Chemistry Test

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Chemistry Test - 3

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Weekly Quiz Competition

Question 1
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A depolarizer used in dry cell batteries is

A
A. ammonium chloride.

B
B. manganese dioxide.

C
C. potassium hydroxide.

D
D. sodium phosphate.

Solution

Depolarizer is a substance, which is used in primary cells to prevent a build-up of hydrogen gas bubble.

A battery depolarizer absorbs electrons during discharge of a cell, so it acts as an oxidising agent.

In dry cells, manganese dioxide act as a depolariser.
Question 2
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Which of the following is the correct order of thermal stability of hydrides?

A
A. NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃

B
B. PH3₃ > NH₃ > AsH₃ > SbH₃ > BiH₃

C
C. BiH₃ > SbH₃ > NH₃ > AsH₃ > PH₃

D
D. NH₃ > BiH₃ > SbH₃ > AsH3₃ > PH₃

Solution

Thermal stability is the stability of the molecule towards heat. Thermal stability of hydrides decreases down the group, as on going down the group, the size of the element E (side element with hydrogen) increases. So, the E−H bond length increases, hence- bond strength becomes weak and breaks easily to liberate hydrogen gas.

Thus, the order of thermal stability of hydrides is NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
Question 3
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Which undergoes catalytic hydrogenation much faster?

A
A.

B
B.

C
C. equally

D
D. cannot be predicted

Solution

he reagent metal/Hydrogen is used in the catalytic hydrogenation. It is a syn addition to the pi-bond.

Hence, the reactivity depends on the steric hindrance of the groups present in the molecule.

The double bond present in

is less sterically hindered than
Question 4
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The temporary hardness of water due to calcium bicarbonate and can be removed by adding

A
A. CaCO₃

B
B. Ca(OH)₂

C
C. CaCl₂

D
D. HCl

Solution

Lime treatment removes bicarbonate hardness by forming insoluble CaCO₃ as,
Question 5
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KO₂ is used in space and submarines because it,

A
A. absorbs CO₂ and increases O2 concentration.

B
B. absorbs moisture.

C
C. absorbs CO₂ .

D
D. produces ozone.

Solution

KO₂ absorbs CO₂ and increases O₂ concentration, so it is used in space and submarines.
Question 6
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Which of the following oxides can act both as an oxidising agent as well as reducing agent?

A
A. N₂ O

B
B. SO₂

C
C. SO₃

D
D. P₂ O₅

Solution

SO₂ oxides can act both as an oxidising agent as well as reducing agent. It behaves as a reducing agent in moist conditions. It converts iron (III) ions to iron (II) ions and decolourises acidified potassium permanganate (VII) solution.

2Fe³⁺ + SO₂ + 2H₂ O → 2Fe²⁺ + SO^2-₄ + 4H⁺

When SO₂ reacts with H₂ S, it produces sulphur and water. In this reaction, ‘SO₂ ’ act as an oxidising agent.
Question 7
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Which among the following is correct for electrolysis of brine solution?

A
A. O₂ gas liberated at cathode

B
B. Sodium metal is collected at anode

C
C. H₂ gas is liberator! at cathode

D
D. Cl₂ gas is Liberated at cathode

Solution

In electrolysis of brine solution, H₂ gas is liberated at cathode. The solution contains four ions Na⁺ , Cl⁺ , H⁺ and OH ions. There occurs a race amongst them for their discharge at their respective electrodes.

Following electrode reactions are possible:

In this electrolysis, H₂ at cathode and Cl₂ , at anode are given off.
Question 8
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The elevation in boiling point of 0.25 molal aqueous solution of a substance is (K_b= 0.52 K kg mol^-1)

A
A. 0.15K

B
B. 0.50K

C
C. 0.13K

D
D. 2.08K

Solution

Key Idea Elevation In boiling is directly proportional to molar concentration of solute in a solution ΔT_b ∝ m, ΔT_b = K_b m

Given, m = 0.25

K_b = 0.52 K kg mol^-1

The elevation in boiling point (ΔT_b ) of 0.25 molal aqueous solution of a substance is

ΔT_b = Kb x m =0.52 x 0.25 = 0.13K
Question 9
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If a metal crystallises in bcc structure with edge length of unit cell 4.29 x 10^-8cm the radius of metal atom is

