NEET CHEMISTRY question bank, MCQ CBSE Notes, Lectures

1 Titrations

Amount of oxalic acid present in a solution can be determined by its titration with KMnO₄ solution in the presence of H₂SO₄. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

1) gets oxidised by oxalic acid to chlorine
2) furnished H⁺ ions in addition to those from oxalic acid
3) reduces permanganate to Mn²⁺
4) Oxidation oxalic acid to carbon dioxide and water

Option 1

Option 2

Option 3 Solution:

Solution

The titration of oxalic acid with KMnO₄ is presence of HCl gives unsatisfactory result because of the fact that KMnO4 can also oxidise HCl along with oxalic acid. HCl on oxidation gives Cl₂ and HCl reduces KMnO4 to Mn²⁺

Option 4

2 Solutions

A certain sample of cuprous sulphide is found to have composition Cu_1.8S, because of incorporation of Cu²⁺ ions in the lattice. What is the mole %of Cu²⁺ in total copper content in this crystal?

1) 99.8%
2) 11.11%
3) 88.88%
4) None

Option 1

Option 2 Solution:

Solution

Let us consider the problem:

Two Cu⁺ replaced 1 Cu²⁺ in Cu₂S sample.

Total loss of Cu=0.2mole

Since of Cu²⁺ in total copper content =0.21.8×100=11.11%

Option 3

Option 4

3 Chemical Equilibrium

PCl₅ (g) at an initial pressure of 1 atm is placed in a closed vessel containing chlorine at a pressure of 0.2 atm. If PCl₅ dissociates to the extent of 20%, then match the following:

Column I	Column II
(A) Partial pressure of PCl₅	p. 0.4 atm
(B) Partial pressure of PCl₃	q. 0.8 atm
(C) Partial pressure of Cl₂	r. 0.10 atm
(D) Equilibrium constant (in terms of partial pressures) for dissociation of PCl₅	s. 0.2 atm.

1) A – r; B – s; C – s; D – p
2) A – q; B – s; C – r; D – p
3) A – q; B – s; C – p; D – r
4) A – p; B – s; C – q; D – r

Option 1

Option 2

Option 3 Solution:

SOLUTION

PCl_5(g) ? PCl_3(g) + Cl_2(g)

(A) Partial pressure of PCl₅ = 1 - (20/ 100) = 0.8 atm

(B) Partial pressure of PCl₃ = 20 / 100 = 0.2 atm

(C) Partial pressure of Cl₂ = 0.2 + 0.2 = 0.4 atm
(D) K_P for dissociation of PCl₅ = $$\cfrac{{{p}_{PC{{l}_{3}}}}\times {{p}_{C{{l}_{2}}}}}{{{p}_{PC{{l}_{5}}}}}\,\,=\,\,\cfrac{0.2\times 0.4}{0.8}\,\,=\,\,0.1\,atm$$

The correct answer is: A – q; B – s; C – p; D – r

Option 4

4 D and F Block Elements

The colour of the precipitate produced by adding NaOH solution to HgCl₂ is :

1) black
2) yellow
3) white
4) brown.

Option 1

Option 2 Solution:

Option 3

Option 4

5 Solid State

The number of atoms in one unit cell of a face centered cubic lattice is :

1) 1
2) 2
3) 3
4) 4

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

The number of atoms per unit cell in a simple cubic, face-centered cubic and body-centred cubic are 1,4,2 respectively.

6 Aldehydes Ketones and Carboxylic Acids

In the Friedel-Crafts acylation reaction, the electrophile is

1) R⁺
2)
3) R−C.=O
4)

Option 1

Option 2 Solution:

Solution

Friedel Crafts Acylation

Option 3

Option 4

7 Chemical Kinetics

On increasing temperature from 200K to 220K, rate constant of reaction A increases by 3 times and rate constant of reaction B increases by 9 times then correct relationship between activation energy of A and B is:

1) E_A=3_EB
2) 3E_A=E_B
3) E_B=2E_A
4) E_A=2E_B

Option 1

Option 2

Option 3 Solution:

