic.i
• How to check CBSE Supply Results 2017

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2.  Click link CBSE XII Compartment Results 2017
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 Central Board of Secondary Education has successfully  conducted 12th Supplementary Exam in July .Hence it is supposed that you must have 

appeared , and done well .   Pl. keep awaiting for results , till Aug .2017. Now , no points in providing you tricks / methodology to score maximum in Chemistry paper..     

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A typing error !  Pl. read   as , 

.......................................CH₃ ** C* H  O   , instead  of  H₃ ** C* HO

  Thanks,

Aldehydes and ketones exhibit the  unusual acidity of hydrogen atoms attached to Carbons - alpha  (   {tex}\alpha{/tex}  )  which is  adjacent  to that of  the carbonyl group. These hydrogens are referred as  {tex}\alpha{/tex} - hydrogens , and the carbon atom to which they  are bonded is {tex}\alpha{/tex}- carbon. In ethanal there is one {tex}\alpha{/tex}  C* & three {tex}\alpha{/tex}. - H  atoms ( marked as H** ).

................................H₃ **  .C*  H O

although weakly acidic ,  ( K_a  =  10^-19  to 10^-20 )  {tex}\alpha{/tex} - H can react  with strong bases to form anion . The unnusual acidity of {tex}\alpha{/tex} - H can be explained by both electron withdrawing ability of the carbonyl ( =CO ) group ,and resonance in anion that is formed . The electron withdrawing ability  of   =CO group is  caused by the groups di[pole nature which results by difference in electronegativities  of   C  and  O atoms.

The anion formed by loss of {tex}\alpha{/tex} - hydrogen can be resonance stabilised because of mobility of {tex}\pi{/tex} - electrons that are adjacent to carbonyl group ( =CO ) . This resonance which stabilises the anion ( commonly termed as  enolate ion ) creates two resonance structures  , an enol & a keto-form . In most cases the keto form is more stable. 

The reaction is very well termed as Aldol condensation . In case of ethanal this reaction , in the simplest form , can be represented as -

2CH₃ CHO  <------dil .NaOH ----------->   CH₃ - CH (OH) -CH₂ -  CHO------------- {tex}\Delta{/tex} /  ( - H₂ O ) ------------->   CH₂ - CH=CH - CHO. -

Ethanal...................................................3- Hydroxybutanal.....................................................................But -2 - enal(

....................................................................( Aldol ).....................................................................( Aldol condensation product ) 

Sea water contains a lot of electrolytes. River contains colloids of sand and clay. When they meet the electrolytes neutralise the charge on colloidal particles and result in the precipitation of sand, clay etc. Thus resulting in a delta formation .

Adsorption is always exothermic. This statement can be explained in two ways.
(i)Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.
(ii)ΔH of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ΔS is negative. Now for a process to be spontaneous, ΔG should be negative.
ΔG= ΔH-TΔS
Since ΔS is negative, ΔH has to be negative to make ΔG negative. Hence, adsorption is always exothermic.

Osmosis:A process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concentrated solution into a more concentrated one.

Osmotic Pressure:The pressure that would have to be applied to a pure solvent to prevent it from passing into a given solution by osmosis, often used to express the concentration of the solution.

Osmosis is a phenomenon where in there happens to be  a movement of solvent particles ( ie. molecules  / ion )  from a solution  of lower concentration  to that of higher concentration when the two are separated from each other by a semipermeable membrane.

while , 

Osmotic pressure  is the pressure built up between the two solution undergoing osmosis . It is  caused due to a difference of vapour pressures  between  both the solutions. 

No.Osmotic pressure is generated due to presence of concentration gradient between the two solutions seperated by a SPM.

No !  a solution of itself has vapour pressure , but not the osmotic pressure . Osmotic pressure comes into consideration only when the phenomenon of osmosis between two solutions with different vapour pressures   & separated by a semipermeable membrane    comes into play . 

Steric hindrance is the stopping of a chemical reaction which might be caused by a molecule's structure. An example of steric hindrance is how rotaxanes are created

Steric hindrance occurs when the large size of groups within a molecule prevents chemical reactions that are observed in related molecules with smaller groups.