A
A. 3.2 x 10^-7 cm

B
B. 1.86 x 10^-8 cm

C
C. 1.07 x 10^-7 cm

D
D. 1.07 x 10^-8 cm

Solution

Key idea The relation between edge length and radius of an atom in a bee structure is related as: r = √3/4d

Given, a = 4.29 x 10^-8 cm

For bcc structure,

= 1.857 x 10^-8 cm = 1.86 x 10^-8 cm
Question 10
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The element which does not belong to group 15 is

A
A. As

B
B. P

C
C. Bi

D
D. Se

Solution

The element which does not belong to group 15 is Se (selenium). It belongs to group 16 of the periodic table. It resides in the 4th period and 16th group having atomic number 34. Electronic configuration of Se is 4s² 4p⁴ .
Question 11
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In leaching of alumina from bauxite by Bayer 's process, then ore is treated with

A
NaOH(aq)

B
NaCN(aq)

C
KCN (aq)

D
Na₂ CO₃ (aq)

Solution

In leaching of alumina from bauxite by Bayer 's process, the ore is treated with NaOH(aq). Concentration is carried out by heating the powdered ore with a concentrated solution of NaOH at 473 - 523 K and 35-36 bar pressure. This process is called digestion. In this reaction, AfeOa Al₂ O₃ (s) + 2NaOH(aq) + 3H₂ O(l) →2Na[Al(OH)₄ ](aq)

Tho sodium aluminate formed is neutralised by passing CO₂ gas and hydrated Al₂ O₃ is precipitated

2Na[Al(OH)₄ ](aq) + CO₂ (g) →Al₂ O₃ .xH₂ O(s) + 2NaHCO₃ (aq)
Question 12
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Carbolic add is oxidised by acidified sodium dichromate to give

A
A. acetone

B
B. ethyl methyl ketone

C
C. benzoquinone

D
D. anthraquinone

Solution

Carbolic acid (also known as phenol) is oxidised by acidified sodium dichromate to give ben 20 quin one. In the presence of air, phenols are slowly oxidised to dark coloured mixtures containing quinones.
Question 13
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In the electrolysis of aqueous sodium chloride with inert electrodes the products obtained at anode and cathode respectively are

A
A. Cl₂ and Na

B
B. O₂ and Na

C
C. Cl₂ and H₂

D
D. Na and Cl₂

Solution

In the electrolysis of aqueous sodium chloride with inert electrodes, the products obtained at anode and cathode respectively are Cl₂ and H₂ .

The solution contains four ions Na⁺ , Cl^- , H⁺ , OH.

Following reaction takes place:
Question 14
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The integrated rate equation for first order reaction, A →product is

A
A.

B
B.

C
C.

D
D.

Solution

The integrated rate equation for first order reaction, A → product is k =

For 1st order reactions, R → P

Rate = -d(R)/dt = k[R]

On integrating this elation, we get

In [R] = -kt + l …(i)

At t = 0, R = [R]₀ , In[R]₀ = -k x 0 + l

In[R]₀ = l

Substituting the value of I in (i) and on rearrangement, we get

In[R] = -kt + InR₀ And, k = 1/t in [R₀ ]/[R]
Question 15
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Which mixture is used for respiration by deep sea divers?

A
A. Ne + O₂

B
B. Kr + O₂

C
C. Ar + O₂

D
D. He + O₂

Solution

Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure, underwater. Increased pressure increases the solubility of atmospheric gases in blood. When divers come towards the surface, the pressure gradually decreases. This blocks capillaries and creates a medical condition called bends. To avoid bends, the tanks are used by scuba divers that are filed with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen). Thus, option (d) is correct.
Question 16
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Which among the following statements is true about the Schottky defect?

A
A. In this defect cation and anion are lacking in stoichiometric proportion

B
B. Formation of metal alloy is example of this defect

C
C. In this cation or anion moves from regularsite to place between lattice site

D
D. in this regular cation is replaced by different cation

Solution

In Schottky defect, cation and anion are lacking in stoichiometric proportion. It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal. This defect can easily be illustrated by the following figure.
Question 17
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Soaps are the sodium or potassium salts of higher fatty acids, containing number of carbon atoms more than,

A
A. 10

B
B. 6

C
C. 8

D
D. 12

Solution

Soaps are the sodium or potassium salts of higher safety acids, containing a number of carbon atoms more than 12, e.g. stearic, oleic and palmitic acids. These are prepared by heating fat with aqueous sodium hydroxide solution. This reaction is known as saponification.

E.g.