Solution

We have Arrhenius equation for calculation of energy of activation of reaction with rate constant $$K$$ and temperature $$T$$ is

$$K=A$$ $$e^{-Ea/RT}$$ $$- (i)$$

where, $$Ea$$= Arrhenius activation energy

$$A$$= pre exponential factor

$$R$$= Ideal gas constant

According to question,

For reaction $$A$$ :-

$$T_1=200K, T_2=220K$$

$$3(r_1)_A=(r_2)_A$$

where, $$(r_1)_A$$= rate of reaction $$A$$ at $$T_1 K$$

$$(r_2)_A=$$ rate of reaction $$A$$ at $$T_2K$$

$$\because$$ Rate of reaction $$\propto$$ Rate constant

$$\Rightarrow$$ $$3(K_1)_A=(K_2)_A$$

where, $$K_1$$ and $$K_2$$ are rate constants at $$T_1$$ and $$T_2K$$

Now, $$\cfrac {(K_1)_A}{(K_2)_A}=\cfrac {1}{3}$$ $$- (ii)$$

Using $$(i)$$ in $$(ii)$$ and putting the values

$$\cfrac {(K_1)_A}{(K_2)_A}=\cfrac {1}{3}$$

$$\Rightarrow \cfrac {1}{3}= \cfrac {A\quad e^{-Ea/RT_1}}{A\quad e^{-Ea/RT_2}}$$

$$\Rightarrow \cfrac {1}{3}=$$$$ e^{-Ea/R \left(\cfrac {1}{T_2}-\cfrac {1} {T_1}\right)}$$

$$= e^{-Ea/R \left(\cfrac {1}{220}- \cfrac {1}{200}\right)}$$

$$\Rightarrow\cfrac {1}{3}= e^{+\cfrac {E_a}{R} \times \cfrac {20}{220\times 200}}$$

$$\Rightarrow \log \left( \cfrac {1}{3}\right)=\cfrac {E_a}{R}\times \cfrac {20}{220\times 200}$$

$$E_a= 2200$$ $$R$$ $$\log (1/3)$$ $$- (A)$$

For reaction $$B$$:-

$$T_1=200K$$, $$T_2=220K$$

$$\cfrac {(K_1)_B} {(K_2)_B}=\cfrac {1}{9}$$

By following the similar procedure as above,

$$\log \left(\cfrac {1}{9}\right)=\cfrac {E_b}{R}\times \cfrac {1}{2200}$$

where, $$E_b=$$ activation energy of reaction $$B$$.

$$\Rightarrow E_b=2200$$ $$R$$ $$\log\left(\cfrac {1}{9}\right)$$ $$- (B)$$

Now, from $$(A)$$ and $$(B)$$:-

$$\cfrac {E_a}{E_b}=\cfrac {log (1/3)}{log (1/9)}=\cfrac {1}{2}$$

$$\Rightarrow 2E_a=E_b$$

$$\Rightarrow 2E_A=E_B$$

Option 4

8 Surface Chemistry

The volume of gases H₂, CH₄, CO₂ and NH₃ adsorbed by 1g of charcoal at 288K are in the order:

1) H₂ >CH₄ > CO₂ > NH₃
2) CH₂> CO₂> NH₃> H₂
3) CO₂> NH₃> H₂> CH₂
4) NH₃> CO₂> CH₄> H₂

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

NH₃> CO₂> CH₄> H₂

9 Chemical Equilibrium

The standard free energy changes for the reactions:
2H_2(g) + O_2(g) ? 2H₂O_(g) and CO_(g) + H₂O_(g) ? CO_2(g) + H_2(g) are - 457.0 kJ and - 28.5 kJ respectively. Standard free energy for the reaction: 2CO₂ ? 2 CO_(g) + O_2(g) will be:

1) – 514.0 kJ
2) – 485.4 kJ
3) 484.4 kJ
4) 514.0 kJ

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

On multiplying second eq. by 2 and adding to the first eqn.:

2CO_(g) + O_2(g) ? 2CO_2(g)
ΔG = - 457.0 + 2 × (- 28.5) = - 514.0 kJ
Hence, for 2CO_2(g) ? 2CO_(g) + O_2(g), ΔG = 514.0 kJ

The correct answer is: 514.0 kJ

10 D and F Block Elements

Bauxite is concentrated by :

1) Chemical method
2) Roasting
3) Magnetic separation
4) Froth floatation process.

Option 1 Solution:

Option 2

Option 3

Option 4

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