Model Paper 2

Silver atoms

Reason — Ag⁺ ions from the electrolyte gain electrons (reduction) and are deposited as silver atoms on the article.

Ag⁺ + e^- ⟶ Ag

Inert gases

Reason — Ionisation potential (or ionisation energy) is the energy required to remove the outermost electron from an atom in the gaseous state. It generally increases from left to right across a period (because effective nuclear charge increases) and decreases down a group (because atomic size and shielding increase).
Noble gases have completely filled valence shells and very small atomic radii within their periods; therefore, much more energy is needed to remove an electron from them, giving them the maximum ionisation energy among the groups listed.

14

Reason —

Given,
The relative molecular mass(RMM) of carbon monoxide = 28.
The vapour density of carbon monoxide = ?

Vapour density = $\frac{\text{ Relative molecular mass}}{2}$

Vapour density = $\frac{28}{2}$ = 14

Hence, the vapour density of carbon monoxide is 14.

Nitrogen

Reason — A triple covalent bond involves three shared pairs of electrons between two atoms. Each nitrogen atom shares 3 pairs of electrons, forming a triple bond.

Both A and R are true and R is the correct explanation of A.

Reason — The molecule CH₃COOH possesses four hydrogen atoms. Three are attached to the methyl group (CH₃) and are non-ionisable, while the single hydrogen in the –COOH (carboxylic) group is ionisable. Because only one hydrogen can be released as an H⁺ ion, the basicity of acetic acid is 1. Thus both the Assertion and the Reason are correct, and the Reason properly explains the Assertion.

Copper hydroxide

Reason — Copper hydroxide reacts with NH₄OH to give pale blue precipitate of Cu(OH₂). When excess NH₄OH is added, Cu(OH)₂ dissolves due to the formation of a soluble deep blue complex.

Copper and zinc

Reason — Alloy brass consist of 60-70% of copper and 40-30 % of zinc.

Calcium oxide

Reason — The drying agent used to dry Ammonia is quicklime or calcium oxide (CaO). Other drying agents like conc. sulphuric acid and sulphurous acid are not used, as ammonia being basic, reacts with them. Calcium hydroxide (slaked lime) is slightly basic, but less effective than calcium oxide in drying. It may also partially react with ammonia.

46.67

Reason —

Molar mass of urea (CON₂H₄) = 12 + 16 + 28 + 4 = 60 g

Molar mass of nitrogen (N₂) = 2 x 14 = 28 g

60 g urea has mass of nitrogen = 28 g

∴ 100 g urea will have mass

= $\frac{28 \times 100 }{60}$

= 46.67%

pH 10

Reason —

Universal indicator colour range.

pH	Colour
0 - 3	Red
4 - 6	Orange to yellow
7	Green
8 - 11	Blue
12 - 14	Purple/ Violet

Option 2 is labelled incorrectly, it should be labelled as blue.

Both A and R are true, and R is the correct explanation of A.

Reason — Hydrogen is neutral (neither acidic nor basic), and while concentrated H₂SO₄ is a drying agent, it cannot be used for drying hydrogen because concentrated H₂SO₄ is a strong oxidizing agent and may oxidize hydrogen, especially under certain conditions, which makes it unsafe and unsuitable. Hence, the assertion (A) is true.
Hydrogen gas reacts with concentrated H₂SO₄ because sulphuric acid is a strong oxidizing agent. Hence, the reason (R) is true and it explains why hydrogen cannot be dried with concentrated H₂SO₄.

Acetylene

Reason — In the laboratory preparation of Acetylene, calcium carbide reacts with water to give acetylene and calcium hydroxide.

$\underset{\text{calcium carbide}}{\text{CaC}_2} + \underset{\text{water}}{2\text{H}_2\text{O}} \longrightarrow \underset{\text{Acetylene}}{\text{C}_2\text{H}_2} + \underset{\text{calcium hydroxide}}{\text{Ca(OH)}_2}$

Ethyne

Reason — Acetylene has the chemical formula C₂H₂. According to IUPAC nomenclature, alkynes are named with the suffix “-yne” and have 2 carbon atoms, so the prefix is “eth-”. Hence. the name 'Ethyne'

Potassium hydroxide

Reason — When dissolved in water, a strong electrolyte entirely dissociates into ions, making it an excellent electrical conductor.

