Chemistry

271. Which one of the following has different number of molecules ? (All ar

Which one of the following has different number of molecules ? (All are kept at normal temperature and pressure)

[amp_mcq option1=”3 gram of Hydrogen” option2=”48 gram of Oxygen” option3=”42 gram of Nitrogen” option4=”2 gram of Carbon” correct=”option4″]

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To determine which option has a different number of molecules, we need to calculate the number of moles for each substance. The number of molecules is directly proportional to the number of moles (Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules; here we are comparing masses).
– A) 3 gram of Hydrogen (H₂): Molar mass ≈ 2 g/mol. Moles = 3 g / 2 g/mol = 1.5 moles.
– B) 48 gram of Oxygen (O₂): Molar mass ≈ 32 g/mol. Moles = 48 g / 32 g/mol = 1.5 moles.
– C) 42 gram of Nitrogen (N₂): Molar mass ≈ 28 g/mol. Moles = 42 g / 28 g/mol = 1.5 moles.
– D) 2 gram of Carbon (C): Molar mass ≈ 12 g/mol. Moles = 2 g / 12 g/mol = 1/6 moles ≈ 0.167 moles.
Options A, B, and C all contain 1.5 moles (and thus the same number of molecules). Option D contains a significantly different number of moles (and thus atoms, as carbon exists as atoms in its elemental solid form).
– The number of molecules in a given mass of a substance is proportional to the number of moles, which is calculated as Mass / Molar Mass.
– Avogadro’s number (approximately 6.022 x 10²³) represents the number of particles (atoms, molecules, etc.) in one mole of a substance.
The condition “All are kept at normal temperature and pressure” (NTP) is relevant for comparing volumes of gases using Avogadro’s law, but here we are comparing masses and thus the number of moles/molecules directly using molar masses. Note that elemental carbon is a solid at NTP, while hydrogen, oxygen, and nitrogen are gases (diatomic molecules H₂, O₂, N₂). The question phrasing “number of molecules” for carbon is slightly imprecise as solid carbon consists of atoms in a lattice, but the underlying principle is comparing the amount of substance (moles).

272. There are six electrons, six protons and six neutrons in an atom of an

There are six electrons, six protons and six neutrons in an atom of an element. What is the atomic number of the element ?

[amp_mcq option1=”6″ option2=”12″ option3=”18″ option4=”24″ correct=”option1″]

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The atomic number of an element is defined by the number of protons found in the nucleus of an atom of that element. The question states that the atom has six protons. Therefore, the atomic number of the element is 6.
– Atomic Number (Z) = Number of Protons.
– In a neutral atom, Number of Protons = Number of Electrons.
– The mass number is the total number of protons and neutrons in the nucleus (Mass Number = Protons + Neutrons).
– In this case, Protons = 6, Electrons = 6, Neutrons = 6.
– Atomic Number = 6.
– Mass Number = 6 + 6 = 12. This describes an atom of Carbon-12.
The number of neutrons can vary for atoms of the same element, leading to different isotopes. However, the number of protons is unique to each element and determines its identity.

273. Identify the element having zero valency

Identify the element having zero valency

[amp_mcq option1=”Sulphur” option2=”Phosphorous” option3=”Lead” option4=”Radon” correct=”option4″]

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Valency is the combining capacity of an element. Elements with zero valency are those that are chemically inert and do not readily form chemical bonds under normal conditions because their outermost electron shell is full. These elements are the noble gases. Among the given options, Radon (Rn) is a noble gas.
– Sulphur (S) typically has valencies of 2, 4, or 6.
– Phosphorous (P) typically has valencies of 3 or 5.
– Lead (Pb) typically has valencies of 2 or 4.
– Radon (Rn) is a noble gas (Group 18 of the periodic table) and its valency is considered zero.
Noble gases (Helium, Neon, Argon, Krypton, Xenon, and Radon) have a stable electron configuration with a full outermost electron shell (except Helium, which has 2 electrons in its single shell), making them very unreactive.

