Solved Sample Paper 2

Only II

Reason — Delivery tube should not dip in the solution in order to test the evolution of hydrogen gas properly.

Zinc

Reason — Oxides and hydroxides of certain metals i.e zinc, aluminium, lead are amphoteric and react with acids and alkalis to give salt and water.

solder

Reason — Solder (fuse metal) is made up of 50% lead, and 50% tin. Hence, it is not an alloy of copper. Brass, bronze and duralumin contain copper.

Nickle ammonium sulphate

Reason — In electroplating, the electrolyte must be a suitable salt of the plating metal. Hence, in nickle plating, nickle ammonium sulphate will be the electrolyte.

C₃H₇

Reason — Molecular formula of hexane is C₆H₁₄

∴ Ratio of C and H is 6 : 14

Simple ratio is 3 : 7

Hence, empirical formula = C₃H₇

acidic salt

Reason — A salt formed by the incomplete neutralisation of an acid molecule by a base radical is called an acid salt.

Fountain experiment

Reason — Fountain experiment demonstrates that hydrogen chloride is highly soluble in water.

a colourless liquid

Reason — Concentrated sulphuric acid is a colourless liquid.

Reason — Silver chloride is a white precipitate that is soluble in excess of Ammonium hydroxide solution.

A and D

Reason — Substances having pH value greater than 7 are considered as basic substances. Hence, A and D in the given table are basic.

Only P

Reason — On moving from left to right in a period, atomic size decreases and electronegativity increases.

lead (II) bromide

Reason — During electrolysis of Lead [II] bromide, Br^-1 ions are discharged at the anode. Reddish brown fumes are due to bromine vapours.

Reaction at anode:
Br^- - e^- ⟶ Br
Br + Br ⟶ Br₂

Ethanol

Reason — Ethanol (ethyl alcohol) is used as an antiseptic in hospitals.

Only P

Reason —

Number of moles = $\dfrac{\text{Given mass}}{\text{Atomic mass}}$

Atomic mass of CO₂ = 12 + 2 (16) = 12 + 32 = 44 g

Atomic mass of SO₂ = 32 + 2 (16) = 32 + 32 = 64 g

Number of moles in 88g of CO₂ = $\dfrac{88}{44}$ = 2 moles

Number of moles in 128 g of SO₂ = $\dfrac{128}{64}$ = 2 moles

Hence, 88g of CO₂ contains equal number of moles in 128 g of SO₂.

lead

Reason — Lead hydroxide dissolves in excess of sodium hydroxide solution to form sodium plumbite.
Pb(OH)₂ + 2NaOH (excess) ⟶ Na₂PbO₂ + 2H₂O

(a) Impure copper

(b) Sulphuric acid [H₂SO₄]

(c) Baeyer's process

(d) Hall-Heroult's process

(e) Ionisation

(a) Hydrogen Chloride (HCl) gas.

$\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$

(b) By passing through conc. H₂SO₄.

(c) Mg + 2HCl ⟶ MgCl₂ + H₂
     NH₃ ↑ + HCl ↑ ⟶ NH₄Cl (s)

(a) The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is called electronegativity

(b) Electrolysis of aqueous copper sulphate will form copper metal at cathode.

(c) Quick lime is not used to dry HCl gas because it is alkaline in nature.

(d) The fusible compound formed by the combination of flux and gangue is called slag .

(e) Hydrochloric acid is used in the digestion of proteins.

(a) Structural formulae are shown below:

(1) 1,1-dichloroethane

(2) Butyne

(3) Ethan-1-al

(b) IUPAC names of the organic compounds are:

(1) 2-propanone

(2) 2,3-dimethyl butane

(a) H₂SO₄

(b) CO₂

(a) When sodium hydroxide (NaOH) solution is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)₂] which is soluble in excess of sodium hydroxide and readily soluble in acetic acid.

Pb(NO₃)₂ + 2NaOH ⟶ Pb(OH)₂ ↓ + 2NaNO₃

Pb(OH)₂ + 2NaOH (excess) ⟶ Na₂PbO₂ + 2H₂O

Whereas a milky white precipitate is formed in case of calcium nitrate which is sparingly soluble in excess of NaOH.

