Subject

Chemistry

Class

CBSE Class 12

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

When chromite ore FeCr2O4 is fused with NaOH in presence of air, a yellow-coloured compound (A) is obtained, which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KCl forms an orange coloured crystalline compound (C).

(i)Write the formulae of the compounds (A), (B) and C.
(ii)Write one use of compound (C).
                                  
                                                OR
Complete space the space following space chemical space equations colon
left parenthesis straight i right parenthesis space 8 MnO subscript 4 superscript minus space plus 3 straight S subscript 2 straight O subscript 3 superscript 2 minus end superscript space plus straight H subscript 2 straight O space rightwards arrow
left parenthesis ii right parenthesis space Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript space plus 3 Sn to the power of 2 plus end exponent space plus 14 straight H to the power of plus rightwards arrow


Chromite space ore comma space on space fusion space with space NaOH space gives space sodium space chromate space which space is space yellow space in space colour colon
4 FeCr subscript 2 straight O subscript 4 space plus 16 NaOH space plus 7 straight O subscript 2 space rightwards arrow stack 8 Na subscript 2 CrO subscript 4 with left parenthesis straight A right parenthesis below space plus 2 Fe subscript 2 straight O subscript 3 space plus 8 straight H subscript 2 straight O
Sodium space chromate space left parenthesis straight A right parenthesis space on space acidification comma space gives space sodium space dichromate space left parenthesis straight B right parenthesis colon
2 Na subscript 2 CrO subscript 4 space plus straight H subscript 2 SO subscript 4 space rightwards arrow stack Na subscript 2 Cr subscript 2 straight O subscript 7 with left parenthesis straight B right parenthesis below space plus Na subscript 2 SO subscript 4 space plus straight H subscript 2 straight O
Sodium space dichromate space left parenthesis straight B right parenthesis space on space reaction space with space KCl space forms space potassium space dichromate space which space is space orange space in space colour colon
Na subscript 2 Cr subscript 2 straight O subscript 7 space plus 2 KCl space rightwards arrow stack straight K subscript 2 Cr subscript 2 straight O subscript 7 space with left parenthesis straight C right parenthesis below plus 2 NaCl
Potassium space dichromate space is space powerful space oxidising space agent. space It space is space used space prepartion space of space azo space compounds.

Or
left parenthesis straight i right parenthesis space 8 MnO subscript 4 superscript minus space plus 3 straight S subscript 2 straight O subscript 3 superscript 2 minus end superscript space plus straight H subscript 2 straight O space rightwards arrow space 8 MnO subscript 2 space plus 6 SO subscript 4 superscript 2 minus end superscript space plus 2 OH to the power of minus
ii right parenthesis space Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript space plus 3 Sn to the power of 2 plus end exponent plus 14 straight H to the power of plus space rightwards arrow 2 Cr to the power of 3 plus end exponent space plus 3 Sn to the power of 4 plus end exponent space plus 7 straight H subscript 2 straight O
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2.

From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell Ans the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii)Which cell is used in automobiles and inverters?
(iv)Which cell does not have long life?


(i) Mercury cell is used in hearing aids.
(ii)Fuel cell was used in the Apollo space programme.
(iii)Lead storage cell is used in automobiles and inverters.
(iv) Dry cell does not have a long life.

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3.

(i) Differentiate between adsorbtion and absorption.

(ii)Out of MgCl2 and AlCl3, which one is more effective in causing coagulation of negatively charged sol and why?

(iii)Out of sulphur sol and proteins, which one form multimolecular colloids?


(i) Adsorption is a surface phenomenon that causes the accumulation of molecules of a substance at the surface of a solid or liquid rather than in the bulk. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when a chalk stick is dipped into an ink solution, only its surface becomes coloured.

On the other hand, the process of absorption is a bulk phenomenon. In absorption, the absorbed substance gets uniformly distributed throughout the bulk of the solid or liquid. For example, when a sponge is dipped in water the whole sponge gets wets.

(ii)According to the Schulze-Hardy rule, the effectiveness of the salt causing flocculation depends on the charge on the ion of opposite sign to the charge on the sol particles. The greater the magnitude of the opposite charge, the higher the ability of a salt to coagulate the sol.
Thus, trivalent salt AlCl3 is more effective in causing the coagulation of a negatively charged sol than divalent salt MgCl2.

(iii)Out of sulphur sol and proteins, sulphur sol forms multimolecular colloids. Sulphur sol consists of particles containing a thousand or more S8 molecules. On the other hand, proteins form macromolecular colloids.

 

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4.