274. Match List I with List II and select the correct answer using the code

Match List I with List II and select the correct answer using the code given below the Lists :

List I
(Process)		 List II
(Type of change)
A. Heating Camphor 1. Chemical
B. Cooling of water vapour up to room temperature 2. Evaporation
C. Cooking an egg 3. Condensation
D. Formation of water vapour at room temperature 4. Sublimation

Code :

[amp_mcq option1=”A-4, B-3, C-1, D-2″ option2=”A-4, B-1, C-3, D-2″ option3=”A-2, B-1, C-3, D-4″ option4=”A-2, B-3, C-1, D-4″ correct=”option1″]

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Matching the process to the type of change:
– A. Heating Camphor: Camphor changes directly from solid to gas upon heating, a process called sublimation. This is a physical change. (A-4)
– B. Cooling of water vapour up to room temperature: Water vapour (gas) changes into liquid water upon cooling, a process called condensation. This is a physical change. (B-3)
– C. Cooking an egg: Cooking an egg involves the denaturation of proteins, which is an irreversible chemical change where new substances with different properties are formed. (C-1)
– D. Formation of water vapour at room temperature: Liquid water changes into water vapour (gas) at room temperature through evaporation. This is a physical change. (D-2)
The correct match is A-4, B-3, C-1, D-2.
– Physical changes alter the form or appearance of a substance but not its chemical composition (e.g., melting, boiling, freezing, condensation, sublimation, dissolution).
– Chemical changes result in the formation of new substances with different chemical properties (e.g., cooking, burning, rusting, digestion).
Sublimation is a less common phase transition where a substance goes directly from solid to gas without passing through the liquid phase. Deposition is the reverse process (gas to solid).

275. The alkali metals have relatively low melting point. Which one of the

The alkali metals have relatively low melting point. Which one of the following alkali metals is expected to have the highest melting point?

[amp_mcq option1=”Li” option2=”Na” option3=”K” option4=”Rb” correct=”option1″]

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The alkali metals are in Group 1 of the periodic table: Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs), and Francium (Fr). Moving down the group, the metallic bonding strength generally decreases due to the increasing atomic size and shielding effect, which weakens the attraction between the positive ions and the delocalized valence electrons. Weaker metallic bonding leads to lower melting points. Therefore, the melting points of alkali metals decrease down the group. Among the options provided (Li, Na, K, Rb), Lithium is at the top of the group and is expected to have the highest melting point.
Melting points of alkali metals decrease down the group (from Li to Fr) due to decreasing metallic bond strength.
The approximate melting points are: Li (180.5 °C), Na (97.8 °C), K (63.5 °C), Rb (39.3 °C), Cs (28.5 °C). As expected, Lithium has the highest melting point among the given options. This trend is a typical example of how physical properties change across a group in the periodic table.

276. The symbol of the element ‘Tungsten’ is:

The symbol of the element ‘Tungsten’ is:

[amp_mcq option1=”Ta” option2=”W” option3=”Tl” option4=”Tc” correct=”option2″]

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The chemical symbol for the element Tungsten is W. The symbol comes from its alternative name, Wolfram, which is derived from the mineral wolframite, a primary source of the element.
The chemical symbol W represents the element Tungsten, derived from the name Wolfram.
Let’s look at the other symbols provided:
A) Ta: Tantalum
C) Tl: Thallium
D) Tc: Technetium
Remembering the symbols for elements is fundamental in chemistry. Tungsten (W) is an important transition metal known for its high melting point.

277. Match List I with List II and select the correct answer using the code

Match List I with List II and select the correct answer using the code given below the Lists:

List I
(Element)		 List II
(Use)
A. Li 1. Time keeper in atomic clocks
B. Na 2. Batteries
C. K 3. Transfer of nerve impulses
D. Cs 4. Control of the water content in the blood

[amp_mcq option1=”2 3 4 1″ option2=”1 2 3 4″ option3=”2 4 3 1″ option4=”1 3 2 4″ correct=”option3″]

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The correct option is C, 2 4 3 1.
Let’s match the elements with their uses:
A. Li (Lithium): Lithium is a highly reactive alkali metal widely used in batteries, particularly rechargeable lithium-ion batteries that power portable electronic devices, electric vehicles, etc. This matches characteristic 2. (A-2)
B. Na (Sodium): Sodium ions (Na+) are essential electrolytes in the body. They play a critical role in regulating blood pressure and the control of water balance in the blood and body tissues through osmosis. This matches characteristic 4. (B-4)
C. K (Potassium): Potassium ions (K+) are also essential electrolytes involved in various bodily functions, including the transmission of nerve impulses and muscle contractions (along with sodium ions). This matches characteristic 3. (C-3)
D. Cs (Cesium): Cesium is an alkali metal, and the isotope Cesium-133 is famously used in atomic clocks due to the highly stable frequency of radiation emitted during electron transitions in its atom, which defines the length of a second. This matches characteristic 1. (D-1)
Thus, the correct matching is A-2, B-4, C-3, D-1, which corresponds to option C (2 4 3 1).
Elements have a wide range of applications based on their unique chemical and physical properties. Lithium, sodium, and potassium are alkali metals known for their reactivity. Cesium is the heaviest stable alkali metal and is known for its very low ionization energy, making it useful in applications like atomic clocks and photocells.

