Inorganic Chemistry

Which one of the following is called dry ice?

[amp_mcq option1=”Solid carbon dioxide” option2=”Liquid carbon dioxide” option3=”Liquid nitrogen” option4=”Liquid ammonia” correct=”option1″]

Dry ice is the solid form of carbon dioxide (CO₂).

– At atmospheric pressure, solid carbon dioxide sublimes directly into a gas without melting into a liquid, hence the name “dry ice”.
– Its sublimation temperature at 1 atm is -78.5 °C.

Dry ice is used as a cooling agent in various applications, such as preserving food, creating fog effects, and cooling materials in laboratories. Liquid carbon dioxide exists under high pressure. Liquid nitrogen and liquid ammonia are different substances with different properties and uses.

Which one of the following reactions will give NO (nitric oxide) gas as one of the products?

[amp_mcq option1=”3Cu + 8HNO₃ (dilute)→” option2=”Cu + 4HNO₃ (conc.)→” option3=”4Zn + 10HNO₃ (dilute)→” option4=”Zn + 4HNO₃ (conc.)→” correct=”option1″]

The reaction of copper (Cu) with dilute nitric acid (HNO₃) produces nitric oxide (NO) gas, copper(II) nitrate (Cu(NO₃)₂), and water. The balanced chemical equation is 3Cu + 8HNO₃ (dilute) → 3Cu(NO₃)₂ + 2NO + 4H₂O.

– Nitric acid is a strong oxidizing agent, and its reaction with metals produces nitrogen oxides rather than hydrogen gas (which is produced by the reaction of active metals with non-oxidizing acids).
– The specific nitrogen oxide produced depends on the concentration of HNO₃ and the reactivity of the metal. Dilute HNO₃ reacts with moderately reactive metals like copper to produce NO. Concentrated HNO₃ reacts with most metals (except very unreactive ones) to produce nitrogen dioxide (NO₂).

– Reaction with concentrated HNO₃: Cu + 4HNO₃ (conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O (produces NO₂).
– Reaction with very dilute HNO₃ and highly reactive metals (like Zn): 4Zn + 10HNO₃ (very dilute) → 4Zn(NO₃)₂ + N₂O + 5H₂O (produces nitrous oxide, N₂O) or even NH₄NO₃. The reaction given in option C with dilute HNO3 producing N2O is also correct for zinc.

What is the formula mass of anhydrous sodium carbonate? (Given that the atomic masses of sodium, carbon and oxygen are 23 u, 12 u and 16 u respectively)

[amp_mcq option1=”286 u” option2=”106 u” option3=”83 u” option4=”53 u” correct=”option2″]

The formula mass of anhydrous sodium carbonate (Na₂CO₃) is 106 u.

– The chemical formula for anhydrous sodium carbonate is Na₂CO₃.
– The formula mass is the sum of the atomic masses of all atoms in the formula unit.
– Atomic masses given: Sodium (Na) = 23 u, Carbon (C) = 12 u, Oxygen (O) = 16 u.
– Formula mass of Na₂CO₃ = (2 × Atomic mass of Na) + (1 × Atomic mass of C) + (3 × Atomic mass of O)
– Formula mass = (2 × 23 u) + (1 × 12 u) + (3 × 16 u)
– Formula mass = 46 u + 12 u + 48 u = 106 u.

Anhydrous means without water. Sodium carbonate also exists as hydrated forms, like sodium carbonate decahydrate (washing soda, Na₂CO₃·10H₂O), which would have a significantly higher formula mass due to the inclusion of water molecules.

The compound C₆H₁₂O₄ contains

[amp_mcq option1=”22 atoms per mole” option2=”twice the mass percent of H as compared to the mass percent of C” option3=”six times the mass percent of C as compared to the mass percent of H” option4=”thrice the mass percent of H as compared to the mass percent of O” correct=”option3″]

We need to analyze the composition of the compound C₆H₁₂O₄ in terms of the relative masses or mass percentages of its constituent elements.

The compound is C₆H₁₂O₄. Using approximate atomic masses (C=12, H=1, O=16):
Mass of C in one molecule = 6 * 12 = 72
Mass of H in one molecule = 12 * 1 = 12
Mass of O in one molecule = 4 * 16 = 64
Total mass of one molecule (or molar mass) = 72 + 12 + 64 = 148.

Let’s evaluate the options:
A) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro’s number of molecules, so one mole contains 22 * Avogadro’s number of atoms, not just 22 atoms. Incorrect.
B) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12/148)*100% vs 2 * (72/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect.
C) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let’s check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element / total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true.
D) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect.

Option C is the only statement that accurately reflects the mass composition of the compound, interpreting “six times the mass percent of C as compared to the mass percent of H” as the mass percent of C being six times the mass percent of H.

Calculating mass percentages:
%C = (72/148) * 100% ≈ 48.65%
%H = (12/148) * 100% ≈ 8.11%
%O = (64/148) * 100% ≈ 43.24%
Check C: 6 * %H = 6 * 8.11% ≈ 48.66%, which is very close to %C (48.65%).

How many moles of hydrogen atom are present in one mole of Aluminium hydroxide?

[amp_mcq option1=”One mole” option2=”Two moles” option3=”Three moles” option4=”Four moles” correct=”option3″]

The chemical formula for Aluminium hydroxide is Al(OH)₃. This formula indicates that one molecule of Aluminium hydroxide contains one atom of Aluminium (Al) and three hydroxide groups (OH). Each hydroxide group (OH) consists of one Oxygen atom (O) and one Hydrogen atom (H). Therefore, one molecule of Al(OH)₃ contains a total of 3 hydrogen atoms (one in each of the three OH groups).