Compound	Type	Electrolyte strength	Explanation
Alcohol	Covalent compound	Non-electrolyte	Does not ionize in water
Potassium hydroxide	Ionic compound (Strong base)	Strong electrolyte	Completely dissociates into K+ and OH- in water
Ammonium hydroxide	Weak base	Weak electrolyte	Partially ionizes in water
Glucose	Covalent organic compound	Non-electrolyte	Dissolves but does not form ions in water

Iron (II) sulphate

Reason — Iron (II) sulphate (FeSO₄) reacts with NaOH to give pale green precipitate of ferrous hydroxide.

$\underset{\text{pale green solution}}{{\text{FeSO}_4}} + 2\text{NaOH} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{colourless in solution}}{\text{Na}_2\text{SO}_4}$

The sulphate ion (SO₄^2-) from FeSO₄ reacts with BaCl₂ to form, white precipitate of BaSO₄

Ba²⁺ + SO₄^2- ⟶ BaSO₄

(a)

A — Concentrated sulphuric acid,

B — Sodium nitrate or potassium nitrate

C — Nitric acid.

(b)

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 ^{\circ}\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

(c) All glass apparatus is used in the laboratory preparation of nitric acid since the vapours of nitric acid being highly corrosive attack rubber, cork, etc.

(d) Pure nitric acid [HNO₃] is colourless and unstable and decomposes slightly even at ordinary temperatures and in the presence of sunlight. The decomposition results in formation of reddish brown nitrogen dioxide [NO₂] which remains dissolved in the acid thus imparting a slight yellowish brown colour.

(e) If dry air or CO₂ is bubbled through the yellow acid, the latter turns colourless because it drives out NO₂ from warm acid which is further oxidised to nitric acid.

(a) HCl in the liquefied form is neutral

(b) Organic compounds are generally soluble in Organic solvents.

(c) An inert electrode used in electrolysis of copper sulphate solution is platinum.

(d) Hydrocarbons having triple bond is alkynes.

(e) An acidic gas gives dense white fumes of NH₄C1 with ammonia.

(a) Hydrocarbon

(b) Non-electrolyte (Covalent compound)

(c) Electron affinity

(d) Hall-Héroult process (electrolytic reduction)

(e) Covalent bond

(a)

2-pentanal

2-methyl butanol

1-butyne

(b)

1. Butanoic acid
2. But-2-yne

(a) K < Na < Si < S < Cl

(b) Li < Be < B < C < N < O < F

(a) Chloride ion (Cl^-)

(b) Sulphite ion (SO₃^2-)

(a) C + 4ΗΝΟ₃ ⟶ CO₂ + 2H₂O + 4NO₂

(b) Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂

(c) 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

(a) $\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170^{\circ}\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethylene}}{\text{C}_2\text{H}_4} + \text{H}_2\text{O}\$

(b)

$\underset{\text{ Bromoethane [ethyl bromide]}}{\text{C}_2\text{H}_5\text{Br}} + \underset{\text{ aqueous}}{\text{NaOH}} \xrightarrow{\text{boil}} \underset{ \text{ethanol}} {\text{C}_2\text{H}_5\text{OH}} + \text{NaBr}$

(c) $\underset{\text{ ethene}}{\text{C}_2\text{H}_4} + \text{Br}_2 \xrightarrow[\text{[inert solvent]}]{\text{CCl}_4} \underset{\text{1,2-dibromooethane [ethylene bromide]}}{\text{C}_2\text{H}_4\text{Br}_2}$

(a) Caustic soda is used in the Bayer purification of bauxite. Bauxite is heated under pressure with conc. caustic soda (NaOH) for 2 to 8 hours. Hot, concentrated NaOH dissolves the amphoteric alumina present in the ore, forming soluble sodium aluminate:

$\underset{\text{Amphoteric oxide}}{\underset{\text{[Impure bauxite]}}{\text{Al}_2\text{O}_3.2\text{H}_2\text{O }}} + \underset{\text{Base}}{\underset{\text{[conc. soln]}}{2\text{NaOH }}} \xrightarrow{150 - 200 ^{\circ}\text{C}} \underset{\text{Salt}}{\underset{\text{[sodium aluminate]}}{2\text{NaAlO}_2}} + \underset{\text{Water}}{ \text{3H}_2\text{O}}$

The insoluble impurities mainly Fe₂O₃ and SiO₂ are left behind and are referred to as Red mud.

(b) Fluorspar is added to the electrolyte mixture as it lowers the fusion point of the mixture i.e., the mixture fuses around 950°C instead of 2050°C.

