Subject

Chemistry

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

Which reducing agent is employed to get copper from the leached low-grade copper ore?


Copper can be obtained from low-grade ore through the process of leaching using acid or bacteria (leaching is a process in which ore is treated with a suitable reagent that dissolves ore but not the impurities).

The solution containing copper can be reduced with the help of reducing agents such as scrap iron or H2 to get copper metal.

Cu2+(aq) + Fe Cu(s) + Fe2+(aq)

 Cu2+ (aq) + H2(g)  Cu(s) + 2 H+(aq) 

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2.

Examine the given defective crystal:
 

Answer the following questions:


(i) What type of stoichiometric defect is shown by the crystal?

(ii) How is the density of the crystal affected by this defect?

 (iii) What type of ionic substances shows such defect? 


(i) Schottky defect is shown by the mentioned crystal, as an equal number of cations and anions are missing in the crystal lattice.

(ii) This defect leads to decrease in density, as an equal number of the cations and anions are missing from the crystal lattice. A number of such defects in ionic solids are quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature.

(iii) This kind of defect is shown by that ionic substance in which the cations and anions are of almost similar sizes. 


Examples:  NaCl, KCl and CsCl.  

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3.

An element with density 11.2 g cm-3  forms a f.c.c. lattice with edge length of 4 x10-8

Calculate the atomic mass of the element. (Given:  NA = 6.022x 10-23 (mol-1


Density, d = 11.2 g cm-3

Edge length, a = 4x10-8 cm

Avogadro number, NA = 6.022x1023 mol-1

Number of atoms present per unit cell, Z (fcc) = 4


We know for a crystal system,

 straight d equals space fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction

straight m space equals fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A over denominator straight Z end fraction

We space get comma

straight M space equals fraction numerator 11.2 space straight x space 64 space straight x space 10 to the power of negative 24 end exponent space straight x space 6.022 straight x space 10 to the power of 23 over denominator 4 end fraction equals space 107.91 space straight g

   

Thus, the atomic mass of the element is 107.91 g.

 

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4.

Define an ideal solution and write one of its characteristics.


Definition:  The solutions that obey Raoult’ s law over the entire range of concentration are called ideal solutions. Examples:  n-hexane and n-heptane.

Characteristics: For ideal solutions: 

 Enthalpy of mixing (mixH) of the pure components to form the solution is zero.

The volume of mixing (mixV) is also equal to zero. 

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5.

Which of the following is a more stable complex and why?

 (i) [Co(NH3)6]3+

 (ii) [Co(en)3]3+


Chelating ligands form more stable complexes compared to non-chelating ligands. Since ethylene diamine is a bidentate ligand and forms stable chelate, [Co(en)3]3+ will be a more stable complex than [Co(NH3)6]3+

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6.

Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 of benzene to lower its freezing point by 0.48 K (Kf= 5.12 kg mol-1 )


We have given:

Tf  = 0.48 K
Kf=5.12K kg mol-1

 W1 = 75 g

 W2 = ?

 M2 = 256 g mol-1  

increment straight T subscript straight f space equals space fraction numerator straight K subscript straight f space straight x space straight w subscript 2 space straight x space 1000 over denominator straight M subscript 2 space straight x space straight w subscript 1 end fraction space

straight w subscript 2 space equals fraction numerator increment straight T subscript straight f space straight x space straight M subscript 2 space xw subscript 1 over denominator straight K subscript straight f space straight x space 1000 end fraction

equals space fraction numerator 0.48 space straight x space 256 space straight x space 75 over denominator 5.12 space straight x space 1000 end fraction space equals space 18 space straight g space

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7.

Write two differences between 'order of reaction' and 'molecularity of reaction?


Order

Molecularity

The overall order of reaction is the sum of all the exponents of all the reactants present in the rate law expression.

It is the number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

It is determined experimentally.

It is a theoretical concept.

It may be equal to zero or have fractional values.

It cannot be equal to zero and it always has integral values( which cannot exceed 3)

 
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8.

Outline the principles behind the refining of metals by the following methods:

Chromatographic method 


Chromatography method:

Chromatography is the technique used for separation of the components of a mixture that are soluble in the same solvent and are differently adsorbed on an adsorbent. This technique was first used for separation of colours and hence, the name. There are two phases in chromatography - the stationary phase and the mobile phase. The components of the mixture are absorbed differently in the stationary phase, thus leading to their separation. There are many different types of chromatography - paper chromatography, thin-layer chromatography, gas chromatography etc. 

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9.

Give one example each of 'oil in water' and 'water in oil' emulsion.


Type of emulsion

Example

Oil in water

 Milk, vanishing cream

Water in oil

 Butter, cold cream, cod liver oil

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10.

Outline the principles behind the refining of metals by the following methods:

Zone refining method


Zone Refining Method:

Zone refining method is based on the principle that impurities are more soluble in the molten state than in the solid state of a metal. This method is used in the purification of elements like Germanium(Ge), Silicon(Si), Indium (In) etc.

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