A mole is a unit of amount of substance, equal to Avogadro’s number (approximately 6.022 × 10²³) of elementary entities (atoms, molecules, ions, etc.). If one molecule contains a certain number of atoms of a particular element, then one mole of molecules will contain the same number of moles of those atoms.

In one mole of Al(OH)₃ molecules, there are Avogadro’s number of Al(OH)₃ molecules. Since each molecule has 3 hydrogen atoms, there are 3 × (Avogadro’s number) of hydrogen atoms. This quantity of hydrogen atoms represents 3 moles of hydrogen atoms.

The setting time of cement is lowered by adding

[amp_mcq option1=”oxides of aluminium” option2=”gypsum” option3=”oxides of magnesium” option4=”silica” correct=”option1″]

The correct answer is A) oxides of aluminium. The question asks what is added to lower the setting time of cement, which means making it set faster. Gypsum (option B) is added to *increase* the setting time (retard setting). While standard accelerators are typically calcium chloride or alkali aluminates, reactive forms of aluminium oxides or phases rich in alumina like Calcium Aluminate Cements (CAC) or specific additives containing activated alumina can act as accelerators, causing rapid setting. Therefore, among the given options, oxides of aluminium (in a suitable form or context as present in cement phases like C₃A) are the most plausible candidates for contributing to or accelerating the setting process compared to the other options which are either retarders (gypsum) or primary components/cause of expansion (magnesium oxide).

– Setting time refers to the time required for cement paste to lose its plasticity.
– Additives are used to control setting time: accelerators decrease setting time, retarders increase it.
– Gypsum is a common retarder, preventing flash set by reacting with C₃A.
– Oxides of aluminium are present in cement clinker primarily as tricalcium aluminate (C₃A), which hydrates rapidly and contributes to early setting. Certain forms of reactive alumina can act as accelerators.

– Calcium chloride (CaCl₂) is a very common and effective accelerator used in cement.
– Oxides of magnesium (MgO), if present in excess in portland cement, can cause delayed expansion and soundness issues, not primarily affecting initial setting time in a controlled manner.
– Silica (SiO₂) is a main component of cement (silicates C₃S and C₂S), which are responsible for strength development, but adding pure silica is not a method to control setting time.
– The phrasing “lowered by adding” implies acceleration. While gypsum is a crucial additive related to setting time, its effect is the opposite (increasing/retarding). This points towards A as the intended answer, assuming a context where an aluminium oxide based additive is used for acceleration.

Which one of the following species is not capable of showing disproportionation reaction ?

[amp_mcq option1=”ClO⁻” option2=”ClO₂⁻” option3=”ClO₃⁻” option4=”ClO₄⁻” correct=”option4″]

ClO₄⁻ (Perchlorate) is not capable of showing disproportionation reaction.

Disproportionation is a redox reaction where an element in a specific oxidation state is simultaneously oxidized to a higher oxidation state and reduced to a lower oxidation state. For this to happen, the element must be in an intermediate oxidation state. In ClO₄⁻, Chlorine is in its highest common oxidation state of +7. It can only be reduced to a lower state (e.g., +5, +3, +1, 0, -1), but it cannot be oxidized further. Therefore, it cannot undergo disproportionation.

In ClO⁻ (+1), ClO₂⁻ (+3), and ClO₃⁻ (+5), Chlorine is in intermediate oxidation states and can be both oxidized and reduced, hence capable of disproportionation.

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

List I (Element)	List II (Highest Valency)
A. Sulfur	1. Five
B. Phosphorous	2. Six
C. Lead	3. Two
D. Silver	4. Four

Code :

	A	B	C	D
(a)	2	4	1	3
(b)	2	1	4	3
(c)	3	1	4	2
(d)	3	4	1	2

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

The correct match is A-2, B-1, C-4, D-3.

The highest common valency (or oxidation state) for the elements are:
– Sulfur (S): Can exhibit +6 (e.g., in sulfates), so highest valency is Six (2).
– Phosphorus (P): Can exhibit +5 (e.g., in phosphates), so highest valency is Five (1).
– Lead (Pb): Can exhibit +4 (e.g., PbO₂), so highest valency is Four (4).
– Silver (Ag): The most common valency/oxidation state is +1. While higher, unstable oxidation states exist, +1 is the standard. The option provides ‘Two’ (3) as the highest valency for Silver. Assuming this option set is intended, the pairings match option B for A, B, and C.

Based on standard chemistry, the highest common valency for Silver is 1. However, matching the provided options, A-2, B-1, C-4 align with common understanding, leading to option B being the most probable intended answer despite the inaccuracy regarding Silver’s listed highest valency.

Combination of one volume of nitrogen with three volumes of hydrogen produces

[amp_mcq option1=”one volume of ammonia” option2=”two volumes of ammonia” option3=”three volumes of ammonia” option4=”one and a half volumes of ammonia” correct=”option2″]

The reaction between nitrogen and hydrogen to produce ammonia is represented by the balanced chemical equation: N₂(g) + 3H₂(g) → 2NH₃(g). According to Avogadro’s law, at the same temperature and pressure, the ratio of the volumes of reacting gases and gaseous products is equal to the ratio of their stoichiometric coefficients in the balanced equation. Thus, 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia.

– The balanced equation for the Haber process is N₂(g) + 3H₂(g) → 2NH₃(g).
– The stoichiometric coefficients are 1 for N₂, 3 for H₂, and 2 for NH₃.
– At constant temperature and pressure, the volume ratio of reacting gases and gaseous products is equal to the mole ratio (and stoichiometric coefficient ratio).

This principle relating volumes of reacting gases to their stoichiometric coefficients is known as Gay-Lussac’s law of combining volumes.

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″]

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).