Model Paper 1

You are provided with the following substances

A. Lead chloride

B. Zinc sulphide

C. Sodium carbonate

D. Lead Nitrate

E. Ammonium carbonate

F. Calcium chloride

G. Calcium sulphite

H. Zinc carbonate

I. Silver Nitrate

Choose one substance in each case which matches the description below :

(i) A normal salt formed when basic oxide and acidic oxide react

(ii) An insoluble binary salt

(iii) Insoluble carbonate

(iv) An insoluble salt formed from two soluble salts.

(v) A salt which on heating leaves metal as residue.

Answer

(i) G. Calcium sulphite

(ii) B. Zinc sulphide

(iii) H. Zinc carbonate

(iv) A. Lead chloride

(v) I. Silver Nitrate

How will you identify the following salts: calcium chloride, lead chloride and zinc chloride?

Answer

In order to test the three salts we add ammonium hydroxide solution to each of the salt solutions first a few drops and then in excess.

We will observe that there is no ppt. formed in case of calcium chloride, whereas chalky white ppt. is formed in case of lead chloride which is insoluble in excess of ammonium hydroxide solution. Zinc chloride reacts to form a gelatinous ppt. which dissolves in excess of ammonium hydroxide solution.

State your observations and write balanced equations.

(i) Sodium chloride solution is mixed with lead nitrate solution.

(ii) Ethylene gas is passed through a solution of bromine in carbon tetrachloride.

(iii) Sodium hydroxide is added to iron (III) sulphate solution.

(iv) Chlorine is passed through a solution of potassium iodide.

(v) Ammonium hydroxide solution is added to copper sulphate solution in excess.

Answer

(i) When sodium chloride solution is mixed with lead nitrate solution, a white precipitate of lead chloride is seen and the soluble salt sodium nitrate is formed.

2NaCl + Pb(NO₃)₂ ⟶ 2Na(NO₃)₂ + PbCl₂↓

(ii) Brown colour of bromine disappears when ethylene is bubbled through a solution of bromine in carbon tetrachloride.

C₂H₂ + Br₂ ⟶ C₂H₂Br₂ + Br₂ ⟶ C₂H₂Br₄

(iii) When sodium hydroxide is added to iron (III) sulphate, reddish brown ppt is formed, which is insoluble in excess of sodium hydroxide.

Fe₂(SO₄)₃ + 6NaOH ⟶ 2Fe(OH)₃ + 3Na₂SO₄

(iv) When chlorine gas is passed over aqueous potassium iodide, the colourless solution changes to yellow brown due to the formation of iodine.

Cl₂ + 2KI ⟶ 2KCl + I₂

(v) When ammonium hydroxide is reacted with a solution of copper sulphate, a pale blue ppt. of copper hydroxide is formed which dissolves in excess of ammonium hydroxide forming a soluble complex salt [Cu(NH₃)₄]SO₄ [tetramine copper [II] sulphate] .

CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂ ↓

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

Give balanced equations for the laboratory preparation of the following gases :

(i) Ammonia

(ii) Ethylene.

Answer

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

(ii)

$\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{ethylene}}{\text{C}_2\text{H}_4} + \text{H}_2\text{O}\$

Give one difference between the following.

(i) Ionisation and electrolytic dissociation.

(ii) Ionisation potential and electron affinity.

(iii) Polar and non-polar molecule.

(iv) Normal salt and acidic salt.

(v) Saturated hydrocarbons and unsaturated hydrocarbons.

Answer

(i) Difference between ionisation and electrolytic dissociation:

(ii) Difference between ionisation potential and electron affinity :

(iii) Difference between polar and non-polar molecule :

(iv) Difference between normal salt and acidic salt :

(v) Difference between saturated hydrocarbons and unsaturated hydrocarbons :

Define :

(i) Electrolysis

(ii) Molar volume

(iii) Halogenation

(iv) Catenation

(v) Electronegativity

Answer

(i) Electrolysis — It is the process of decomposition of a chemical compound [electrolyte] in the aqueous or fused [molten] state by the passage of direct electric current resulting in discharge of ions as neutral atoms at the respective electrodes.

(ii) Molar volume — It is the volume occupied by 1 mole of a gas at STP.

(iii) Halogenation — Halogenation refers to a chemical reaction in which a halogen element, such as chlorine, bromine, or iodine, is added to a molecule or compound.