Ca(NO₃)₂ + 2NaOH ⟶ Ca(OH)₂ ↓ + 2NaNO₃

(b) On reaction with ammonium hydroxide (NH₄OH), ferrous chloride forms a dirty green ppt whereas ferric chloride forms a reddish brown ppt. Hence, the two can be distinguished easily.

FeCl₂ + 2NH₄OH ⟶ Fe(OH)₂ ↓ + 2NH₄Cl

FeCl₃ + 3NH₄OH ⟶ Fe(OH)₃ ↓ + 3NH₄Cl

The ions are present are:

(a) hydrogen chloride — H⁺ and Cl^-

(b) lead sulphate — Pb²⁺ and SO₄^2-

(c) ammonium hydroxide — NH₄⁺ and OH^-

(a) Neon [Ne]

(b) Magnesium [Mg]

(c) Silicon [Si]

(a) Lead nitrate

(b) Manganese dioxide

(a) Na₂CO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + CO₂ ↑
The products formed are sodium sulphate, water and carbon dioxide gas.

(b) Fe + H₂SO₄ ⟶ FeSO₄ + H₂ ↑
The products formed are ferrous sulphate and hydrogen gas.

(a) Calcium is the most reactive metal.

(b) Silicon is a metalloid.

(c) Nitrogen is a non-metal.

(a) Covalent compounds have low melting and boiling points because they have weak forces of attraction between the binding molecules, thus less energy is required to break the force of bonding.

(b) When both atoms have high ionisation, the energy required to remove electrons from them is very large. Thus, transfer of electrons is not possible. The atoms share electrons to complete their octet. Hence they form covalent bonds.

(c) The atoms that have four or more electrons in outermost shell do not favour the loss of their electrons due to energy considerations and thus transfer of electrons is not possible. So, they form covalent bonds to achieve stable electronic configuration.

Percentage of particular element = $\dfrac{\text{Mass}}{\text{Total mass}}$ x 100

C₆H₁₂O₆ = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g per mole

Percentage of C = $\dfrac{72}{180}$ x 100 = 40%

Percentage of H = $\dfrac{12}{180}$ x 100 = 6.67%

Percentage of O = $\dfrac{96}{180}$ x 100 = 53.33%

Hence, Percentage of C = 40%, H = 6.67% and O = 53.33%

(a) Butene — carbon carbon double bond (C=C)

(b) Propanoic acid — Carboxylic acid (-COOH)

(a) (i) True

Electrolysis of molten lead bromide involves oxidation and reduction reactions and hence is a redox reaction.

Dissociation of lead bromide

PbBr₂ ⇌ Pb²⁺ + 2Br^1-

Reduction reaction at cathode:
Pb²⁺ + 2e^- ⟶ Pb

Oxidation reaction at the anode:
Br^1- - 1e^- ⟶ Br
Br + Br ⟶ Br₂

(a) (ii) True

Copper does not undergo chemical decomposition due to flow of electric current through it. Hence, copper is a good conductor of electricity but it is a non-electrolyte.

(b) Reaction at anode [active copper] : Cu - 2e^- ⟶ Cu²⁺

Reaction at anode [inert platinum] : OH^1- - 1e^- ⟶ OH x 4

4OH ⟶ 2H₂O + O₂

(a) NO₂

(b) C₂H₂

(c) H₂S

(a) Duralumin — Al [95%], Mg [0.5%], Mn [0.5%], Cu [4%]

It is used for making bodies of aircraft.

(b) Solder — Pb [50%], Sn [50%]

It is used for soldering purposes.

Simplest ratio of whole numbers = O : N = 2 : 1

Hence, empirical formula is NO₂

Molecular weight = 92

Empirical formula weight = 14 + 2(16) = 14 + 32 = 46

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{92}{46} = 2$

Molecular formula = n[E.F.] = 2[NO₂] = N₂O₄

Hence, molecular formula is N₂O₄.

