Solved Sample Paper 2

Section A

Question 1(i)

Which of the following setups is/are most appropriate for the evolution of hydrogen gas and its identification?

1. I
2. IV
3. III
4. Only II

Answer

Only II

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

Question 1(ii)

The oxides and hydroxides of which metal are amphoteric?

1. Sodium
2. Zinc
3. Iron
4. Magnesium

Answer

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.

Question 1(iii)

The substance which is not an alloy of copper is ...............

1. brass
2. bronze
3. solder
4. duralumin

Answer

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.

Question 1(iv)

Which of the following electrolyte is used in the process of nickle plating?

1. Silver nitrate
2. Potassium argentocyanide
3. Nickle ammonium sulphate
4. Carbon rods

Answer

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.

Question 1(v)

The empirical formula of hexane is

1. C₃H₇
2. C₆H₁₄
3. C₃H₈
4. C₄H₁₂

Answer

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

Question 1(vi)

The salt that is formed by the incomplete neutralisation of an acid molecule by a basic radical is ...............

1. acidic salt
2. basic salt
3. neutral salt
4. none of the above

Answer

acidic salt

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

Question 1(vii)

The experiment which demonstrates that hydrogen chloride is highly soluble in water is ...............

1. Contact process
2. Fountain experiment
3. Haber's process
4. Ostwald's process

Answer

Fountain experiment

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

Question 1(viii)

Concentrated sulphuric acid is

1. yellow in colour
2. blue in colour
3. a colourless liquid
4. green in colour

Answer

a colourless liquid

Reason — Concentrated sulphuric acid is a colourless liquid.

Question 1(ix)

The white precipitate that is soluble in excess NH₄OH is ...............

1. lead chloride
2. copper sulphate
3. silver chloride
4. All of these

Answer
silver chloride

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

Question 1(x)

The table shown below gives information about four substances A, B, C and D

Identify the basic compounds from the above given substances.

1. A and B
2. B and C
3. A and D
4. C and D

Answer

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.

Question 1(xi)

Which of the following is correct about atomic size and electronegativity along a period from left to right?

P. decreases, Increases
Q. increases, Increases
R. increases, Decreases

1. Only P
2. Only Q
3. Only R
4. Both P and Q

Answer

Only P

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

Question 1(xii)

A compound which liberates reddish brown gas around the anode during electrolysis in its molten state is

1. lead (II) nitrate
2. lead (II) sulphate
3. lead (II) oxalate
4. lead (II) bromide

Answer

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

Question 1(xiii)

Which of the following is used as an antiseptic in hospitals?

1. Methanol
2. Propanol
3. Ethanol
4. Vinegar

Answer

Ethanol

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

Question 1(xiv)

The atomic mass of sulphur (S), oxygen (O) and carbon (C) are approximately 32, 16 and 12 respectively. Which of the following statements regarding the number of mole in 88 g of CO₂ and 128 g of SO₂ is correct?

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

Q. 88g of CO₂ contains double the number of moles in 128g of SO₂

1. Only P
2. Only Q
3. Both P and Q
4. Neither P nor Q

Answer

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.

Question 1(xv)

The metal whose hydroxide is soluble in excess of sodium hydroxide solution is

1. magnesium
2. lead
3. silver
4. copper

Answer

lead

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

Question 2(i)

Identify the following.

(a) Anode used in the electro refining of copper.

(b) A strong but non-volatile mineral acid.

(c) Name the process by which bauxite ore is purified.

(d) Name the process by which purified bauxite undergoes electrolytic reduction.

(e) The process of formation of positive and negative charge ions from molecules which are not initially in the ionic state.

Answer

(a) Impure copper

(b) Sulphuric acid [H₂SO₄]

(c) Baeyer's process

(d) Hall-Heroult's process

(e) Ionisation

Question 2(ii)

Match the following Column I with Column II.

Answer

Question 2(iii)

The following diagram shows an experimental set up for the laboratory preparation of pungent choking gas. The gas is acidic in nature.

(a) Name the gas collected in the jar and write the balanced chemical equation for the above preparation.

(b) How is the gas purified?

(c) Write the reaction of the gas produced with magnesium and ammonia.

Answer

(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)

Question 2(iv)

Complete the following by choosing the correct answers from the bracket.

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

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

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

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

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

Answer

(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.

Question 2(v)

(a) Draw the structural formula for the following.

(1) 1,1-dichloroethane

(2) Butyne

(3) Ethan-1-al

(b) Name the following organic compounds in IUPAC system.

Answer

(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

Section B

Question 3(i)

Draw the electron dot structure for the following.

(a) H₂SO₄

(b) CO₂

Answer

(a) H₂SO₄

(b) CO₂

Question 3(ii)

Distinguish between the following as directed.

(a) Calcium nitrate and lead nitrate using NaOH.

(b) Ferrous chloride and ferric chloride using NH₄OH.

Answer

(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

Question 3(iii)

Name the ions present in

(a) hydrogen chloride

(b) lead sulphate

(c) ammonium hydroxide

Answer

The ions are present are:

(a) hydrogen chloride — H⁺ and Cl^-

(b) lead sulphate — Pb²⁺ and SO₄^2-

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

Question 3(iv)

These questions are based upon the periodic properties of the elements. Answer the following questions in one or two lines.

(a) Element that has completely filled two shells.

(b) Element that has electronic configuration 2, 8, 2.

(c) Element that has total of three shells with four electrons in its valence shell.