(iv) Catenation — Carbon atoms possess a unique property to link themselves together to form very long chains of different sizes which may be straight, branched or cyclic and this property is called catenation.

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

Write a balanced equation by taking an example of your own.

(i) An acid and an alkali.

(ii) An ammonium salt and an alkali

(iii) A nitrate and a mineral acid

(iv) A hydrocarbon and oxygen in excess

(v) A metal hydrogen carbonate and an acid.

Answer

(i) HCl + NaOH ⟶ H₂O + NaCl

(ii) NH₄Cl + NaOH $\xrightarrow{\Delta}$ NaCl + H₂O + NH₃↑

(iii) KNO₃ + H₂SO₄ (conc.) $\xrightarrow{\lt 200 \degree\text{C}}$ KHSO₄ + HNO₃

(iv) C₂H₄ + 3O₂ ⟶ 2CO₂ + 2H₂O + heat

(v) NaHCO₃ + HCl ⟶ NaCl + H₂O + CO₂

5.72 grams of hydrated sodium carbonate is heated strongly. The white residue left over is 2.12 g. Calculate:

(i) the formula of hydrated sodium carbonate

(ii) the percentage of carbon in the compound.

[Na = 23, C = 12, O = 16, H = 1]

Answer

(i) Hydrated sodium carbonate = Na₂CO₃.xH₂O

Mass of hydrated sodium carbonate = 5.72g

Mass of white residue = 2.12g

Mass of x water = 5.72 - 2.12 = 3.6 g

Molecular wt. of sodium carbonate Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106

Molecular wt. of H₂O = 2(1) + 16 = 18

$\dfrac{\text{Mass of water}}{\text{Mass of anhydrous sodium carbonate}}$ = $\dfrac{18\text{x}}{106}$ = $\dfrac{3.6}{2.12}$

∴ x = $\dfrac{3.6 \times 106}{2.12 \times 18}$ = $\dfrac{381.6}{38.16}$ = 10

Hence, formula of hydrated sodium carbonate = Na₂CO₃.10H₂O

(ii) Molecular wt. of hydrated sodium carbonate Na₂CO₃.10H₂O = 2(23) + 12 + 3(16) + 10(18) = 46 + 12 + 48 + 180 = 286g

286 g of sodium carbonate contains 12 g of carbon

100 g of sodium carbonate will contain $\dfrac{12}{286}$ x 100 = 4.19 g of carbon.

Hence, percentage of carbon in the compound = 4.19%

Give equations for the purification of bauxite. Name the process involved.

Answer

Al₂O₃.2H₂O + 2NaOH ⟶ 2NaAlO₂ + 3H₂O

NaAlO₂ + 2H₂O ⟶ NaOH + Al(OH)₃

2Al(OH)₃ $\xrightarrow{1000 \degree\text{C}}$ Al₂O₃ + 3H₂O

The process is called 'Bayer's process'.

Within a Group, where would you expect to find the element with :

(i) the greatest metallic character ?

(ii) the largest atomic size ?

Answer

(i) Within a group, the metallic character increases as we move down, hence, we can expect the last element of the group to be most metallic.

(ii) Within a group, the atomic size increases as we move down, hence, we can expect the last element of the group to have the largest atomic size.

Define ionisation potential ? State whether it increases on going down a group.

Answer

The energy required to remove an electron from a neutral isolated gaseous atom and convert it into a positively charged gaseous ion is called ionisation potential.

On moving down a group ionization potential decreases.

(i) Which concentrated acid will oxidize sulphur directly to sulphuric acid ? Write the equation for the same.

(ii) Complete the following : concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is — (less volatile/stronger) in comparison to these two acids.

(iii) Write the observation and the balanced equation, when a paper dipped in acidified potassium dichromate solution is put on the mouth of a test tube containing sulphur dioxide gas.

Answer

(i) Hot and conc. nitric acid.

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(ii) Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is less volatile in comparison to these two acids.

(iii) Acidified potassium dichromate solution changes to green colour from orange colour.

K₂Cr₂O₇ + H₂SO₄ + 3SO₂ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Give equations for the following :

(i) Preparation of acetylene

(ii) Preparation of ethyl alcohol

(iii) Addition of chlorine to ethylene

(iv) Preparation of ethane

(v) Sulphurous acid reacts with bromine.

