Chemistry Test-10

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Chemistry Test-10

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Weekly Quiz Competition

Question 1
4 / -1

What is the product of the reaction?

A
\(\mathrm{CH}_{3} \mathrm{CHO}\)

B

Racemic \((2 \mathrm{R}, 3 \mathrm{R})\) and \((2 \mathrm{~S}, 3 \mathrm{~S})-2,3-\) butanediol

C
Meso\(-2,3-\)butanediol

D
Cis\( -2 - 3 -\)epoxybutane

Solution

The two oxygens from the \(\mathrm{OsO}_{4}\) interact with one face of the double bond. This results in a 5-membered ring (called an osmate ester) and generates the Cis stereochemistry. The osmate ester is broken up into the Vicinal diol by use of a reducing agent. The reaction leads to the formation of syn diols.
The given alkene is symmetrical and the product of reaction \(\mathrm{Cis}-2,3-\) Butanediol has an internal mirror plane, therefore, it forms Meso \(-2,3-\) butanediol.
The two oxygens from the \(\mathrm{OsO}_{4}\) interact with one face of the double bond. This results in a 5-membered ring (called an osmate ester) and generates the Cis stereochemistry. The osmate ester is broken up into the Vicinal diol by the use of a reducing agent. The reaction leads to the formation of syn diols.
The given alkene is symmetrical and the product of reaction \(\mathrm{Cis}-2,3-\) Butanediol has an internal mirror plane therefore it forms Meso \(-2,3-\) butanediol.
Question 2
4 / -1

The number of water molecules is maximum in:

A
\(18\) grams of water

B
\(18\) moles of water

C
\(18\) molecules of water

D
\(1.8\) grams of water

Solution

According to Avogadro's law, The number of water molecules is maximum in 18 moles of water.

\(1\) mole of every substance occupies \(22.4 {~L}\) at STP and contains Avogadro's number \(6.023 \times 10^{23}\) of particles.

18 grams of water:

Number of moles \(=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{18 g}{18 g / m o l}=1\) mole

1 mole of \(H_{2} O\) contains \(=6.023 \times 10^{23}\) molecules of water.

18 moles of water:

Since, 1 mole of \(H_{2} O\) contains \(=6.023 \times 10^{23}\) molecules of water. 18 moles of \(H_{2} O\) contains \(=\frac{6.023 \times 10^{23}}{1} \times 18=108.4 \times 10^{23}\)molecules of water.

18 molecules of water:

It already has 18 molecules so, it is not needed to calculate.

1.8 gram of water:

Number of moles \(=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{1.8 g}{18 q / {mol}}=0.1\) mole

1 mole of \(H_{2} O\) contains \(=6.023 \times 10^{23}\) molecules of water.

0.1 mole of \(H_{2} O\) contains \(=\frac{6.023 \times 10^{23}}{1} \times 0.1=6.023 \times 10^{22}\)molecules of water.
Question 3
4 / -1

In analogy to O₂⁺[PtF₆]^–, a compound N₂⁺[PtF₆]^– will not be formed because-

A
The I.E. of N₂ gas is higher than that of N⁻atom

B
The I.E. of N₂ gas is lower than that of O₂ gas

C
The I.E. of N₂ gas is higher than that of O₂ gas

D
None of these

Solution

In analogy to O₂⁺[PtF₆]⁻ a compound N₂⁺[PtF₆]⁻ will not be formed because the I.E. of N₂gas higher than that of O₂gas. The I.E. of N₂ gas is higher than that of O₂. The I.E. is decreasing from left to right in the periodic table hence the N₂ cannot lose its electrons so easily as O₂ in forming O₂⁺[PtF6]⁻.
Question 4
4 / -1

Which one of the following regions of the atmosphere contains ozone?

A
Troposphere

B
Thermosphere

C
Mesosphere

D
Stratosphere

Solution

Ozone is present in the stratosphere (about 20 km above the surface of the earth).
Most ozone resides in the lower stratosphere in what is commonly known as the “ozone layer.” The remaining ozone, about 10%, is found in the troposphere, which is the lowest region of the atmosphere, between Earth's surface and the stratosphere.
Question 5
4 / -1