(b) Empirical formula of the compound = CHO

Empirical formula mass = 12 + 1 + 16 = 29

Molecular formula mass = 2 x vapour density
= 2 x 29 = 58 g

Molecular formula = (Empirical formula)_n

∴ Molecular formula = (CHO)₂ = C₂H₂O₂

(a) Contact process

(b) Sulphuric acid has a great affinity for water. It readily removes elements of water from other compounds. Hence, it is used as a dehydrating agent.

(c) Concentrated sulphuric acid as an Oxidising agent:

Cu + 2H₂SO₄ [conc.] ⟶ CuSO₄ + 2H₂O + 2SO₂ ↑

Concentrated sulphuric acid as a non-volatile acid:

NaNO₃ + H₂SO₄ [conc.] $\xrightarrow{\lt 200 \degree\text{C}}$ NaHSO₄ + HNO₃

(a) Duralumin

(b) Solder

(c) Brass

(a) Ethyne being a hydrocarbon contains carbon hence, all the carbon particles do not oxidise completely. Thus, it burns with sooty flame.

(b) $\underset{\text{calcium carbide}}{\text{CaC}_2} + \underset{\text{water}}{2\text{H}_2\text{O}} \longrightarrow \underset{\text{ethyne [acetylene]}}{\text{C}_2\text{H}_2} + \text{Ca(OH)}_2$

(a) Ethyne gas (C₂H₂)

(b) Ethane gas (C₂H₆)

(a) 2KOH + H₂SO₄ ⟶ K₂SO₄ + 2H₂O

(b) BaCl₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaCl

(c) Cu + 2H₂SO₄ ⟶ CuSO₄ + 2H₂O + SO₂

(a) Yellow particles of Sulphur precipitate.
Na₂S₂O₃ + 2HCl ⟶ 2NaCl + SO₂ + S + H₂O

(b) Pungent smelling ammonia gas is evolved.
NH₄Cl + NaOH ⟶ NaCl + H₂O + NH₃ ↑

(c) Ammonia burns in the atmosphere of excess oxygen with a green or greenish yellow flame, forming nitrogen and water.
4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O

(a) Given, 40% of air contains oxygen

∴ 40% of 1000 cm³ = $\dfrac{40}{100}$ x 1000 = 400 cm³

[By Lussac's law]

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 &\longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} & : & 4\text{ vol.} \end{matrix}$

To calculate the volume of propane consumed :

$\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 400 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{1}{5} \times 400 = 80 \text{ cm.}^3$

Hence, volume of propane consumed is 80 cm³

(b) The mass of 22.4 L of a gas at S.T.P. is equal to its gram molecular mass.

360 cm³ of N₂ at S.T.P. weighs 0.45 g

∴ 22,400 cm³ of N₂ will weigh $\dfrac{0.45}{360} \times 22,400$ = 28 g

Hence, Gram Molecular Mass of N₂ = 28 g.

(a) (1) methyl chloride converts to dichloromethane under diffused sunlight, ultraviolet light or high temperature [400°C]

$\underset{\text{methylchloride} }{\text{CH}_3\text{Cl}} + \underset{\text{[excess]}}{\text{Cl}_2} \longrightarrow \underset{\text{dichloromethane}}{\text{C}\text{H}_2\text{Cl}_2}+ \text{HCl}$

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

$\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170\degree\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)

$\underset{\text{methyl alcohol}}{\text{CH}_3\text{OH}} + \underset{\text{acetic acid}}{\text{CH}_3\text{COOH}} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{methyl acetate}}{\text{CH}_3-\text{COO}-\text{CH}_3} + \text{H}_2\text{O}$

The steps required to convert hydrogen chloride to anhydrous iron (III) chloride are:

1. Prepare chlorine by reacting oxidising agents (like MnO₂, PbO₂, Pb₃O₄) with conc. HCl.
   MnO₂ + 4HCl (conc.) ⟶ MnCl₂ + Cl₂ +2H₂O
2. Now, pass the chlorine (prepared) over conc. H₂SO₄ for drying.
3. Then the dry chlorine is heated over with iron to give anhydrous iron (III) chloride FeCl₃.
   2Fe + 3Cl₂ ⟶ 2FeCl₃