Answer

(i) Preparation of acetylene

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

(ii) Preparation of ethyl alcohol

$\text{C}_2\text{H}_5\text{Cl} + \underset{\text{aqueous}}{\text{KOH}} \xrightarrow{\text{boil}} \underset{\text{ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} + \text{KCl}$

(iii) Addition of chlorine to ethylene

(iv) Preparation of ethane

$\underset{\text{sodium propionate}}{\text{C}_2\text{H}_5\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[300\degree \text{C}]{\text{CaO}} \underset{\text{ethane}}{\text{C}_2\text{H}_6} ↑ + \text{Na}_2\text{CO}_3$

(v) Sulphurous acid reacts with bromine

$\text{H}_2\text{SO}_3 + \text{Br}_2 + \text{H}_2\text{O} \longrightarrow \text{H}_2\text{SO}_4 + 2\text{HBr}$

Complete and balance the following:

(i) Potassium nitrate + sulphuric acid ⟶

(ii) Ferrous chloride + Ammonium hydroxide ⟶

(iii) Red lead + conc. HCl ⟶

(iv) Sodium thiosulphate + dilute Hydrochloric acid ⟶

(v) Copper with dil nitric acid ⟶

Answer

(i) Potassium nitrate + sulphuric acid ⟶ potassium bisulphate + nitric acid

$\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{KHSO}_\text{4} + \text{HNO}_\bold{3}$

(ii) Ferrous chloride + Ammonium hydroxide ⟶ ammonium chloride + ferrous hydroxide

$\text{FeCl}_2 + 2\text{NH}_4\text{OH} \longrightarrow 2\text{NH}_4\text{Cl} + \text{Fe(OH)}_2$

(iii) Red lead + conc. HCl ⟶ lead chloride + water + chlorine

$\text{Pb}_3\text{O}_4 + 8\text{HCl (conc.)} \xrightarrow{\Delta} 3\text{PbCl}_2 + 4\text{H}_2\text{O} + \text{Cl}_2 ↑$

(iv) Sodium thiosulphate + dilute Hydrochloric acid ⟶ sodium chloride + sulphur + sulphur dioxide + water

$\text{Na}_2\text{S}_2\text{O}_3 + 2\text{HCl (dil.)} \longrightarrow 2\text{NaCl} + \text{SO}_2 + \text{S} + 2\text{H}_2\text{O}$

(v) Copper with dil nitric acid ⟶ copper nitrate + water + nitrogen monoxide

$3\text{Cu} + 8\text{HNO}_3 \text{( dil.)} \longrightarrow 3\text{Cu(NO}_3)_2 + 4\text{H}_2\text{O} + 2\text{NO}↑$

1 gram mixture of NaCl and NaNO₃ is dissolved in water and treated with an excess of AgNO₃ solution. 1.43 g of precipitate is formed. Calculate percentage of sodium chloride in the mixture. [5]

[Ag = 108, Na = 23, N = 14, O = 16, Cl = 35.5].

Answer

$\underset{58.5 \text{g}}{\text{NaCl}} + \text{AgNO}_3 \longrightarrow \underset{143.5 \text{ g}}{\text{AgCl}}\downarrow + \text{NaNO}_3$

Molecular mass of AgCl = 108 + 35.5 = 143.5

Molecular mass of NaCl = 23 + 35.5 = 58.5

143.5 g AgCl is formed by 58.5 g NaCl

∴ 1.43 g of AgCl will be formed by $\dfrac{58.5}{143.5}$ x 1.43 = 0.58 g

Percentage of NaCl = $\dfrac{0.58}{1}$ x 100 = 58.29%

Hence, percentage of NaCl is 58.29%

[Dilute sulphuric acid, copper, iron, sodium, zinc, copper carbonate and sodium carbonate]

Choosing only from the list of substances mentioned above, write an equation for the reaction which you would use in the lab to obtain :

(i) Zinc carbonate

(ii) Copper sulphate

(iii) Sodium sulphate

(iv) Iron sulphate

Answer

(i) Zinc carbonate

Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂↑

ZnSO₄ + Na₂CO₃ ⟶ Na₂SO₄ + ZnCO₃

(ii) Copper sulphate

CuCO₃ + H₂SO₄ (dil.) ⟶ CuSO₄ + H₂O + CO₂

(iii) Sodium sulphate

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

(iv) Iron sulphate

Fe + H₂SO₄ (dil.) ⟶ FeSO₄ + H₂

A volatile chloride of metal M contains 34.5% of the metal. If the density of metal chloride relative to hydrogen is 162.5. Calculate the molecular formula of metal chloride.

