NEB Grade 12 Chemistry Paper 2081

NEB Grade 12 Chemistry Paper 2081 Complete Solutions

Group ‘A’

Rewrite the correct option of each question in your answer sheet.

1. A chemical reaction occurs as follows:

NaOH+H_{3}PO_{4} \rightarrow NaH_{2}PO{4}+H_{2}O

What is the equivalent weight of H₃PO₄?

a. 25
b. 49
c. 58
d. 98

2. If 0.01M solution of acetic acid is 0.01% ionized, what will be the dissociation constant of acetic acid?

a. 1×10^-3
b. 1×10^-4
c. 1×10^-8
d. 1×10^-10

3. If the rate of reaction is equal to the rate constant, what will be the order of reaction?

a. Zero
b. First
c. Second
d. Third

4. Standard electrode potentials of four metals: P, Q, R and S are +0.34V, -0.25V, -2.93V and +0.85V respectively. Which is the correct arrangement in the order of decreasing reactivity?

a. S > P > Q > R
b. Q > R > P > S
c. R > Q > P > S
d. R > S > Q > P

5. What features of transition metals make them suitable to act as catalysts?

a. Large ionic charge
b. Variable oxidation state
c. Highly reactive nature
d. Large surface area for the reactant to be absorbed

6. Which of the following metal ions show a green colour in its salt?

a. Ti³⁺
b. Cr³⁺
c. Mn²⁺
d. Fe²⁺

7. If the compound Y is heated with acid anhydride in presenceof sodium acetate, what will it give?

\text{Benzene}\xrightarrow[anhy. \ AlCl_{3}]{CH_{3}Cl} \text{X}\xrightarrow[O]{CeO_{2}/H^{+}}\text{Y}

a. Cinnamic acid
b. Picric acid
c. Benzoic acid
d. Phthalic acid

8. Which of the following compounds gives positive Tollen’s test as well as iodoform test?

a. Propanone
b. Ethanol
c. Ethanal
d. Methanal

9. Which one of the following is most basic in nature?

a. NH₃
b. CH₃NH₂
c. (CH₃)₂NH
d. C₆H₅NH₂

10. Which of the following compounds feacts with chlorobenzene to form DDT?

a. Carbonyl chloride
b. Acetone
c. Chloral
d. Chloroform

11. If you are asked to prepare primary alcohol using Grignard’s reagent, what will you start with?

a. Methanal
b. Ethanal
c. Propanone
d. Acetyl chloride

Group ‘B’

12. Standard reduction potentials of Cu²⁺/Cu and Fe²⁺/Fe are +0.34V and -0.44V, respectively.

a) Write down the cell notation indicating anode and cathode.
b) Calculate the standard emf of the cell.
c) Write the complete cell reaction. 

Answers:

a. Fe(s)/Fe⁺⁺(aq.)//Cu⁺⁺(aq.)/Cu(s)
Fe(s) /Fe⁺⁺ is Anode and Cu⁺⁺/Cu is cathode in Galvanic cell.

(Electrode with lower RP acts as anode while that with higher RP acts as cathode).

b. E°Cell = E° cathode – Eº anode = 0.34 – (-0.44) = 0.78V

c.

At Anode: Fe(s) → Fe⁺⁺(aq.) + 2e^–
At Cathode: Cu⁺⁺(aq) + 2e^– → Cu(s)
Overall reaction. Fe(s) + Cu ⁺⁺ (aq.) → Fe ⁺⁺(aq.) + Cu(s)

OR

Define heat of combustion. Heat of combustion of carbon (s), sulphur (s) and carbon disulphide (i) are -395 KJ/mol, -295 KJ/mol, and -1110 KJ/mol, respectively. Calculate the heat of formation of CS2(l).

Answers: It is the quantity of heat required to burn 1 mole of hydrocarbon in excess oxygen.

C + O₂ → CO₂, ∆H = -395 KJ/mol
S + O₂ → SO₂, ∆H = -295 KJ/mol
CS₂ + 3O₂ → CO₂ + 2SO₂, ∆H = -1110 KJ/mol

∆H_f = ∑∆H_f (Product) – ∑∆H_f (Reactant)
or, -1110 = [∆H_f (CO₂) + 2 ∆H_f (SO₂)] – [∆H_f (CS₂) + 3 [∆H_f (O₂)]
or, -1110 = [-395 + 2 x (-295)] – [∆H_f (CS₂) + 3 x 0]
or, -1110 = [-395 – 590] – ∆H_f (CS₂)
or, -1110 = -985 – ∆H_f (CS₂)
or, ∆H_f (CS₂) = -985 + 1110

so, ∆H_f (CS₂) = 125 KJ/mol

13. Define the term pH of a solution. Calculate the hydroxyl ion concentration in mole/litre of a solution whose pH is 4.7. Also determine the weight of NaOH required to produce these ions in one litre of the solution.