278. To weld metals together, high temperature is required. Such a high tem

To weld metals together, high temperature is required. Such a high temperature is obtained by burning:

[amp_mcq option1=”Acetylene in oxygen” option2=”LPG in oxygen” option3=”Methane in oxygen” option4=”Acetylene in nitrogen” correct=”option1″]

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The correct option is A, Acetylene in oxygen.
Welding requires a high-temperature flame to melt and fuse metals. The combustion of acetylene (C₂H₂) with oxygen (O₂) produces an oxy-acetylene flame, which can reach temperatures of around 3500 °C, making it suitable for welding. Other fuel gases like LPG and methane produce lower temperatures when burned with oxygen. Burning acetylene in nitrogen would not sustain combustion as nitrogen is inert and would not provide the necessary oxidant.
Oxy-acetylene welding is a common process used for welding, cutting, and heating metals. The high temperature is achieved by using pure oxygen as the oxidant instead of air, which contains about 78% nitrogen and would limit the flame temperature. Different ratios of acetylene to oxygen can produce oxidizing, neutral, or carburizing flames, used for various applications.

279. Which of the following statements regarding heavy water are correct?

Which of the following statements regarding heavy water are correct?

  • 1. It is extensively used as a moderator in nuclear reactors
  • 2. It cannot be used in exchange reaction to study reaction mechanism
  • 3. Viscosity of heavy water is relatively smaller than that of ordinary water
  • 4. The dielectric constant of heavy water is smaller than that of ordinary water

Select the correct answer using the code given below:

[amp_mcq option1=”1 and 2″ option2=”2 and 3″ option3=”3 and 4″ option4=”1 and 4″ correct=”option4″]

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Statement 1 is correct: Heavy water (D₂O) is an excellent moderator in nuclear reactors because deuterium has a low neutron absorption cross-section and can effectively slow down fast neutrons to thermal energies. Statement 2 is incorrect: Heavy water and deuterated compounds are commonly used in isotopic exchange reactions to study reaction mechanisms and kinetics. Statement 3 is incorrect: The viscosity of heavy water is slightly higher than that of ordinary water at typical temperatures. Statement 4 is correct: The dielectric constant of heavy water is slightly smaller than that of ordinary water.
This question tests specific properties of heavy water compared to ordinary water.
Heavy water is chemically similar to ordinary water but has different physical properties due to the heavier deuterium isotope replacing hydrogen. It is essential for some types of nuclear reactors (like CANDU reactors) that use unenriched uranium fuel.

280. Graphite is a much better conductor of heat and electricity than diamo

Graphite is a much better conductor of heat and electricity than diamond. This is due to the fact that each carbon atom in graphite:

[amp_mcq option1=”undergoes sp² hybridization and forms three sigma bonds with three neighbouring carbon atoms” option2=”undergoes sp³ hybridization” option3=”is tetrahedrally bonded” option4=”is free from van der Waals force” correct=”option1″]

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The correct answer is A) undergoes sp² hybridization and forms three sigma bonds with three neighbouring carbon atoms.
Graphite’s structure consists of layers of carbon atoms arranged in hexagonal lattices. Each carbon atom in graphite is sp² hybridized and forms three strong sigma bonds with three adjacent carbon atoms within the same layer. The remaining unhybridized p-orbital on each carbon atom overlaps sideways with p-orbitals of neighbouring atoms, forming a delocalized pi electron system across the layer. These delocalized electrons are free to move within the layers, making graphite a good conductor of heat and electricity.
In contrast, diamond has a tetrahedral structure where each carbon atom is sp³ hybridized and forms four strong sigma bonds with four neighbouring carbon atoms. All valence electrons are localized in these covalent bonds, so there are no free electrons to conduct electricity, making diamond an excellent electrical insulator. The strong covalent network also makes diamond an excellent thermal conductor, but the question specifically asks about electrical conductivity compared to diamond, and graphite’s delocalized electrons are key. Graphite’s thermal conductivity is also generally higher than diamond in some directions due to strong in-plane bonds and vibration modes. However, the primary reason for its *better* electrical conductivity than diamond is the presence of delocalized electrons, which stems from its sp² hybridization and bonding structure described in option A.
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