The IUPAC name for the complex [Co(NO₂)(NH₃)₅]Cl₂ is

A
Nitrito -N- pentaamminecobalt(III)chloride

B
Nitrito-N-pentaamminecobalt(II)chloride

C
Pentaamminenitrito-N-cobalt(II)chloride

D
Pentaamminenitrito-N-cobalt(III)chloride

Solution

[Co(NO₂)(NH₃)₅]CI₂
The IUPAC name of the given coordination complex according to the nomenclature rules is Pentaamminenitrito-N-cobalt(III)chloride.
Question 6
4 / -1

Consider the following reaction:
\(\mathrm{xMnO}_{4}^{-}+\mathrm{y} \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{zH}^{+} \rightarrow \mathrm{xMn}^{2+}+2 \mathrm{y} \mathrm{CO}_{2}+\frac{\mathrm{z}}{2} \mathrm{H}_{2} \mathrm{O}\)
The values of \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z}\) in the reaction are, respectively:

A
\(2,5,16\)

B
\(5,3,8\)

C
\(1,3,6\)

D
\(2,6,6\)

Solution

Reaction:
\(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)
The above reaction is the balanced chemical reaction. Thus the value of \(\mathrm{x}, \mathrm{y}\), and \(\mathrm{z}\) is \(2,5,\) and \(16 .\)
Question 7
4 / -1

Direction: Choose the correct option according to the given Assertion[A] and Reason[R].
Assertion [A]: KMnO₄ is a stronger oxidizing agent than K₂Cr₂O₇.
Reason [R]: This is due to the increasing stability of the lower species to which they are reduced.

A
Both [A] and [R] are correct and [R] is the correct explanation of [A]

B
Both [A] and [R] are correct but [R] is not the correct explanation of [A]

C
[A] is correct but [R] is incorrect

D
[A] is incorrect but [R] is correct

Solution

Both [A] and [R] are correct and [R] is the correct explanation for [A].
In KMnO₄, the oxidation state of Mn is +7 (highest oxidation state) and is reduced to Mn²⁺, whereas in K₂Cr₂O₇, the oxidation state of Cr is +6 (highest oxidation state) and is reduced to Cr³⁺.
Mn²⁺ is more stable than Cr³⁺. Therefore, KMnO₄ is a stronger oxidizing agent than K₂Cr₂O₇.
Question 8
4 / -1

If the solubility of \({Ag}_{2} \mathrm{CrO}_{4}\) is \({s}\) moles/litre, its solubility product will be ___________.

A

\({s}^{2}\)

B

\({s}^{3}\)

C

\({4s}^{3}\)

D

\({2s}^{3}\)

Solution

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as \(K_{s p}\).
The equation for the ionization of the silver chromate is given as:
\(A g_{2} C r O_{4} \leftrightharpoons 2 A g^{+}+C r O_{4}^{2-}\)
We are given:
Solubility of \(A g_{2} C r O_{4}=s \mathrm{~mol} / \mathrm{L}\)
By stoichiometry of the reaction:
\(1\) mole of \(A g_{2} C r O_{4}\) gives 2 moles of \(A g^{+}\) and 1 mole of \(C r O_{4}^{2-}\).
When the solubility of \(A g_{2} C r O_{4}\) is \(s\) moles/liter, then the solubility of \(A g^{+}\) will be \(2 {~s}\) moles/liter and solubility of \(\mathrm{Cr} \mathrm{O}_{4}^{2-}\) will be \({s}\) moles/liter.
Expression for the equilibrium constant of \(A g_{2} C r O_{4}\) will be:
\(\begin{array}{l}
K_{s p}=\left[A g^{+}\right]^{2}\left[C r O_{4}^{2-}\right] \\
K_{s p}=[2 s]^{2}[s]=4 s^{3}
\end{array}\)
Thus, the solubility product of \(A g_{2} C r O_{4}\) is \(4 s^{3}\).
Question 9
4 / -1

For the reversible reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) 500^{\circ} \mathrm{C},\) the value of \(\mathrm{Kp}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmosphere. The corresponding value of \(\mathrm{K}_{\mathrm{c}},\) with concentration in mole litre \(^{-1}\), is -

A

\(\frac{1.44 \times 10^{-7}}{[(0.082 \times 773)^{-2}]}\)

B

\(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

C

\(\frac{1.44 \times 10^{-5}}{[(8.314 \times 773)^{-2}]}\)

D

\(\frac{1.44 \times 10^{-5}}{[(0.082\times 500)^{-2}]}\)

Solution

\(\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}\left\{\begin{array}{l}\mathrm{n}_{\mathrm{p}}=\text { number ot moles of products (gaseous) } \\ \mathrm{n}_{r}=\text { number of moles of reactants (gaseous) } \\ \qquad \Delta \mathrm{n}_{\mathrm{g}}=2-4=-2\end{array}\right.\)
\(\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{(\mathrm{RT})^{\Delta n_{\mathrm{g}}}}\)
\(\mathrm{Kc}=\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\left(\mathrm{R}\right.\) in \(\mathrm{L}\) atm mole \(\left.{ }^{-1} \mathrm{~K}^{-1}\right)\)
Question 10
4 / -1

Which of the following is NOT a property of amino acids?