Answer: It is defined as the negative logarithim of hydrogen ion concentration present in any solution.
pH = -log[H⁺]

Given data:
pH = 4.7
pOH = 14 – 4.7 = 9.3

We have,
pOH = -log[OH^–]

On taking antilog, we get,
[OH^–] = 10^-9.3 = 5.011 x 10^-10M

Again,
Volume (V) = 1000 ml
Molarity(M) = Normality (N) = 5.011 x 10^-10M [for NaOH, M=N]
Eq. wt. (E) = 40
Weight (w) = ?

We know that,

\begin{align*}
 W &=\frac{VEN}{1000}\\ 
&=\frac{1000 \times 40\times 5.0111\times 10^{-10}}{1000}\\ 
&=2.004\times 10^{-8} \ gm
\end{align*}

14. A coinage metal (M) of electronic configuration (Ar) 3d¹⁰ 4s¹ belongs to group IB in the periodic table.
a) Draw a blast furnace for the smelting process during the extraction of (M) using its chief ore.
b) Explain the different chemical reactions involved during the formation of matte in the furnace.  

The different chemical reactions involved during the formation of matte in the furnace are as follow:

i. Unreacted FeS gets oxidized to FeO.

2FeS+3O_{2}\overset{\Delta }{\rightarrow} 2FeO+2SO_{2}

ii. Cu₂O formed by oxidation of Cu₂S reacts with FeS to form Cu₂S.

2Cu_{2}S+3O_{2}\overset{\Delta }{\rightarrow}2Cu_{2}O+2SO_{2}\\
Cu_{2}O + FeS \overset{\Delta }{\rightarrow}Cu_{2}S+FeO

iii. FeO is converted into slag by the action of sand.

FeO+\underset{Flux}{SiO_{2}}\overset{\Delta }{\rightarrow}\underset{Slug}{FeSiO_{3}}

Slag being light and molten is removed from the upper layer. At the hearth of the furnace, molten mass is obtained containing about 50% copper known as copper matte. It consists of a mixture of sulphide copper and iron.

15. a) Give a reaction for the preparation of corrosive sublimate. 

Hg+Cl_{2}\overset{\Delta }{\rightarrow}HgCl_{2}

b. What is the action of corrosive sublimate with

i) KI solution

KI+HgCl_{2}\rightarrow HgI_{2}+2KCl\\ 
HgI_{2}+\underset{excess}{HI}\rightarrow \underset{Nessler's\ reagent}{K_{2}[HgI_{4}]} 

ii. SnCl₂ solution

2HgCl_{2}+SnCl_{2}\rightarrow Hg_{2}Cl_{2}+SnCl_{4}\\ 
Hg_{2}Cl_{2}+SnCl_{2}\rightarrow 2Hg+SnCl_{4} 

iii) NH₄OH solution ?

HgCl_{2}+2NH_{4}OH\rightarrow \underset{white\ ppt.}{Hg(NH_{2})Cl}+NH_{4}Cl+2H_{2}O

c) Mention any two important uses of corrosive sublimate.
Answer: Any two important uses of corrosive sublimate are:

i. Used as antiseptic.
ii. To prepare Nessler’s reagent.
iii. Used as an antidote for poison.
iv. To preserve timbers and leathers.

16. Starting from benzene, how would you prepare nitrobenzene? 

How would you convert nitrobenzene into:
i) azoxybenzene

ii) p-aminophenol

iii. aniline

iv. azobenzene

17. i) Identify the compounds (A) and (B).

 A\xrightarrow[]{aq. \ NaOH} B\xrightarrow[O]{K_{2}Cr_{2}O_{7}/H^{+}}Propane

ii) Starting from compound (A), how would you obtain 2,3-dimethylbutane?

iii) Convert compound (B) into propane.

iv) Predict the product, when compound (A) is heated with sodium methoxide.

OR

Make a correct sequence of reactions using the suitable conditions from the following compounds. Benzoin, Toluene, Benzene diazonium chloride, Phenol, Benzaldehyde and Benzene

18. Write down a chemical equation of each of the following. 

i) Carbonylation reaction

ii) Fehling’s test

iii) Aldol condensation

iv) Williamson’s ether synthesis

v) 2, 4- DNP test

19. Write an example of each of primary (1°) and secondary (2°) alcohol. How is Victor-Meyer’s test applied to distinguish between them?

Primary alcohol = CH₃CH₂CH₂OH → Propan-1-ol

Secondary alcohol:

Victor-Meyer’s test:

i. Given alcohol is treated with PI₃ (red P₄ + I₂) to get corresponding Iodoalkane.
ii. The iodoalkane thus formed is treated with an alcoholic AgNO₂ solution to get the corresponding nitroalkane.
iii. The nitroalkane thus formed is treated with nitrous acid(HNO₂) and then with aqueous alkali.

