2012_CBSE_XIIScience_5_2_SET1_sectionB

8/11/2019 2012_CBSE_XIIScience_5_2_SET1_sectionB

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Grade XII ALL INDIA SET 3

Chemistry (Theory)

[Time allowed: 3 hours] [Maximum marks:70]

General Instructions:

(i) All questions are compulsory.(ii) Marks for each question are indicated against it.

(iii) Question number 1 to 8are very short-answer questions and carry 1mark each.

(iv) Question number 9to 18are short-answer questions and carry 2 marks each.(v) Question number 19to 27are also short-answer questions and carry 3 marks each.

(vi) Question number 28to 30are long-answer questions and carry 5marks each.

(vii) Use Log Tables, if necessary. Use of calculators is not allowed.

Q9. 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? 2

Ans. 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

2

R 2k a

= 4ka2

= 4 R

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

(ii)If the concentration of the reactant is reduced to half, i.e. 1

A2

a , then the rate of the

reaction would be

2

2

1R

2

1

4

1

4

k a

ka

R

Therefore, the rate of the reaction would be reduced to

th1

.4


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Q10. Express the relation among cell constant, resistance of the solution in the cell and conductivity

of the solution. How is molar conductivity of a solution related to its conductivity?

OR

The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2mol

1.Calculate the conductivity of the conductivity of this solution. 2

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

1 l

R a

The quantityl

ais cell constant.

l Distance between 2 electrodes

a Area of cross section

R Resistance

Also, conductivity,Molarconductivity

Volumeof solution

Thus, ^m= V

OR

m

m

2 1

3

1

ConductivityMolar Conductivity =

Concentration

=

Conductivity, =

138.9(Scm mol )1.5(mol/L)=

1000(cm /L)

= 0.208 Scm

k

c

k c

Q11. Explain the role of each of the following:

(i) NaCN in the extraction of silver(ii) SiO2in the extraction of copper 2

Ans. (i) The roasted ore of gold is leached with a solution of sodium cyanide in the presence ofoxygen 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.

2 2 2

2 2 4

4Au 8NaCN 2H O O 4 Na Au(CN) 4KOH

2Na Au(CN) + Zn Na [Zn(CN) ] + 2 Au


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(ii) Copper mattecontains Cu2S and FeS. In the blast furnace, copper matteis added with

powdered coke and silica. The oxidation of ore takes place in this process. As a result, cuprousoxide 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).

2 3FeO SiO FeSiO

Flux Slag

Q12. Complete the following chemical reaction equations: 2

(i) 22 7Cr O I H

(ii) 4 2MnO NO H

Ans. (i) 2 + 3+2 7 2 2Cr O +14H +6I 2Cr +3I +7H O

(ii) + 2+4 2 3 2

2MnO +5NO +6H 2Mn + 5NO +3H O

Q13. Explain the following facts giving appropriate reason in each case:(i) NF3is an exothermic compound whereas NCl3is not.

(ii) All the bonds in SF4are not equivalent. 2

Ans. (i)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 NF3is quite strong and it is an exothermic

compound.

(ii)SF4has sp3d hydridisation and thus have trigonal bipyramid structure in which one axial

position is occupied by a lone pair of electrons. This results in two types of angles.

Equatorial bond angle F S FBPBPrepulsion Axial bond angle F S F


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Q14. Explain the following behaviours:

(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular

masses.(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol. 2

Ans. Alcohols undergo intermolecular H-bonding while hydrocarbons do not.

Hence, they are more soluble in water than the hydrocarbons of comparable molecular masses.

(ii)

The nitro-group is an electron-withdrawing group. The presence of this group in the ortho

position decreases the electron density in the OH bond. As a result, it is easier to lose a proton.Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance.

Hence, ortho nitrophenol is a stronger acid.

On the other hand, methoxy group is an electron-releasing group. Thus, it increases the electrondensity in the OH bond and hence, the proton cannot be given out easily.


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For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

Q15. Explain the mechanism of acid catalysed hydration of an alkene to form corresponding

alcohol. 2

Ans. Some reactive alkenes like 2methyl propene undergo direct hydration in the presence of

mineral acids which act as catalysts. The addition of water to the double bond takes place in

accordance with Markonikoffs rule.

Q16. Complete the following chemical reaction equations: 2

(i) 6 5 2 3 2 2C H N Cl+H PO H O

(ii) 6 5 2 2C H NH +Br aq.

Ans. (i) C6H5N2Cl + H3PO2+ H2OC6H6+ N2+ H3PO3+ HCl(ii)

Q17. Describe the following giving the relevant chemical equation in each case: 2

(i) Carbylamine reaction

(ii) Hofmanns bromamide reaction.

Ans. (i) Carbylamine reaction

Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic

and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide,

carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours.Secondary and tertiary amines do not respond to this test.


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2 3 2R NH CHCl 3KOH alc. R NC 3KCl 3H O

Primary Chloroform Potassium Carbylamine

amine hydroxide

For example,

3 2 3 3 2CH NH CHCl 3KOH alc. CH NC 3KCl 3H O

Methanamine Methyl carbylamine

or methyl isocyanide

Hoffmann bromamide reaction

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium

hydroxide, a primary amine with one carbon atom less than the original amide is produced. Thisdegradation reaction is known as Hoffmann bromamide reaction. This reaction involves the

migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen

atom.

For example,

Q18. Differentiate between disinfectants and antiseptics. Give one example of each group. 2

Ans. Antiseptics and disinfectants are effective against micro-organisms. However, antiseptics are

applied to the living tissues such as wounds, cuts, ulcers, and diseased skin surfaces, while

disinfectants are applied to inanimate objects such as floors, drainage system, instruments, etc.Disinfectants are harmful to the living tissues.

Iodine is an example of a strong antiseptic. Tincture of iodine (23 percent of solution of

iodine in alcoholwater mixture) is applied to wounds. 1 percent solution of phenol is used as

a disinfectant.

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2012_CBSE_XIIScience_5_2_SET1_sectionB