Model Paper 1

Only C

Reason — During the electrolysis of molten lead bromide, the cathode and anode are both made of graphite, an inert conductor. Graphite anode is preferred to other inert electrodes such as platinum since graphite is unaffected by the reactive bromine vapours.

Fluorine

Reason — In period 2 elements (Li, Be, B, C, N, O, F, Ne), Fluorine (F) has the highest electron affinity. As, electron affinity increases across a period due to decreasing atomic radius and increasing nuclear charge. Atoms more readily accept electrons to complete their octet.

Ammonium chloride

Reason —

• Covalent bonds: Within the NH₄⁺ ion, the N–H bonds are covalent.
• Coordinate bond: The lone pair from nitrogen forms a coordinate (dative) bond with a proton (H⁺) to form the NH₄⁺ ion.
• Ionic bond: The electrostatic attraction between the NH₄⁺ cation and the Cl^- anion constitutes the ionic bond in the crystal.

Sulphur dioxide

Reason — Concentrated sulphuric acid acts as an oxidising agent. Its oxidising property is due to the fact that on thermal decomposition it yields nascent oxygen [O].

H₂SO₄ ⟶ H₂O + SO₂ + [O]

Nascent oxygen oxidises non-metals like carbon, sulphur dioxide (SO₂) is evolved.

C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂ ↑

1 mole of CO₂

Reason — Since each molecule is 1 mole, the weight depends entirely on molar mass.

So, 1 mole of CO₂ has molecular mass of 44 g/mol, hence, weigh the most.

Liquid carbon tetrachloride

Reason — Liquid carbon tetrachloride (CCl₄) is a non-polar covalent compound; it does not ionise in water (and in its pure liquid state contains no ions), so it does not conduct electricity. In contrast, acetic acid is a weak electrolyte (partially ionises), while aqueous solutions of sodium hydroxide and potassium chloride are strong electrolytes because they produce a high concentration of ions in solution.

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

Reason — Carbon monoxide (CO) is classified as a neutral oxide; it does not react with either acids or bases and, consequently, does not form an acidic or basic solution in water. Under ordinary conditions there is no reaction:

CO + H₂O ⟶ no acid formed

Therefore the assertion (A) is true, and the reason (R) correctly explains why CO fails to produce an acid with water.

Oxidising agent

Reason — Concentrated sulphuric acid acts as a oxidising agent, it is due to the fact that on thermal decomposition it yields nascent oxygen [O].

H₂SO₄ ⟶ H₂O + SO₂ + [O]

Nascent oxygen oxidises non-metals, metals and inorganic compounds.

C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂ ↑

Ammonium nitrite

Reason — When ammonium nitrite (NH₄NO₂) is heated, it decomposes to form nitrogen gas (N₂) and water vapour (H₂O). This is a thermal decomposition reaction.

NH₄NO₂ $\xrightarrow{\Delta}$ N₂ + 2H₂O

The other substances do not furnish nitrogen gas on simple heating at laboratory temperatures:

• Ammonium nitrate mainly produces nitrous oxide (N₂O) and water.
• Magnesium nitride does not decompose to N₂; it yields ammonia when treated with water.
• Ammonium chloride sublimes rather than decomposes to nitrogen.

Reason — Electrolytic refining is a process by which metals containing impurities are purified electrolytically to give a pure metal. Option 2 shows the set-up used for electrolytic refining of copper:

• Anode – impure copper block (loses electrons and goes into solution).
  Cu - 2e^- ⟶ Cu²⁺
• Cathode – thin sheet of pure copper (gains mass as metal is deposited).
  Cu²⁺ + 2e^- ⟶ Cu
• Electrolyte – acidified aqueous CuSO₄ solution.

As current passes, pure copper is transferred from the impure anode to the cathode, while insoluble impurities collect below the anode as "anode mud".

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

Reason — Vapour density (V.D.) is related to relative molecular mass by:

$\text{V.D.} = \dfrac{\text{ Relative molecular mass}}{2}$

Substituting the given value:

$\text{V.D.} = \dfrac{62}{2} = 32$

Thus the assertion is correct. Because the reason states the same relationship—“The RMM is twice the vapour density”—it correctly explains why the vapour density is 32.

Bronze

Reason — Bronze is made up of 18% of tin and 2% of zinc, remaining 80% percent will be copper.

Ethane

Reason — Acetylene undergoes two step addition reaction with excess of hydrogen in the presence of nickel (catalyst). In the first step of reaction it gives ethene.

C₂H₂ + H₂ $\xrightarrow{\text{Ni}}$ C₂H₄

In the second step of reaction it gives ethane.

C₂H₄ + H₂ $\xrightarrow{\text{Ni}}$ C₂H₆

Ethanal

Reason — Acetaldehyde is a 2-carbon aldehyde with the formula CH₃CHO. It contains the aldehyde ( –CHO ) group. According to IUPAC naming, for 2 carbon atoms prefix "eth-" and for aldehyde group suffix "-al" is added. So, the IUPAC name of acetaldehyde will be Ethanal.