-If blood red colour is formed, the alcohol is primary.
-If a blue colour is formed, the alcohol is secondary.

Group ‘C’

20. The concentration of the H₂SO₄ solution can be determined by the titration with standard Na₂CO₃ solution.
i) Is the Na₂CO₃ solution a primary standard? Why?

Answer: Yes, Na₂CO₃ is the primary standard because of the following:

a. It is easily available in the pure and dry state.
b. It is not highly reactive and highly hygroscopic.
c. The concentration of its solution remains constant for a long time.
d. It has a high equivalent weight which causes a minimum error during weighing.

ii) Differentiate between the equivalent point and end point.

Answer: Differences between the equivalent point and end point are as follows:

iii) 100 ml of Na₂CO₃ solution contains 0.53 g of Na₂CO₃. If 10 ml of this solution is added to ‘x’ ml of water to obtain 0.01M Na₂CO₃ solution, calculate the value of x.

solution:

For 100 ml of Na₂CO₃ solution containing 0.53 g of Na₂CO₃,

Volume(V) = 100 ml
Mass (w) = 0.53 g
Eq. wt. = 53
Normality (N) = ?
We know that,

 \begin{align*}
\text{Normality} &=\frac{W}{V}\times \frac{1000}{Eq. wt.}\\ 
&=\frac{0.53}{100} \times \frac{1000}{53}\\ 
&= 0.1\ \text{N}
\end{align*}

So, the normality of 0.01M Na₂CO₃ solution = 2 x 0.01 = 0.02 N.

From the Normality equation,

 \begin{align*}
V_{1}N_{1} &= V_{2}N_{2}\\ 
\text{or,} \ 10 \times 0.1 &= (10+x)\times 0.02\\ 
\text{or,} \ 1 &= 0.2 + 0.02x\\ 
0.02x &= 1-0.2\\ 
0.02x &= 0.8\\ 
\therefore x &= 40\ \text{ml.}

\end{align*}

OR,

a. Define the half-life of a reaction. Deduce the relation that the half-life for a reaction is directly proportional to the initial concentration of the reactant.

Answer: The half-life of a reaction is the time during which half of the initial concentration of reactant is converted into a product. It is denoted by t_1/2.

Proof:
Let us consider a general zero-order reaction. Let a be the initial concentration of a reactant at time t = 0 and after some time t, x amount of product is formed.

A → Product

The rate of the above reaction is given by:

  \begin{align*}
Rate &= K_{o}[A]^{o}\\ 
or, \ \frac{dx}{dt} &=K_{o}\\ 
or, \ \frac{dx}{dt}&= K_{o}(a-x)^{o}\\ 
or, \ dx &= K_{o}dt\\ 
Now,\\ 
\text{Integrating above equation,}\\ 
\int dx &=K_{o}\int dt\\
or, \ x &= K_{o}t + I
\end{align*}

where I is integration constant. To know its value, let us apply the initial condition i.e. when time = 0, the amount of product is also 0 i.e. x = 0. Substituting this value in the above equation, we get

  \begin{align*}
0 &= K_{o}\times 0 + I\\ 
so, \ I &= 0\\ 
\text{Hence,}\\ 
x &=K_{o}t\\ 
K_{o}&=\frac{x}{t} ---\text{eqn. i.}
\end{align*}

This is the integrated rate law equation for the zero-order reaction.

Half-life period: According to the definition, at half-life period (t_1/2), x = a/2, where a is the initial concentration of the reaction.
Substituting this value in eqn. i, we get,

  \begin{align*}
K_{o}&=\frac{a/2}{t_{1/2}}\\ 
or, \ t_{1/2} &= \frac{a}{2K_{o}}\\ 
or, \ t_{1/2}\ \alpha \ &a
\end{align*}

Hence, the half-life period of an order reaction is directly proportional to the initial concentration of the reactant.

b. For a reaction: 2NO(g) + Cl₂(g) → 2NOCl(g)