A
Colorless

B
Crystalline Solids

C
Water Insoluble

D
It behave like salts

Solution

Water Insolubleis NOT a property of amino acids.
Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule.
The equal number of amino and carboxyl groups makes it neutral; more number of amino than carboxyl groups makes it basic and more carboxyl groups as compared to amino groups makes it acidic.
The amino acids, which can be synthesized in the body, are known as nonessential amino acids.
On the other hand, those which cannot be synthesized in the body and must be obtained through diet, are known as essential amino acids.
Amino acids are usually colourless and crystalline solids.
These are water-soluble, high melting solids and behave like salts rather than simple amines or carboxylic acids.
Question 11
4 / -1

A bee-sting leaves an acid that causes pain and irritation. The injected acid is ________.

A
Acetic acid

B
Sulphuric acid

C
Citric acid

D
Methanoic acid

Solution

A bee-sting leaves an acid that causes pain and irritation. The injected acid is Methanoic acid.
Nettle sting and bee-sting have methanoic acid commonly called formic acid (HCOOH).
The leaves and young stems of this herbaceous plant are fitted with stinging hairs tipped with formic acid and other irritants.
If touched, these needle-like hairs inject the stinging acid into the skin, triggering a burning, tingling sensation and an itchy rash.
Acetic Acid:-
Chemical formula:- CH₃COOH.
IUPAC name:- Ethanoic acid
Anhydrous Acetic acid is called Glacial Acetic acid.
Its crystals are obtained at a temperature slightly lower than 16.7°C.
Sulphuric Acid:-
It is commonly known as King of Chemicals.
it is used in the preparation of a very large number of other useful chemicals such as hydrochloric acid, nitric acid, dyes, drugs
Sulfuric acid is used in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate.
It is also used as an electrolyte in lead-acid batteries.
Citric acid:-
The fruits containing citric acid are termed citrus fruits. eg: lemon, orange, grapefruit, etc.
These are sour in taste and are a good source of Vitamin C, thus help in preventing the disease caused by the deficiency of Vitamin C, i.e., Scurvy.
Question 12
4 / -1

The material used for a fuse wire should have ________.

A
High resistivity

B
High melting point

C
Low melting point

D
High ductility

Solution

The material used for a fuse wire should have Low melting point.

The material used for fuse elements must be of low melting point, low ohmic loss, high conductivity, low cost, and free from detraction.

The material used for making fuse element has a low melting point such as tin, lead, or zinc.

An alloy of lead and tin has a low melting point and a very high resistance.

Fuse is a safety device consisting of a wire that melts and breaks an electric circuit if the current exceeds a safe level.

It is made of zinc, copper, silver, aluminium, or alloys.
Question 13
4 / -1

Which among the following liquids are conductors of electricity?

A. Vinegar

B. Sugar solution

C. Salt solution

D. Distilled water

A
C and D

B
B and D

C
A, C and D

D
A and C

Solution

The only Vinegar and Salt solutions are capable to conduct electricity.

The conduction of electricity is possible through any liquid if it has dissolved ions in the liquid.

Vinegar: Acetic acid is commonly known as Vinegar. It is a weak acid. When it dissolves in water, it breaks into its ions. So make a conductive solution for electricity.

CH₃COOH + H₂O ⇔ CH₃COO- + H₃O+

Sugar solution: Sugar is an organic compound. When it dissolved in water it does not break into the ions. So the sugar solution does not conduct electricity.

Salt solution: Salts are made up of ions, so when they dissolved in water breaks into ions and made a conductive solution with water.

Distilled water: Distilled water has no free ions, So it is not conductive.
Question 14
4 / -1

Which of the following chemical compound is also known as Lunar Caustic?

A
Silver Nitrate

B
Sodium Hydroxide

C
Hydrochloride Acid

D
Calcium Oxide

Solution

Silver Nitrate is an inorganic compound also known as Lunar Caustic with the formula of AgNO₃.