Given,
At same temperature and pressure Mass of hydrogen = 50 g
Mass of gas X = 200 g
Mass of gas Y = 500 g

V.D. of gas X = $\frac{\text{mass of certain volume of gas X }}{\text{mass of equal volume of H}_2}$ = $\frac{200}{50}$ = 4

V.D. of gas Y = $\frac{\text{mass of certain volume of gas Y}}{\text{mass of equal volume of H}_2}$ = $\frac{500}{50}$ = 10

Molecular weight = 2 x V.D. = 2 x 10 = 20 a.m.u.

∴ The vapour density of gas X is 4 and Molecular mass of gas Y is 20 a.m.u.

(a) HCl gas is covalent in nature and does not ionise in non-polar solvents like toluene. Since, there is no ionisation of HCl, so no hydronium (H₃O⁺) ions are produced. As, acidity depends on the presence of these ions, neither blue nor red litmus shows any colour change; the solution is neutral to litmus.

(b) The mouth of the inverted funnel provides a larger surface area over which ammonia gas can come into contact with water, so the gas dissolves rapidly. At the same time, keeping the funnel rim just below the water surface allows air to enter if the pressure inside the apparatus falls, preventing water from being sucked back into the flask that contains the gas.

(c) Liquor ammonia (aqueous NH₃) is stored under considerable internal pressure owing to the high volatility of ammonia. The bottle is first cooled in ice-cold water. The cooling lowers the vapour pressure, so when the stopper is removed the contents do not spurt out and there is no sudden escape of ammonia gas.

(a) 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃↑

Finely grounded mixture of ammonium chloride and calcium hydroxide in excess is taken in a round-bottom flask fitted in a slanting position and gently heated to liberate ammonia gas.

(c) Ammonia gas is collected in inverted gas jars by the downward displacement of air.

(a) Fe²⁺ (ferrous ion)

(b) Zn²⁺ (zinc ion)

Preparation of lead sulphate from lead chloride:

1. Dissolve lead chloride (PbCl₂) in hot water as the salt is only sparingly soluble in cold water but dissolves better when heated.
2. While the solution is still hot, slowly add a dilute solution of sulphuric acid with constant stirring.
   $\text{PbCl}_2 + \text{H}_2\text{SO}_4 [\text{aq}] \longrightarrow\text{PbSO}_4 \downarrow + 2\text{HCl}$
3. Allow the mixture to cool, then filter to collect the precipitate.
4. Wash the precipitate with distilled water and finally dry it.

(a) When Ammonia gas is passed through an aqueous solution of Zinc Nitrate:

• A white, gelatinous precipitate of zinc hydroxide, Zn(OH)₂, first appears.
  Zn(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Zn(OH)₂ ↓
• On passing excess ammonia, the precipitate dissolves and the solution becomes clear and colourless owing to the formation of the soluble tetra-amminezinc(II) complex, [Zn(NH₃)₄]²⁺.

(b) The blue crystals of hydrated copper sulphate (CuSO₄·5H₂O) lose their water of crystallisation, giving off steam and turning into a white powder of anhydrous CuSO₄:
$\underset{\text{(blue crystal)}}{\text{CuSO}_4.5\text{H}_2\text{O}} \xrightarrow{\text{Heat}} \underset{\text{(white powder)}}{\text{CuSO}_4} + 5\text{H}_2\text{O}$

(c) Copper(II) sulphate reacts with NaOH to form a blue precipitate of copper(II) hydroxide (Cu(OH)₂), which is insoluble in excess NaOH.

CuSO₄ + 2NaOH ⟶ Cu(OH)₂ ↓ + Na₂SO₄

(a) Electrodes A and B are made of graphite (carbon).

(b) Aluminium is formed at cathode i.e., electrode A.

(c) Two aluminium compounds in the electrolyte C are alumina [Al₂O₃] and Cryolite [Na₃AlF₆]

(a) The amount of energy released while converting a neutral gaseous isolated atom into a negatively charged gaseous ion (anion) by the addition of electron is called Electron Affinity (E.A.).

(b) The property of self linking of atoms of an element through covalent bonds in order to form straight chains, branched chains and cyclic chains of different sizes is known as catenation.

$$\begin{matrix} 2\text{C}_2\text{H}_6 & + & 7\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 6\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$$

[By Gay Lussac's law]

2 Vol. of C₂H₆ requires 7 Vol. of oxygen

∴ 300 cc C₂H₆ will require $\frac{7}{2}$ x 300

= 1050 cc of Oxygen

Hence, unused oxygen = 1250 - 1050 = 200 cc

Similarly,

2 Vol. of C₂H₆ produces 4 Vol. of carbon dioxide

∴ 300 cc C₂H₆ produces $\frac{4}{2}$ x 300

= 600 cc of Carbon dioxide

Hence, carbon dioxide produced = 600 cc.