Mg

Reason — Alkaline earth metals are the elements in Group 2 of the periodic table. These include Be, Mg, Ca, Sr, Ba, Ra.

So, the alkaline earth metal of period 3 is Mg ( Magnesium )

(a)

1. Hydroxide ions (OH^-) (from water)
2. Sulphate ions (SO₄^2-) (from H₂SO₄)

(b)

The anion discharged: Hydroxide ion (OH^-)
Product at anode: Oxygen gas (O₂)
Reaction at anode:
4OH^- - 4e^- ⟶ 4OH
4OH ⟶ 2H₂O + O₂

(c)

Product at cathode: Hydrogen gas (H₂)
The reaction at cathode:
4H⁺ + 4e^- ⟶ 4H
2H + 2H ⟶ 2H₂

(d) No change in colour is observed. The gases produced hydrogen and oxygen are colourless and the solution remains colourless since H₂SO₄ and water are both colourless.

(e) 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.

(a) Coordinate bond

(b) Acid salt

(c) Substitution reaction

(d) Water of crystallisation

(e) Ostwald process

(a) A solution M turns blue litmus red, so it must contain hydronium ions; another solution O turns red litmus blue and hence, must contain hydroxide ions;

(b) When solution M and O are mixed together, the products will be salt and water

(c) If a piece of magnesium was put into a solution M, Hydrogen gas would be evolved.

(a) Ethane gas [C₂H₆]

(b) Sulphur dioxide gas [SO₂]

(c) Nitrogen dioxide gas [NO₂]

(d) Oxygen gas [O₂]

(e) Chlorine gas [Cl₂]

(i) Diethyl ether

(ii) 1-propanal

(iii) Propanone [acetone]

(iv) 1, 2, dichloroethane

(v) ethanoic acid

(a) The element belongs to 3^rd period.

Reason — The element Z has atomic number 16 and so the electronic configuration will be 2, 8, 6. The number of shells present in an atom determines it's period. Hence, Z will belong to 3^rd period as it has three shells.

(b) H₂Z

(a) Electrovalent bond exits between M and O because the bond is formed between a metal and non-metal due to oppositely charged ions.

(b) Number of electrons in the outer most shell of M is 1. It is so because the valency of O is -2 and as 2 atoms of M combine with O to form M₂O, hence we can say that M has 1 valence electron.

(a) Pb²⁺ + 2e^- ⟶ Pb

(b) The anode is the oxidising electrode because bromide ions give up electrons there, producing bromine gas:
Br^1- - 1e^- ⟶ Br
Br + Br ⟶ Br₂

(a) Cryolite (Na₃AlF₆)

(b) Zinc blende (ZnS)

The first step is to convert insoluble lead carbonate into soluble lead nitrate by treating lead carbonate with dilute nitric acid.

PbCO₃ (s) + 2HNO₃(dil) ⟶ Pb(NO₃)₂ (aq) + H₂O (l) + CO₂

In the second step, the resulting solution is then treated with sulphuric acid or sodium sulphate solution and a white precipitate of lead suphate is obtained.

Pb(NO₃)₂ + 2NaCl ⟶ 2NaNO₃ + PbCl₂

(a) Na₂CO₃ + H₂SO₄(dil.) ⟶ Na₂SO₄ + H₂O + CO₂

(b) CuCO₃ + 2HCl (dil) ⟶ CuCl₂ + H₂O + CO₂

(c) 2Fe + 3Cl₂ ⟶ 2FeCl₃

(i) $\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

(ii) Na₂SO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + SO₂

(iii) $\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}$

(a) A Mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.

(b) Gay-Lussac's Law of combining volumes — When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(a) Ethane.

(b) Sodium ethoxide and Hydrogen.

(a) Add silver nitrate solution to both the solution, Sodium chloride gives white precipitate of silver chloride. However sodium nitrate shows no reaction with silver nitrate.

(b) On using moist lead acetate paper, hydrogen sulphide reacts to form a black precipitate of lead(II) sulphide, whereas HCl gas does not react with lead salts and show no change.

(c) On adding ammoniacal silver nitrate, no change is seen in ethene whereas white ppt. of silver acetylide is formed in case of ethyne.

(a) Substance B is fused calcium chloride.

(b) Iron (III) chloride is highly deliquescent so it is kept dry with the help of B [fused calcium chloride (drying agent)]

(c) Iron[III] chloride is to be stored in a closed container because iron (III) chloride is highly deliquescent and it absorbs moisture from the surrounding air to form a saturated solution.

(a)

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

(b)

$\underset{\text{ monochloroethane}}{\text{C}_2\text{H}_5\text{-Cl}} + \underset{ \text{ aq. sodium hydroxide}}{\text{NaOH[aq]}} \xrightarrow{\text{boil}} \underset{ \text{ Ethanol [ethyl alcohol]} }{\text{C}_2\text{H}_5\text{-OH}} + \text{NaCl}\$

(a)

Given, V.D. = 8

Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.