It is used in photography. It is also less light-sensitive than halides.

It was once called lunar caustic, since the ancient alchemists, who linked silver with the moon, called silver luna.

In solid silver nitrate, in a trigonal planar structure, silver ions are three-coordinated.

Silver nitrate can be treated with nitric acid by reacting with silver, such as silver bullion or silver foil, resulting in silver nitrate, water, and nitrogen oxides. The by-products of the reaction depend on the nitric acid concentration used.

3 Ag + 4 HNO₃ (cold and diluted) → 3 AgNO₃+ 2 H₂O + NO

Ag + 2 HNO₃ (hot and concentrated) → AgNO₃ + H₂O + NO₂

Question 15

4 / -1

Which of the following polymer is used to make Bullet proof glass?

Glyptal

PMMA

Teflon

None

Solution

The PMMA polymer is used to make Bullet proof glass.

It is a synthetic polymer.
Its full name is Polymethyl methacrylate (PMMA).
It is rigid and transparent.
It is used as a substitute for glass in products such as shatterproof windows, skylights, illuminated signs, and aircraft canopies.
Kevlar is a heat-resistant and strong synthetic fiber.

Usage of other polymers:

Polymer	Uses of Polymer
BUNA – S	Synthetic rubber
BUNA – N	Synthetic rubber
Teflon	Non-stick cookware – plastics
Terylene	Fabric
Glyptal	Fabric
Bakelite	Plastic switches, Mugs, buckets
PVC	Tubes, Pipes
Melamine Formaldehyde Resin	Ceramic plastic material
Nylon-6	Fabric

Polymer	Uses of Polymer
BUNA – S	Synthetic rubber
BUNA – N	Synthetic rubber
Teflon	Non-stick cookware – plastics
Terylene	Fabric
Glyptal	Fabric
Bakelite	Plastic switches, Mugs, buckets
PVC	Tubes, Pipes
Melamine Formaldehyde Resin	Ceramic plastic material
Nylon-6	Fabric

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Chemistry Test-10

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Question 1

\(\mathrm{CH}_{3} \mathrm{CHO}\)

Racemic \((2 \mathrm{R}, 3 \mathrm{R})\) and \((2 \mathrm{~S}, 3 \mathrm{~S})-2,3-\) butanediol

Meso\(-2,3-\)butanediol

Cis\( -2 - 3 -\)epoxybutane

Solution

Question 2

\(18\) grams of water

\(18\) moles of water

\(18\) molecules of water

\(1.8\) grams of water

Solution

Question 3

The I.E. of N2 gas is higher than that of N−atom

The I.E. of N2 gas is lower than that of O2 gas

The I.E. of N2 gas is higher than that of O2 gas

None of these

Solution

Question 4

Troposphere

Thermosphere

Mesosphere

Stratosphere

Solution

Question 5

Nitrito -N- pentaamminecobalt(III)chloride

Nitrito-N-pentaamminecobalt(II)chloride

Pentaamminenitrito-N-cobalt(II)chloride

Pentaamminenitrito-N-cobalt(III)chloride

Solution

Question 6

\(2,5,16\)

\(5,3,8\)

\(1,3,6\)

\(2,6,6\)

Solution

Question 7

Both [A] and [R] are correct and [R] is the correct explanation of [A]

Both [A] and [R] are correct but [R] is not the correct explanation of [A]

[A] is correct but [R] is incorrect

[A] is incorrect but [R] is correct

Solution

Question 8

\({s}^{2}\)

\({s}^{3}\)

\({4s}^{3}\)

\({2s}^{3}\)

Solution

Question 9

\(\frac{1.44 \times 10^{-7}}{[(0.082 \times 773)^{-2}]}\)

\(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

\(\frac{1.44 \times 10^{-5}}{[(8.314 \times 773)^{-2}]}\)

\(\frac{1.44 \times 10^{-5}}{[(0.082\times 500)^{-2}]}\)

Solution

Question 10

Colorless

Crystalline Solids

Water Insoluble

It behave like salts

Solution

Question 11

Acetic acid

Sulphuric acid

Citric acid

Methanoic acid

Solution

Question 12

High resistivity

High melting point

Low melting point

High ductility

Solution

The I.E. of N₂ gas is higher than that of N⁻atom

The I.E. of N₂ gas is lower than that of O₂ gas

The I.E. of N₂ gas is higher than that of O₂ gas