  \begin{align*}
\text{Rate} &= k[NO]^{m}[Cl_{2}]^{n}--(eqn. \ i)\\ 
0.60 &= k[0.15]^{m}[0.15]^{n}--(eqn. \ ii)\\ 
1.20 &= k[0.15]^{m}[0.30]^{n}--(eqn. \ ii)\\ 
2.40 &= k[0.30]^{m}[0.15]^{n}--(eqn. \ iii)\\ 
&\text{Dividing (i) by (ii), we get,}\\ 
\frac{0.60}{1.20}&= \frac{k[0.15]^{m}[0.15]^{n}}{k[0.15]^{m}[0.30]^{n}}\\ 
or,\ \left ( \frac{1}{2} \right )^{1} &= \left ( \frac{1}{2} \right )^{1}\\
so,\ n &= 1. \\
Now, \\
&\text{Dividing (i) by (iii), we get,}\\ 
\frac{0.60}{2.40}&= \frac{k[0.15]^{m}[0.15]^{n}}{k[0.30]^{m}[0.15]^{n}}\\ 
or,\ \left ( \frac{1}{4} \right ) &= \left ( \frac{1}{2} \right )^{m}\\
or,\ \left ( \frac{1}{2} \right )^{2} &= \left ( \frac{1}{2} \right )^{m}\\
\therefore m &= 2. \\
\end{align*}

(a) Write the expression for rate law. 

Answer: Rate = k[NO]²[Cl₂]²
Order with respect to NO is 2 and Cl₂ is 1.

(b) Calculate the value of K and specify its unit.
solution:

  \begin{align*}
\text{Rate} &= k[NO]^{2}[Cl_{2}]^{2}\\ 
or,0.60 &= k[0.15]^{2}[0.15]^{2}\\
or, k &= 177.77\  L^{2}mol^{-2}min^{-1}
\end{align*}

(c) Find the value of x. 
solution:

  \begin{align*}
\text{Rate} &= k[NO]^{2}[Cl_{2}]^{2}\\ 
or, x &= 177\times [0.25]^{2}[0.25]^{2}\\
or, x &= 2.777\  molL^{-1}min^{-1}
\end{align*}

21. a) The vapour of organic compound (A) if inhaled causes the loss of consciousness and if heated with conc. HNO₃ forms a component of tear gas.
i) How would you prepare compound (A) by using one of the isomers of C₃H₆O?

Answer: A is chloroform (CHCl₃), Isomer of C₃H₆O is acetone. 

• Preparation of bleaching powder paste.
  CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

• Chlorination of acetone.

• Hydrolysis of trichloroacetone into chloroform

Overall reaction:
CH₃COCH₃ + CaOCl₂ + H₂O → CHCl₃ + (CH₃COO)₂Ca + H₂O

ii) What happens when compound (A) is:

(a) exposed in the air?
Answer: Chloroform reacts with air in the presence of sunlight to give carbonyl chloride or phosgene which is a highly poisonous gas.

(b) heated with Ag powder
Answer: Chloroform is heated with silver powder to give acetylene gas.

(c) condensed with acetone
Answer: Chloroform reacts with acetone to give chloretone which is a hypnotic (sleep-inducing) drug.

(d) heated with aniline in the presence of alc. KOH?
Answer: When chloroform is heated with aniline in the presence of alc. KOH, Phenyl isocyanide or carbylamine is formed which is a bad-smelling compound.

B. Give reasons:

i) Ether is stored in a bottle containing iron wire.
Answer: When ether is exposed to air in the presence of sunlight, ether peroxide is formed. Ether peroxide is thermally unstable and explodes on heating. So, ether is stored in a bottle containing iron wire to prevent the formation of ether peroxide.

ii) Though ethanol is an organic compound, it is soluble in water.
Answer: Ethanol is soluble in water because it forms intermolecular hydrogen bonds with water molecules.

‘OR‘

A sequence of reactions in general form is expressed as:

Compound A is a carboxylic acid that produces ethanoic anhydride on being heated with P₂O₅.

a) Identify A, B, C, D and E with reactions involved. 

b) If the compound A undergoes the Hell-Volhard-Zelinsky reaction, write the reactions leading to the final product.

c) What product will be formed if compound B is treated with ethanol?

d. How would you convert compound E into methanol?

 \underset{Methanol}{CH_{3}OH}\xrightarrow[O]{PCC}\underset{Methanal}{HCHO}

22. a) What is Portland cement? List out the major constituents of Portland cement.
Answer: The lightweight product consisting of lime with siliceous fine material such as sand, slag or fly ash mixed with water to form a paste that has some homogeneous void or cell structure is called Portland cement.

Major constituents of Portland cement are: CaO, SiO₂, CaSO₄.2H₂O, Al₂O₃. Fe₂O₃

b)Give the structural formula of the monomer and use of each of the following:

i) Nylon 6,6
Answer: Monomers are: Adipic acid [COOH-(CH₂)₄-COOH]
Hexamethylene diamine [NH₂-(CH₂)₆-NH₂]

• Used to manufacture refrigerator and TV cabinets.

ii) Polystyrene
Answer: Monomer is Styrene: [C₆H₅-CH=CH₂]

• Used to manufacture carpets and brushes.

c) Differentiate between:

i) Nuclear reactions and chemical reactions.

ii) Fibrous and non-fibrous raw materials for paper production.