Subject

Chemistry

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

Define ‘peptization’.


Peptization can be defined as a process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte.The electrolyte used in this process is known as a peptizing agent.

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2.

Express the relation among cell constant, the resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?


The conductivity (k) of the solution in a cell is the reciprocal of its resistivity.

 straight k equals space 1 over straight R open parentheses 1 over straight a close parentheses

The quantity  1/2  is cell constant.

L --> Distance between 2 electrodes

a --> Area of cross section

R -->  Resistance

Also comma space conductivity comma space straight k equals space fraction numerator Molar space conductivity over denominator Volume space of space solution end fraction

thus comma space molar space conductivity space equals space kV

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3.

The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9S cm2mol-1. Calculate the conductivity of the conductivity of this solution.


Molar space conductivity comma space straight k equals space conductivity over Concentration

straight capital lambda subscript straight m space equals straight k over straight c

conductivity comma

straight k equals space straight capital lambda subscript straight m space equals straight k over straight c

conductivity comma space straight k equals space straight capital lambda subscript straight m space straight x space straight C

equals space fraction numerator 138.9 space left parenthesis Scm squared space mol to the power of negative 1 end exponent right parenthesis space straight x space 1.5 space left parenthesis mol divided by straight L right parenthesis over denominator 1000 space left parenthesis cm cubed divided by straight L right parenthesis end fraction space equals 0.208 space straight S space cm to the power of negative 1 end exponent
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4.

A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?


Let the concentration of the reactant be [A] = a

Rate of reaction, R = k [A]2 = ka2

(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

R = k (2a)2

= 4ka

= 4R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e.,left square bracket A right square bracket equals 1 half a then the rate of the reaction would be
Rate space of space reaction comma space straight R space equals space straight k open parentheses 1 half straight a squared close parentheses

equals 1 fourth ka squared

equals 1 fourth straight R

 

Therefore, the rate of the reaction would be reduced to 1 to the power of th over 4

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5.

How is copper extracted from a low-grade ore of it?


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

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6.

Explain the role of each of the following:

(i) NaCN in the extraction of silver

(ii) SiO2 in the extraction of copper


(i) The roasted ore of gold is leached with a solution of sodium cyanide in the presence of oxygen for many days. The role of NaCN in this process is to dissolve the gold to form an aurocyanide complex, from which the metal is obtained by displacement.

4Au + 8NaCN + 2H2O + O2  --->  4 Na [Au(CN)2] + 4KOH

2Na[Au(CN)2] + Zn ---> Na2[Zn(CN)4] + 2Au

(ii) Copper matte contains Cu2S and FeS. In the blast furnace, copper matte is added with powdered coke and silica. The oxidation of ore takes place in this process. As a result, cuprous oxide and ferrous oxide are produced. The role of silica in this process is to remove the iron oxide obtained as ‘slag’. FeO combines with silica (flux) to form iron silicate, FeSiO3(slag).

FeO + SiO2 rightwards arrow with space space space increment space space space space on top FeSiO2

          Flux            slag

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7.

Explain the following facts giving appropriate reason in each case:

(i) NF3 is an exothermic compound whereas NCl3 is not.

(ii) All the bonds in SF4 are not equivalent.


(i) As we move down the group 17, the size of the atom increases from fluorine to chlorine. The instability of NCl3 is due to the weak NCl bond. This is due to the large difference in the size of nitrogen and chlorine atoms. On the other hand, atoms of both nitrogen (75 pm) and fluorine (72 pm) are small sized. Thus, bonding in NF3 is quite strong and it is an exothermic compound.

(ii)SF4 has four bonded atoms and one lone pair so SF4 has sp3d hydridisation and thus have trigonal bipyramid structure in which one axial position is occupied by a lone pair of electrons. This lone pair finds a position that minimizes the number of 900 repulsion it has with bonding electron pairs. This results in two types of angles.

Equatorial bond angle F - S - F° (LP-BP repulsion >BP – BP repulsion)

Axial bond angle F - S - F<90°


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8.

What is the basicity of H3PO2 acid and why?


Basicity of H3PO2 depends on upon the number of ionizable -OH groups present in the molecule. That is the number of hydrogen attached to the electronegative atom oxygen.

H3PO2 has one ionizable –OH group, thus its basicity is 1.

 The structure of H3PO2 is as follows:

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9.

What are n-types semiconductors?


The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. These are generated when the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As.

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10. Complete the following chemical equations:


bold i bold right parenthesis bold space bold Cr subscript 2 bold O subscript 7 superscript 2 minus end superscript space plus space bold H to the power of plus space plus space bold I to the power of minus space rightwards arrow

ii right parenthesis space bold MnO subscript 4 superscript minus space plus space bold NO subscript 2 superscript minus space plus space bold H to the power of plus rightwards arrow

bold i bold right parenthesis bold space bold Cr subscript 2 bold O subscript 7 superscript 2 minus end superscript space plus space 14 bold H to the power of plus space plus space 6 bold I to the power of minus space rightwards arrow space 2 bold Cr to the power of 3 plus end exponent space plus space 3 bold I subscript 2 space plus space 7 bold H subscript 2 bold O

bold ii bold right parenthesis bold space bold 2 bold MnO subscript bold 4 superscript bold minus bold italic space bold plus bold italic space bold 5 bold NO subscript bold 2 superscript bold minus bold italic space bold plus bold italic space bold 6 bold italic space bold H to the power of bold plus bold rightwards arrow bold italic space bold 2 bold Mn to the power of bold 2 bold plus end exponent bold italic space bold plus bold italic space bold 4 bold NO subscript bold 3 superscript bold minus bold italic space bold plus bold italic space bold 3 bold H subscript bold 2 bold O
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