Steric hindrance is the stopping of a chemical reaction which might be caused by a molecule's structure. An example of steric hindrance is how rotaxanes are created

When a solid solute is added  to the solvent like water , some solute dissolves  and its concentration increases in solution . This proces is known as as ' dissolution ' . Some solute particles in solution  collide with solid solute particles and get separated out of solution . This process is known as ' crystallisation '. A stage is reached when the ttwo processes occur at the same rate and thus a dynamic equilibrium is reached ,

..............................................Dissolution <----------->  crystallisation

Factors affecting the rate of dissolution :

(i)  Effect  of temperature -  

The  rate of dissolution of a solid in a liquid is significantly affected  by temperature changes  as  the above dynamic equilibria must follow LeChatelier's Principle. In general , if the dissolution process is endothermic  ( {tex}\Delta H{/tex} _sol < 0 ) , the solubility increases with rise in temperature. and if it is exothermic ( {tex}\Delta H{/tex} _sol > 0 )   the rate of dissolution   decreases . These trends have been experimentally observed governing the concerned rate law too..

(ii) Effect of pressure -

Since both solids as well as liquids are highly incompressible , variations of pressure does not have any significant effect on tthe process or rate of dissolution of a solid in water.

Graphite is odd in its categorisation and does by no means rigorously follow the conventions of either a covalent solid, or a molecular solid (for example it can conduct electricity). Graphite has properties of both molecular solids (it is soft), and covalent solids (it has a very high melting point. In fact it will break down before it melts.).

For this reason, the categorisation of some materials, such as graphite, can be somewhat complicated, and the idea of graphite being a covalent solid really just comes from its similarity with other covalent solids.

Dopant: A substance used to produce a desired electrical characteristic in a semiconductor.

Dopant increses the conductivity of semiconductor by creating new donor level near to conduction band or acceptor level near to valence band.

Doping is nothing but adding N or P type impurities to a pure semiconductor. So these impurities will contain electrons or holes based on their impurity. When these kind of doping takes place in semiconductors and voltage is applied, these electrons will conduct electricity. In case of holes, electrons due to potential excitation jump from their original positions to these holes and form immobile ions at that point, whereas equal amount of holes are created from where these electrons jumped. Thus conductivity increases due to doping.

Aromatic compounds are substances that consist of one or more rings that contain alternating single and double bonds in its chemical structure and follows the Huckels (4n+2)Pie electrons rule.

Aromatic compounds, also known as arenes or aromatics, are chemical compounds that contain conjugated planar ring systems with delocalized pi electron clouds instead of discrete alternating single and double bonds. Typical aromatic compounds are benzene and toluene.

The question refers to electrolyte - concentration cells which consist of identical electrodes but are dipped in solutions  of  same  electrolyte having different concentrations. Electrical energy given out by the cell is due to tendency of the electrolyte to diffuse from a solution of higher  concentration to that of lower concentration. Thus two concentrations of the same electrolyte tend to become equal with the expiry of time.  An equilibrium is established at this  stage. This results into a cell emf which happens to be maximum at the start  and then gradually falls to zero.The cell   representation of this cell is  ,

M | M⁺ (C₁) || M⁺  (C₂) | M   ,  

where ,  C₁   &  C₂  are the concentrations  of  the same electrolyte in the two half cells  ,  and  also    C₂ >  C₁

A simple example of such a cell  is  

( Zn|Zn²⁺ (C₁) /  Anode || Zn²⁺  (C₂ )  | Zn / Cathode

The emf  of such a cell is given by using Nernst equation 

E Cell   =  {tex}\frac{0.0591}{n}{/tex} log ( C₁ / C₂  )

When C₁   becomes equal to C₂   E cell  = 0

So , for every concentration cell  E^o   =   0

Thus  ,  the cell will not generate emf  when concentrations  become equal

Number of atoms of A per unit cell = contribution from corners + contribution from alternate face centres (these are 2) =  1/8 X 8 + 1/2 X 2 = 2

Number of atoms of B per unit cell = contribution from body centre + contribution from edge centres

= 1 + 1\4 X 12 = 4

> Formula of compound = A₂B₄

or exactly it is  AB₂

_{Since charge on B is -1 therefore Charge on A will be +2}

A occupy all the corners and alternate face centres.

Therefore, No. of atoms of A = {tex}(8\times {1\over 8} )+ (3\times {1\over 2}) = 1 + {3\over 2}{/tex}

= {tex}5\over 2{/tex}

Now, B occupy all the body centre and the edge centre.