For a reaction :2 NH subscript 3 left parenthesis straight g right parenthesis space rightwards arrow with Pt on top space straight N subscript 2 space left parenthesis straight g right parenthesis space plus 3 straight H subscript 2 left parenthesis straight g right parenthesis
Rate space equals space straight k

(i)Write the order and molecularity of this reaction.
(ii)Write the unit of k.


(i) This reaction is catalysed by Pt at high pressure. So, it is a zero-order reaction with molecularity 2.
(ii) The rate law expression for this reaction is Rate = k
Hence, the unit of k is mol L−1 s−1.

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5.

An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contain 2 × 1024 atoms.


We space have space given colon
straight a equals space 250 space pm space equals space 250 space straight x space 10 to the power of negative 12 end exponent straight m
straight z equals space 4
straight m equals 300 straight g
straight m equals space straight M over straight N subscript straight A space left parenthesis where space straight M space is space molar space mass right parenthesis

straight M space equals space fraction numerator 300 space straight x space 6.02 space straight x space 10 to the power of 23 over denominator 2 space straight x space 10 to the power of 24 end fraction space equals space 90.3 space straight g divided by mol

Density comma space straight d equals fraction numerator space zM over denominator straight a cubed straight N subscript straight A end fraction

straight d space equals space fraction numerator 4 space straight x space 90.3 over denominator left parenthesis 250 right parenthesis cubed space straight x space 10 to the power of negative 36 end exponent space straight x space 6.02 space straight x space 10 to the power of 23 space end fraction space equals space 38.4 space straight g divided by space cm cubed

density space equals space 38.4 space straight g divided by cm cubed
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6.

What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction?


When the magnetic moments of domains are arranged in the same direction then, the substance shows ferromagnetism.

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7.

The rate constant for the first-order decomposition of H2O2 is given by the following equation:
log space k space equals space 14.2 space minus space fraction numerator 1.0 space x space 10 to the power of 4 over denominator T end fraction K
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given: R = 8.314 JK–1mol–1).


Given:
Order of the reaction = First order
t1/2 = 200 minutes = 200 × 60 = 12,000 seconds The relation between t1/2
and k is given by t1/2 = 0.693/k

k = 0.693/12000 = 5.7 × 10−5

The rate constant for the first-order decomposition of H2O2 is given by
log space k equals space 14.2 space minus fraction numerator 1.0 space x space 10 to the power of 4 over denominator T end fraction space.... left parenthesis i right parenthesis space

B y space A r r h e n i u s space e q u a t i o n

l o g space k equals space l o g space A space minus space fraction numerator E subscript a over denominator 2.303 R T end fraction space.... left parenthesis i i right parenthesis

C o m p a r i n g space left parenthesis i right parenthesis space a n d space left parenthesis i i right parenthesis comma space w e space g e t

E subscript a space equals 1.91 space x space 10 to the power of 5

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8.

On  adding  NaOH  to  ammonium  sulphate,  a  colourless  gas  with  pungent  odour  is evolved, which forms a blue-coloured complex with Cu2+ ion. Identify the gas.


On adding NaOH to ammonium sulphate, ammonia gas is evolved. It has a pungent odour and forms a blue-coloured complex with the Cu2+ ion. The chemical reactions are as:

left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4 space plus 2 NaOH space rightwards arrow space 2 NH subscript 3 space plus Na subscript 2 SO subscript 4 space plus 2 straight H subscript 2 straight O
4 space NH subscript 3 space plus Cu to the power of 2 plus end exponent space rightwards harpoon over leftwards harpoon space left square bracket Cu left parenthesis NH subscript 3 right parenthesis subscript 4 right square bracket to the power of 2 plus end exponent space left parenthesis aq right parenthesis

2883 Views

9.

When a co-ordination compound CrCl3.6H2O is mixed with AgNO3, 2 moles of AgCl are precipitated per mole of the compound. Write
(i)Structural formula of the complex.
(ii)IUPAC name of the complex.


When a coordination compound CrCl3.6H2O  is mixed with AgNO3 it gives tow moles of AgCl, therefore, the structural formula would contain two Cl ions satisfying the primary valencies, while five H2O molecules and one Cl− ion are present in the coordination sphere, making the coordination number 6. One H2O molecule will be present as the molecule of hydration.

Structural formula: [CrCl(H2O)5]Cl2.H2O

IUPAC name of the compound: Pentaaquachloridochromium(III) chloride

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10.

Write the main reason for the stability of colloidal sols.


There are two main reasons for the stability of colloidal sols:

Solvation: Colloidal particles are covered by a sheath of liquid in which they are extensively solvated, thereby providing stability.

Electrostatic stabilisation: Presence of equal and similar charges on the colloidal particles prevents coagulation of the colloidal sol.

9162 Views