NEB Grade 12 Chemistry 2082 Question Paper Solutions

20. i) Deduce the normality equation, N₁V₁ = N₂V₂.
Ans: According to the law of equivalence, substances react in their equivalent proportion. This means that a certain number of grams equivalent of acid neutralizes the same number of grams equivalent of alkali.

i.e. Number of gram equivalent of acid = number of gram equivalent of alkali
We know that,

Normality = \frac{\text{number of gram equivalent}}{\text{Vol. \ of solution in Ltr}}

So, the number of grams equivalent = volume of solution normality.

At the equivalence point,

Number of gram equivalent of acid = Number of gram equivalent of alkali

Normality of acid x volume of acid = Normality of alkali x volume of alkali
S_AV_A = S_BV_B or N₁V₁ = N₂V₂

This is called the normality equation.

ii) Experimental data obtained by titrating a decinormal solution of oxalic acid with potassium permanganate solution are given below.

a) Name the above titration.
Ans: Redox titration

b) Calculate the equivalent weight of KMnO₄.
Ans: Equivalent weight = 158/5 = 31.5

c) Calculate the normality of KMnO4 from the above data.

From the Normality Equation,

V_{1}N_{1} = V_{2}N_{2}\\
11 \times N_{1} = 10 \times \frac{1}{10}\\
N_{2} = 0.0909N\\
\therefore\ Normality = 0.0909 N 

d) Why is dil. H₂SO₄ is added to the conical flask containing standard oxalic acid before titrating with the KMnO₄ solution?
Ans: KMnO₄ only acts as a strong oxidizing agent in an acidic medium. In an acidic solution, the half-reaction for KMnO4 is:

MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 5H₂O

In this reaction, purple MnO₄^– is reduced to a colourless Mn₂⁺. Without sufficient H⁺ ions (acid), this reduction won’t occur properly, and the titration won’t proceed accurately.

e) Identify the titrant and titrand in this titration.
Ans: Titrant: Oxalic acid
Titrand: KMnO₄

OR

a) Write the applications of the solubility product principle and the common ion effect in qualitative salt analysis.
Ans: Application of the common ion effect and the solubility product principle are:

1. Application in Qualitative analysis:

a. Precipitation of group II metal cation ( Hg⁺⁺ , Pb⁺⁺ , Bi⁺⁺ , Cu⁺⁺, As⁺⁺, Sb⁺⁺ , Sn⁺⁺):

In group II, H₂S is passed through the salt solution acidified with dil. HCl. The dissociation takes place as

H₂S ⇌ 2H⁺ + S^—
HCl → H⁺ + Cl^–

Here, the H+ ion suppresses the ionization of H₂S due to the common ion effect, and hence the concentration of the sulphide ion is lowered. As a result of this, the solubility product of the metal sulphide group is less than their ionic product, and as a result, the radicals of group II are precipitated as their sulphides.

If HCl is not added, the precipitation of group II as well as group IIIB metal cations occurs at the same time as their sulphides, and selective precipitation cannot be achieved.

b. Precipitation of group IIIA metal cation as their hydroxide (Fe⁺⁺⁺, Al⁺⁺⁺, Cr⁺⁺⁺):

In this group, H₂S is passed through the salt solution containing NH₄Cl and NH₄OH. The dissociation takes place as:

NH₄OH ⇌ NH₄⁺ + OH^–
NH₄Cl → NH₄⁺ + Cl^–

Here, the common ion NH₄⁺ reduces the concentration of OH^– in the solution. The concentration of OH^– gets reduced to such an extent that the solubility product becomes less than its ionic product, and these hydroxides get precipitated out.

2. Purification of common salt:

The impure salt, NaCl, can be made pure by passing HCl gas into its saturated solution.

NaCl ⇌ Na⁺ + Cl^–
HCl → H⁺ + Cl^–

Due to the common ion (Cl^–), the new product of ionic concentration will be greater than the solubility product of NaCl. Hence, NaCl gets precipitated while all the impurities remain in the solution.

b) Differentiate between the Bronsted-Lowry and Lewis concepts of bases.

c) Calculate the degree of ionization of HCN having concentration 0.01M (Ka of HCN = 4.8×10^-10). Also, calculate the H⁺ concentration and pH of the solution.
Solution:

From ostwald dilution law,

\begin{align*}
\alpha = \sqrt{\frac{K_{a}}{c}} &= \sqrt{\frac{{4.8\times 10^{-10}}}{0.01}}\\ 
&= 2.2\times 10^{-4}\\ 
\left [ H^{+} \right ] = c\alpha &= 0.01\times 2..2\times 10^{-4}\\ 
&= 2.2\times 10^{-6}\\ 
\therefore pH = &-log[2.2\times10^{-6}] = 5.65
\end{align*}

21. An organic compound (X) reacts with Grignard’s reagent to give (Y), which on oxidation gives (Z). All of the compounds (X), (Y) and (Z) give a positive iodoform test.

a) Identify the compounds (X), (Y) and (Z) with reactions and conditions involved.

b) Why do these compounds give a positive iodoform test?
Ans: Y is alcohol. It has a CH₃-CH(-OH)- unit. X and Z have a CH3-C(=O)- unit because they are carbonyl compounds ( aldehyde and ketone). So, they give the iodoform test.

c) How would you convert compound (Y) into pseudonitrol?

d) Write any two methods for the preparation of compound (Z).

OR

How will you carry out the following conversions?

a. Ethanol to chloroform

b. Phenol into DDT

c. Ethoxyethane into methoxymethane

d. Aniline into chlorobenzene

22. Answer the following questions:

a. What are the raw materials for the production of paper?
Ans:

• Softwood: eg, pines, firs, spruces
• Hardwood: eg, eucalyptus, mulberry, rubber plant wood
• Grass and reeds: eg, lemon, situ, panni, bamboo
• Straw: eg, rice, wheat, barley

b. Draw the flow sheet diagram for the manufacture of Portland cement.

c. Nuclear fusion reaction is also called a thermonuclear reaction. Why?
Ans: The nuclear reaction in which two or more lighter nuclei fuse together to form a heavier nucleus, releasing energy, is called nuclear fusion.

These types of reactions are difficult to occur. Here, two lighter nuclei should come close where positively charged nuclei repel each other. Hence, a huge amount of energy is required to overcome such repulsion. Therefore, a large amount of energy from the source, like the sun, is required to initiate these reactions. So, these reactions are also called thermonuclear reactions.

d. How are addition and condensation polymers formed? Give an example of it.
Ans: Addition polymers are formed by the addition of monomers without the elimination of any molecule. eg, polyethene, PVC, Teflon, etc.

Condensation polymersare formed by the addition of monomers with the elimination of smaller molecules like water. eg, nylon-6,6, bakelite, etc.

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