16 g occupies 22.4 lit.

∴ 24 g. will occupy = $\dfrac{22.4}{16}$ x 24 = 33.6 lit.

Hence, volume occupied by gas = 33.6 lit.

(b) According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N₂ = Number of molecules of H₂ = X

(a) Dry HCl (Hydrogen chloride) gas is the gas Y

(b) High solubility of HCl in water

(c) Ammonia (NH₃) gas also demonstrates high solubility in water.

(a) It is a mixture of molten alumina (Al₂O₃) 20%, cryolite (Na₃AlF₆) 60% and fluorspar (CaF₂) 20%.

(b) Reaction at the cathode:
2Al³⁺ + 6e^- ⟶ 2Al

(c) The anodes are continuously replaced during the electrolysis because:

1. The oxygen evolved at the anode escapes as a gas or reacts with the carbon anode.
2. The carbon anode is thus oxidized to carbon monoxide which either burns giving carbon dioxide or escapes out through an outlet.
   2C + O₂ ⟶ 2CO
   2CO + O₂ ⟶ 2CO₂
3. The carbon anode is hence consumed and renewed periodically after a certain period of usage.

(a) Hydroxyl ion [OH^-].

(b) The red litmus paper turns blue due to the presence of hydroxyl ion in the solution.

(c) Nitric oxide (NO).

(a)

(b) $\underset{\text{ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} + \underset{\text{acetic acid}}{\text{CH}_3\text{COOH}} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{ethyl acetate}}{\text{CH}_3-\text{COO}-\text{C}_2\text{H}_5} + \text{H}_2\text{O}$

(c)

$\underset{\text{ Chloroethane [ethyl chloride]}}{\text{C}_2\text{H}_5\text{-Cl}} + \underset{\text{ alcoholic KOH}}{\text{KOH [aq.]}} \xrightarrow{\text{boil}} \underset{\text{ Ethene [ethylene]} }{\text{C}_2\text{H}_4} + \text{KCl} +\text{H}_2\text{O}$

(a) When dilute hydrochloric acid is added to calcium hydrogen carbonate, carbon dioxide gas is released with effervescence, and a solution of calcium chloride is formed along with water.

(b) Copper, a brownish pink metal is deposited at the cathode when acidified aqueous copper sulphate solution (CuSO₄) is electrolysed with copper electrodes.

(c) The moist starch iodide paper turns blue-black when exposed to chlorine gas.

(d) When calcium hydroxide is heated with ammonium chloride crystals, the pungent smelling gas (ammonia) is given out.

Hydronium ion is the stable positive ion formed when sulphuric acid dissolves in water. Its structure is shown below:

(a) When ammonium hydroxide (NH₄OH) is added to zinc nitrate solution [Zn(NO₃)₂], a gelatinous white ppt of zinc hydroxide [Zn(OH)₂] is obtained which is soluble in excess of NH₄OH.

$\underset{\text{colourless solution}}{{\text{ZnSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

(With excess NH₄OH ppt. dissolves)

$\text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

On the other hand, calcium nitrate solution [Ca(NO₃)₂] does not give any ppt. even when excess of ammonium hydroxide is added.

(b) When dil. sulphuric acid is added to sodium carbonate and heated, colourless, odourless gas is evolved which turns lime water milky and has no effect on KMnO₄ or K₂Cr₂O₇ solutions.

Na₂CO₃ + 2HCl ⟶ 2NaCl + H₂O + CO₂ ↑

When dil. sulphuric acid is added to sodium sulphite and heated, colourless gas with suffocating odour is evolved which turns lime water milky. It turns acidified K₂Cr₂O₇ from orange to clear green and pink coloured KMnO₄ to clear colourless.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ ↑

Hence, the two compounds can be distinguished.

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(a) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.

17 g. occupies 22.4 lit. of vol.

∴ 68 g will occupy = $\dfrac{22.4}{17}$ x 68 = 89.6 lit.

Hence, volume occupied by this gas = 89.6 lit.

(b) 17 g = 1 mole

∴ 68 g = $\dfrac{1}{17}$ x 68 = 4 moles.

1 mole = 6 × 10²³

∴ 4 moles = 4 x 6.023 × 10²³ molecules.

Hence, Moles = 4

(c) Molecules = 4 x 6.023 × 10²³ = 2.4 x 10²⁴ molecules

(a) A, B, and C are isomers of each other as they all have the molecular formula C₄H₈

(b) A, B and C are all alkenes with 4 carbon atoms (C₄H₈), D is an alkene with 5 carbon atoms (C₅H₁₀). Therefore, D(1-pentene) is a member of homologous series of alkenes but is not an isomer of A, B, or C.

(c) 2-Methylpropene