(a) C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂ ↑

(b) NaCl + H₂SO₄ ⟶ NaHSO₄ + HCl

(c) $\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170^{\circ}\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} ↑ + \text{H}_2\text{O}$

(a) Fe(OH)₂

(b) ZnO

(c) CO

(a) Hydrogen and oxygen are obtained in the volume ratio 2 : 1 (H₂ : O₂).

(b) Water in pure state consists almost entirely of molecules. It is a polar covalent compound and can form ions when traces of dilute sulphuric acid is added. As dilute sulphuric acid catalyses this ionisation, hence this electrolysis of acidified water is considered as an example of catalysis.

(c) Reaction at anode:
4OH^- - 4e^- ⟶ 4OH
4OH ⟶ 2H₂O + O₂

(a) Vanadium pentoxide (V₂O₅)

(b) Platinum

(c) Finely divided Iron (Fe)

Given:

Percentage of H = 6.32%

Percentage of N = 17.76%

Percentage of carbon = 100 - (6.32 + 17.76) = 75.92 %

Atomic weights: H = 1, N = 14

(a)

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Hydrogen	6.32	1	$\frac{6.32 }{1}$ = 6.32	$\frac{6.32 }{1.27}$ = 5
Nitrogen	17.76	14	$\frac{ 17.76}{14}$ = 1.27	$\frac{1.27}{1.27 }$ = 1
Carbon	75.92	12	$\frac{75.92 }{12}$ = 6.32	$\frac{ 6.32 }{1.27}$ = 5

Simplest ratio of whole numbers = H : N : C = 5 : 1 : 5

Hence, empirical formula is C₅H₅N

Empirical formula weight = (5 × 12) + (5 × 1) + (1 × 14) = 60 + 5 + 14 = 79 g/mol

V.D. = 39.5

Molecular weight = 2 x V.D. = 2 x 39.5 = 79

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{79}{79} = 1\end{aligned}$$

∴ Molecular Formula = n[E.F.] = 1[C₅H₅N] = C₅H₅N

(b) From the formula C₅H₅N, there are 5 hydrogen atoms per molecule.

In 1 mole of the compound the number of molecules = 6.022 × 10²³

∴ Total hydrogen atoms = 5 × (6.022 × 10²³) = 3.011 × 10²⁴ hydrogen atoms.

(a) Atomic number 12 corresponds to the electronic configuration 2, 8, 2.

• Period: three occupied shells → Period 3.
• Group: two valence electrons → Group 2 (alkaline earth metals).

(b) Second member of halogen group is chlorine. Element X has valency 2, and Chlorine has valency 1. So, the formula will be MgCl₂.

The ion formed is Hydroxyl ion. Its electron dot diagram is shown below:

(a) When Sodium sulphite reacts with HCl, effervescence occurs and a pungent gas (SO₂) is evolved that turns acidified potassium dichromate paper from orange to green.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ ↑

Whereas, Sodium sulphate (Na₂SO₄) gives no visible reaction with HCl and no gas is evolved.

(b) Ammonium Chloride (NH₄Cl) reacts with Ca(OH)₂ to release ammonia gas (NH₃) which has pungent smell, turns red litmus blue.

2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃↑

Whereas, sodium chloride shows no reaction with Calcium hydroxide.

(c) Lead Nitrate (Pb(NO₃)₂) reacts with HCl, forms a white precipitate of lead chloride (PbCl₂) which is soluble in hot water.

Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

However, when Silver nitrate reacts with HCl, it forms a white precipitate of silver chloride (AgCl) which is insoluble in hot water

AgNO₃ + HCl ⟶ AgCl ↓ + HNO₃

(a) By heating ethyl alcohol with concentrated H₂SO₄ at 170°C.

$\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170^{\circ}\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} + \text{H}_2\text{O}\$

(b) Ethyl ethanoate

(c) $\underset{\text{ sodium acetate}}{\text{CH}_3\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[300^{\circ}\text{C}]{\text{CaO}} \underset{\text{methane}}{\text{C}\text{H}_4}↑ + \text{Na}_2\text{CO}_3$