Chemistry Test 262

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Chemistry Test 262

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Weekly Quiz Competition

Question 1
4 / -1

Name the end product in the following series of reactions

A
Aniline

B
Benzylamine

C
Acetonitrile

D
Acetamide

Solution
Question 2
4 / -1

Which of the following does not illustrate the anomalous properties of lithium?

A
Li is much softer than the other alkali metals

B
The melting point and boiling point of Li are comparatively high

C
Li forms a nitride Li₃N unlike the other alkali metals

D
The ions of L_i and its compounds are more heavily hydrated than those of the rest of the alkali metal elements

Solution

The strength of the metallic bond decreases down the group. The hardness of these alkali metals decreases down the group. Hence, 'Li' is harder than other alkali metals.
Question 3
4 / -1

The vapour of Hg absorb some electrons accelerated by a potential difference of 4.5 volts from rest as a result of which light is emitted. If the full energy of a single incident electron is supposed to be converted into light emitted by electron in a single Hg atom, find the wave number (1/λ) of the light

A
3.63 × 10⁶ m⁻¹

B
5.93 × 10⁶m⁻¹

C
5.93 × 10⁶cm⁻¹

D
5.62 × 10⁶m⁻¹

Solution

Potential difference 4.5 V

∴ Potential energy = 4.5 eV energy absorbed and this energy is emitted by electron of Hg atom.
Question 4
4 / -1

Identify the correct statement:

A
Corrosion of iron can be minimised by forming a contact with another metal with a higher reduction potential

B
Iron corrodes in oxygen-free water

C
Corrosion of iron can be minimised by forming an impermeable barrier at its surface

D
Iron corrodes more rapidly in salt water because its electrochemical potential is higher

Solution

Corrosion of iron can be minimised by forming a contact with another metal with a lower reduction potential. Corrosion of iron can be minimised by forming an impermeable barrier at its surface.
Question 5
4 / -1

A crystalline hydrated salt on being heated loses 45.6% of its weight. Percentage composition of anhydrous salt is Al - 10.5 %, K - 15.1 %, S - 24.8 %, O - 49.6 %. Find molecular formula of salt (hydrated) (M = 948 g/mol for hydrated salt).

A
K₂SO₄.Al₂(SO₄)₃.24H₂O

B
K₂SO₄.Al₂(SO₄)₃.12H₂O

C
K₂SO₄.Al₂(SO₄)₃.6H₂O

D
K₂SO₄.Al₄(SO₄)₆.48H₂O

Solution
Question 6
4 / -1

Which one of the following ores is best concentrated by froth flotation method?

A
Galena

B
Cassiterite

C
Magnetite

D
Malachite

Solution

Froth flotation method is used for concentration of sulphide ores.

Galena → PbSGalena → PbS

Malachite is Cu₂(CO₃)₂(OH)₂

Magnetite is Fe₃O₄

Siderite is FeCO₃
Question 7
4 / -1

Which of the following is a biradical?

A
Phenyl

B
Propenyl

C
Vinylidene

D
Ethylidyne

Solution
Question 8
4 / -1

At 500 K, the half life period of a gaseous reaction at an initial pressure of 80 kPa is 350 s350 s. When the pressure is 40 kPa, the half life period is 175 s; the order of the reaction is :

A
zero

B
one

C
two

D
three

Solution
Question 9
4 / -1

Consider the following three halides :

(1) CH₃ − CH₂ − Cl

(2) CH₂ = CH − Cl

(3) C₆H₅ − Cl

Arrange C − Cl bond length of these compounds in decreasing order -

A
1 > 2 > 3

B
1 > 3 > 2

C
3 > 2 > 1

D
2 > 3 > 1
Solution
The C–Cl bond length depends on the hybridization of the carbon atom bonded to the chlorine atom. The greater the s-character of the hybrid orbital in the carbon, the shorter the C–Cl bond length.
1. Compound (1): CH₃–CH₂–Cl
  
  The carbon bonded to chlorine is sp³-hybridized (single bond).
  
  An sp³-hybridized carbon has the least s-character (25%), resulting in the longest bond length.
2. Compound (2): CH₂=CH–Cl
  
  The carbon bonded to chlorine is sp²-hybridized (double bond).
  
  An sp²-hybridized carbon has 33% s-character, leading to a shorter C–Cl bond than sp³.
3. Compound (3): C₆H₅–Cl (chlorobenzene)
  
  The carbon bonded to chlorine is sp²-hybridized, but it is part of the benzene ring.
  
  Due to resonance in the benzene ring, the C–Cl bond has partial double bond character, making it the shortest bond length among the three.
Conclusion: The bond length order is:

Compound (1) > Compound (2) > Compound (3).

Thus, the correct answer is (a) 1 > 2 > 3.
Question 10
4 / -1

Ratio of ΔT_b/K_b of 10 g AB₂ and 14 g A₂B per 100 g of solvent in their respective, solution (AB₂ and A₂B both are non-electroytes) is 1 mol/kg in both cases. Hence, atomic weight of A and B are respectively :

A
100, 40

B
60, 20

C
20, 60

D
None of these

Solution