Answer

(a) Neon [Ne]

(b) Magnesium [Mg]

(c) Silicon [Si]

Question 4(i)

Identify the salts X and Y from the following observations.

(a) Salt X is white in colour. On strong heating, it produces buff yellow residue and liberates reddish brown gas. Solution of salt X produces chalky white insoluble precipitate with excess of ammonium hydroxide.

(b) Salt Y is black in colour. On reacting with conc. HCl, it liberates a pungent greenish yellowish gas which turns moist starch iodide paper blue black.

Answer

(a) Lead nitrate

(b) Manganese dioxide

Question 4(ii)

Write the products and balance the equation.

(a) Na₂CO₃ + H₂SO₄ ⟶

(b) Fe + H₂SO₄ ⟶

Answer

(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.

Question 4(iii)

Calcium, cobalt, nitrogen, magnesium, and silicon are some elements. Using this information complete the following.

(a) ............... is the most reactive metal.

(b) ............... is a metalloid.

(c) ............... is a non-metal.

Answer

(a) Calcium is the most reactive metal.

(b) Silicon is a metalloid.

(c) Nitrogen is a non-metal.

Question 4(iv)

Give reasons:

(a) Covalent compounds have low melting point.

(b) For covalent bond formation both atoms should have high ionisation.

(c) Covalent bond formation takes place when both atoms should have four or more electron in outermost shell.

Answer

(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.

Question 5(i)

Glucose or blood sugar has the molecular formula C₆H₁₂O₆. Determine % of each element.

Answer

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%

Question 5(ii)

Identify the functional group in the following organic compounds.

(a) Butene

(b) Propanoic acid

Answer

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

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

Question 5(iii)

(a) State whether the following statements are true or false. Justify your answer.

(1) Electrolysis of molten lead bromide is considered to be a redox reaction.

(2) Copper is a good conductor of electricity but it is a non-electrolyte.

(b) State the reaction occurring at the anode during electrolysis of copper sulphate solution using a copper anode and platinum anode respectively.

Answer

(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

Question 5(iv)

Choose the answer from the list which fits the description.

[CO₂, NO₂, H₂S, C₂H₂, KBr]

(a) A colourless gas liberated on decomposition of nitric acid.

(b) Water is added to calcium carbide.

(c) Dilute hydrochloric acid is added to zinc sulphide.

Answer

(a) NO₂

(b) C₂H₂

(c) H₂S

Question 6(i)

For each of the substance given below, write the constituents and one use of the following alloys.

(a) Duralumin

(b) Solder

Answer

(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.

Question 6(ii)

Calculate:

(a) A compound contains 69.5% oxygen, 30.5% nitrogen and its molecular weight is 92. Then what will be the molecular formula of the compound?

(b) If the empirical formula of a compound is CHO and its vapour density is 29, find the molecular formula of the compound.

Answer

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

Question 6(iii)

Rahul performed an experiment for the preparation of sulphuric acid. Answer the following questions pertaining to it.

(a) Name the process which is involved in this experiment used for the large scale manufacturing of sulphuric acid.

(b) Which property of sulphuric acid accounts for its use as a dehydrating agent?

(c) Concentrated sulphuric acid acts as both an oxidising agent and a non-volatile acid. Write one equation each to illustrate the above mentioned properties of sulphuric acid.

Answer

(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

Question 6(iv)

Name an alloy of:

(a) Aluminium used in aircraft construction

(b) Lead, used in electrical circuits in joining metals.

(c) Copper, in electrical appliances or household vessels.

Answer

(a) Duralumin

(b) Solder

(c) Brass

Question 7(i)

(a) State why ethyne burns in oxygen with a sooty flame?

(b) Write the balanced chemical equation when water is added to calcium carbide.

Answer

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

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

Question 7(ii)

Name the gas evolved in the following reactions:

(a) when calcium carbide is hydrolysed.

(b) when ethene undergoes hydrogenation.

Answer

(a) Ethyne gas (C₂H₂)

(b) Ethane gas (C₂H₆)

Question 7(iii)

Give the balanced chemical equations for the following reactions:

(a) Potassium sulphate from potassium hydroxide and sulphuric acid.

(b) Barium sulphate from barium chloride and sodium sulphate.

(c) Copper sulphate from copper and concentrated sulphuric acid.

Answer

(a) 2KOH + H₂SO₄ ⟶ K₂SO₄ + 2H₂O

(b) BaCl₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaCl

(c) Cu + 2H₂SO₄ ⟶ CuSO₄ + 2H₂O + SO₂

Question 7(iv)

State one relevant observation for each of the following.

(a) Dilute hydrochloric acid is added to sodium thiosulphate.

(b) Mixture of ammonium chloride and sodium hydroxide is heated.

(c) Burning of ammonia in air.

Answer

(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

Question 8(i)

(a) Propane burns in air according to the following equation.

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

What volume of propane is consumed on using 1000 cm³ of air, considering only 40% of air contains oxygen?

(b) Calculate the gram molecular mass of N₂, if 360 cm³ at S.T.P. weighs 0.45 g.

Answer

(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.

Question 8(ii)

(a) State the condition required for the following reactions.

(1) Conversion of methyl chloride to dichloromethane.

(2) Conversion of ethanol to ethene.

(b) Give the reaction involved in the formation of methyl acetate (ester).

Answer

(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)

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

Question 8(iii)

Write the steps required to convert hydrogen chloride to anhydrous iron (III) chloride. Also, write the equations for the reactions.

Answer

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