So, No. of atoms of B : {tex}1+({1\over 4}\times 12) = 1+3=4{/tex}

Therefore, formula of compound will be

{tex}A_{5\over 2} B_4 = A_5B_8{/tex}

Charge on B = -1

Let charge on A = x

Then

5x +8(-1) = 0

=> 5x -8=0

=> 5x = 8

=> x = {tex}8\over 5{/tex}

The volatility of a substance refers to the readiness with which it vaporizes. Generally speaking, substances with a boiling point below 100 °C are considered volatile and all others are called nonvolatile. Ethyl alcohol and pentane are examples of volatile substances;

The free electrons trapped in the anion vacancies are called as F-centers. Solids containing F-centres are paramagnetic because the electrons occupying the vacant sites are unpaired.

Dichloridobis-ethylenediamine Chromium(III)Chloride.

Kolbe's reaction: When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product.

This reaction is known as Kolbe's reaction.

Williamson ether synthesis: Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.

This reaction involves SN₂ attack of the alkoxide ion on the alkyl halide. Better results are obtained in case of primary alkyl halides.

Vapour pressure of pure water, p^o = 360 mm Hg

Lowered vapour pressure, p = 360 –0 .5% of 360

= 360 – 1.8 = 358.2 mm Hg

Now, Weight of water, w1= 200ml x 1g.ml = 200g

Weight of urea w2 = ?

Molecular weight of water, M1 = 18 g/mol Molecular weight of urea, M2 = 60 g/mol

According to Roult’s law:

{tex}{p^o -p \over p} = {n_2\over n_1 + n_2}{/tex}

=> {tex}{360 - 358.2\over 358.2} = { {w_2 \over M_2}\over {{w_1\over M_1 } + {w_2 \over M_2}}}{/tex}

=> {tex}0.005 = { {w_2 \over M_2}\over {{w_1\over M_1 } + {w_2 \over M_2}}}{/tex}

=> {tex}0.005 {\left ({w_1\over M_1 } + {w_2 \over M_2}\right )}= {w_2 \over M_2}{/tex}

=> {tex}0.005 \times {w_1\over M_1 } = {w_2 \over M_2} - 0.005\times {w_2 \over M_2}{/tex}

=> {tex}0.005 \times {w_1\over M_1 } =0.995\times {w_2 \over M_2}{/tex}

=> {tex}{0.005 \over 0.995} \times {w_1\times M_2\over M_1 } = {w_2 }{/tex}

=> {tex}{0.005 \over 0.995} \times {200\over 18 }\times 60 = {w_2 }{/tex}

=> w₂ =  3.33g

Hence, 3.33g urea should be added

Benzophenone It is a ketone, so it does not undergo Canniyzaro's reaction. Without {tex}\alpha{/tex} -hydrogen, it cannot participate in aldol condensation.

Full cylinder is not filled with nitrogen .

A mixture of gases are mixed 

Freezing point depression is the phenomena that describes why adding a solute to a solvent results in the lowering of the freezing point of the solvent. When a substance starts to freeze, the molecules slow down due to the decreases in temperature, and the intermolecular forces start to take over. The molecules will then arrange themselves in a pattern, and thus turn into a solid. For example, as water is cooled to the freezing point, its molecules become slower and hydrogen bonds begin to "stick" more, eventually creating a solid. If salt is added to the water, the Na+ and Cl- ions attract to the water molecules and interfere with the formation of the large network solid known as ice. In order to achieve a solid, the solution must be cooled to an even lower temperature.

The freezing point depression can also be explained in terms of vapor pressure. Adding solute to a solvent will essentially dilute the solvent molecules, and according to Raoult's law, this leads to a decrease in vapor pressure. Considering the fact that the vapor pressure of the solid and liquid forms must be the same at freezing point, because otherwise the system would not be at equilibrium, the lowering of the vapor pressure leads to the lowering of the temperature at which the vapor pressures of the liquid and frozen forms of the solution will be equal.

 The freezing point depression can be calculated by the formula:

{tex}\Delta T_f=i\times K_ f\times molality{/tex}

In this equation,

{tex}\Delta T_f{/tex} is the freezing point depression,

K_f is the freezing point depression constant,  and i is the van 't Hoff factor.

The freezing point depression constant changes depending on the solvent, and the van 't Hoff factor accounts for the number of particles that a dissolving solute creates in solution.

2 tetrahedral voids on each body diagonal in ccp unit

Total 8 tetrahedral voids in 1 ccp unit

Solution: 

r_c + r_a = {tex}\sqrt 3a\over 2{/tex}

=> {tex}r_c + r_a = {\sqrt 3 \times 4.3\over 2}{/tex}= {tex}3.72 A^°{/tex}