[M = 56, Cl = 35.5].

Answer

Simplest ratio of whole numbers between M : Cl = 1 : 3

Hence, empirical formula is MCl₃

Empirical formula weight = 56 + 3(35.5) = 56 + 106.5 = 162.5

Vapour density (V.D.) = 162.5

Molecular weight = 2 x V.D. = 2 x 162.5 = 325

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{325}{162.5} = 2$

Molecular formula = n[E.F.] = 2[MCl₃] = M₂Cl₆

Hence, Molecular formula = M₂Cl₆

(i) What type of reaction takes place between ethane and chlorine to form monochloroethane?

(ii) The reaction between ethene and chlorine forms only one product. Name the type of this reaction.

(iii)

1. Draw the structural formula of ethene.
2. What is the feature of the ethene structure which allows ethene to react with chlorine in the way it does?

Answer

(i) Substitution reaction

$\underset{\text{ ethane} }{\text{C}_2\text{H}_6} + \text{Cl}_2 \underset{\Delta}{\xrightarrow{\text{diffused sunlight}}} \underset{\text{monochloroethane}}{\text{C}_2\text{H}_5\text{Cl}} + \text{HCl}$

(ii) Halogenation Reaction

$\text{C}_2\text{H}_4 + \text{Cl}_2 \longrightarrow \text{C}_2\text{H}_4\text{Cl}_2$

(iii) Structural formula of Ethene:

2. The carbon-carbon double bond allows ethene to participate in an addition reaction with chlorine.

The list of some organic compounds is given below.

Ethanol, ethane, methanol, methane, ethyne and ethene.

From the list above, name a compound :

(i) formed by the dehydration of ethanol by conc. H₂SO₄.

(ii) which will give red precipitate with ammonical cuprous chloride solution.

(iii) which forms methanoic acid on oxidation in the presence of copper at 200°C.

(iv) which has vapour density 14 and turn alkaline potassium permanganate green.

(v) which form chloroform on halogenation in the presence of sunlight.

(vi) which decolourises bromine solution in carbon tetrachloride.

Answer

(i) Ethene

(ii) Ethyne

(iii) Methane

(iv) Ethene

(v) Methane

(vi) Ethene and Ethyne

Explain the electrolysis of copper sulphate solution using copper electrodes with

(i) a labelled diagram

(ii) equations

(iii) observation

Answer

(i) Below labelled diagram shows the electrolysis of copper sulphate solution using copper electrodes:

(ii) Dissociation of aq. CuSO₄ :

CuSO₄ ⇌ Cu²⁺ + SO₄^2-

H₂O ⇌ H¹⁺ + OH^1-

Reaction at cathode [pure thin sheet of Cu]:
Cu²⁺ + 2e^- ⟶ Cu [Cu deposited on thin sheet]

Reaction at anode [impure block of active Cu]:
Cu - 2e^- ⟶ Cu²⁺ [product nil - Cu²⁺ ions]

(iii) The observations at the anode & at the cathode are:
At cathode — Brownish pink copper metal is deposited at cathode during electrolysis of copper sulphate.
At anode — Copper ions are formed. Copper anode diminishes in mass.

Blue colour of CuSO₄ remains unchanged.

By choosing appropriate words, complete the statements given below.

Nickel, hydrogenation, hydration, dehydration, Conc. H₂SO₄, Conc. HNO₃, Conc. HCl, homologues.

(i) The conversion of ethanol to ethene is an example of ............... .

(ii) Converting ethanol into ethene requires the use of............... .

(iii) Conversion of ethene to ethane is an example of ............... and the catalyst employed is ............... .

(iv) The successive members of ............... series differ by 14 amu.

Answer

(i) The conversion of ethanol to ethene is an example of dehydration.

(ii) Converting ethanol into ethene requires the use of conc. H₂SO₄

(iii) Conversion of ethene to ethane is an example of hydrogenation and the catalyst employed is nickle .

(iv) The successive members of homologues series differ by 14 amu.