Solved 2016 Question Paper ICSE Class 10 Chemistry

Fill in the blanks with the choices given in brackets.

(i) Metals are good ............... (oxidizing agents / reducing agents) because they are electron ............... (acceptors/ donors).

(ii) Electrovalent compounds have ............... (high/low) melting points.

(iii) Higher the pH value of a solution, the more ...............(acidic/alkaline) it is.

(iv) ............... (AgCl / PbCl₂), a white precipitate is soluble in excess NH₄OH.

(v) Conversion of ethene to ethane is an example of ...............(hydration/hydrogenation).

Answer

(i) Metals are good reducing agents because they are electron donors.

(ii) Electrovalent compounds have high melting points.

(iii) Higher the pH value of a solution, the more alkaline it is.

(iv) AgCl, a white precipitate is soluble in excess NH₄OH.

(v) Conversion of ethene to ethane is an example of hydrogenation.

An element with the atomic number 19 will most likely combine chemically with the element whose atomic number is:

1. 17
2. 11
3. 18
4. 20

Answer

17

Reason — The element with atomic number 19 has electronic configuration of (2, 8, 8, 1) and that with atomic number 17 has electronic configuration of (2, 8, 7). This element with atomic number 17 needs 1 electron to complete its octet whereas element with atomic number 19 has an extra electron. Hence, an element with atomic number 19 is most likely to combine with the element with atomic number 17.

The ratio between the number of molecules in 2g of hydrogen and 32g of oxygen is:

1. 1:2
2. 1:0.01
3. 1:1
4. 0.01:1

[Given that H=1, O=16]

Answer

1:1

Reason — 1 mole of any substance contains same (Avogardo No.) of molecules. Hence, ratio is same.

The two main metals in Bronze are:

1. Copper and zinc
2. Copper and lead
3. Copper and nickel
4. Copper and tin

Answer

Copper and tin

Reason — The two main metals in Bronze are Copper and tin .

The particles present in strong electrolytes are:

1. only molecules
2. mainly ions
3. ions and molecules
4. only atoms.

Answer

mainly ions

Reason — The particles present in strong electrolytes are mainly ions.

The aim of the Fountain Experiment is to prove that:

1. HCI turns blue litmus red
2. HCI is denser than air
3. HCl is highly soluble in water
4. HCl fumes in moist air.

Answer

HCl is highly soluble in water

Reason — The aim of the Fountain Experiment is to prove that HCl is highly soluble in water.

Write balanced chemical equations for each of the following:

(i) Action of warm water on AlN.

(ii) Action of hot and concentrated Nitric acid on copper.

(iii) Action of Hydrochloric acid on sodium bicarbonate.

(iv) Action of dilute Sulphuric acid on Sodium Sulphite.

(v) Preparation of ethanol from Ethyl Chloride.

Answer

(i) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃ [g]

(ii) Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

(iii) NaHCO₃ + HCl ⟶ NaCl + H₂O + CO₂ ↑

(iv) Na₂SO₃ + H₂SO₄ (dil) ⟶ Na₂SO₄ + H₂O + SO₂

(v) $\underset{\text{ Chloroethane [ethyl chloride]}}{\text{C}_2\text{H}_5\text{-Cl}} + \text{NaOH [aq.]} \xrightarrow{\text{boil}} \underset{\text{ Ethanol [ethyl alcohol]}}{\text{C}_2\text{H}_5\text{-OH}} + \text{NaCl}$

State your observations when:

(i) Dilute Hydrochloric acid is added to Lead nitrate solution and the mixture is heated.

(ii) Barium chloride solution is mixed with Sodium Sulphate Solution.

(iii) Concentrated Sulphuric acid is added to Sugar Crystals.

(iv) Dilute Hydrochloric acid is added to Copper carbonate.

(v) Dilute Hydrochloric acid is added to Sodium thiosulphate.

Answer

(i) White precipitate of PbCl₂ is formed which is soluble in hot water.
Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

(ii) White coloured precipitate of barium sulphate is formed when sodium sulphate is mixed with barium chloride.
Na₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2NaCl + BaSO₄ ↓ [white ppt. formed]

(iii) Black spongy charred mass of carbon is formed. Steam is seen to be evolved.

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}$

(iv) Effervescence of CO₂ seen which turns lime water milky.
CuCO₃ + 2HCl ⟶ CuCl₂ + H₂O + CO₂

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

Identify the term/substance in each of the following:

(i) The tendency of an atom to attract electrons to itself when combined in a compound.

(ii) The method used to separate ore from gangue by preferential wetting.

(iii) The catalyst used in the conversion of ethyne to ethane.

(iv) The type of reactions alkenes undergo

(v) The electrons present in the outermost shell of an atom.

Answer

(i) Electronegativity

(ii) Froth flotation process

(iii) Nickel or platinum or palladium.

(iv) Addition reaction

(v) Valence Electrons

(i) A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram molecular weight of the gas.

(ii) How much Calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogen dioxide evolved:

2Ca(NO₃)₂ ⟶ 2CaO + 4NO₂ + O₂

[Ca = 40, N = 14, O = 16]

Answer

(i) 20 lit. of gas weighs = 32 g.

1 mole = 22.4 lit. of gas will weigh = $\dfrac{32}{20}$ x 22.4 = 35.84 g.

Hence, gram mol. weight of the gas = 35.84 g.

(ii)

$\begin{matrix} 2\text{Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[(40) + & & 2(40 & & 4[22.4] \\ 2(14 +3(16))] & & + 16) & & \text{ lit.} \\ = 328 \text{ g} & & 112 \text{g} \\ \end{matrix}$

(a) 328 g of Ca(NO₃)₂ produces 112 g of calcium oxide.

∴ 82 g of Ca(NO₃)₂ will produce = $\dfrac{112}{328}$ x 82 = 27.99 g = 28 g

Hence, 28 g of calcium oxide is produced.

(b) 328 g of Ca(NO₃)₂ produces 4(22.4) lit of nitrogen dioxide.

∴ 82 g of Ca(NO₃)₂ will produce = $\dfrac{4 \times 22.4}{328}$ x 82 = 22.4 lit.

Hence, vol of nitrogen dioxide evolved = 22.4 lit.

Match the salts given in Column I with their method of preparation given in Column II:

Answer

(i) Write the IUPAC names of each of the following:

(ii) Rewrite the following sentences by using the correct symbol > (greater than) or < (less than) in the blanks given:

1. The ionization potential of Potassium is .............. that of Sodium.

2. The electronegativity of Iodine is .............. that of Chlorine.

Answer

(i) 1. Propene

2. 2-butyne

3. ethanal

(ii) 1. The ionization potential of potassium is < (less than) that of Sodium.

2. The electronegativity of iodine is < (less than) that of chlorine.

Use the letters only written in the Periodic Table given below to answer the questions that follow:

(i) State the number of valence electrons in atom J.

(ii) Which element shown forms ions with a single negative charge?

(iii) Which metallic element is more reactive than R?

(iv) Which element has its electrons arranged in four shells?

Answer

(i) Atom J has 5 valence electrons as it belongs to the fifth group in the periodic table.

(ii) M forms ions with a single negative charge as it has 7 electrons in the valence shell and obtaining one more electron completes it's octet.

(iii) Metal T is more reactive than R because elements at the bottom of the group are more reactive.

(iv) T has it's electrons arranged in 4 shells as it belongs to 4^th period.

Fill in the blanks by selecting the correct word from the brackets:

(i) If an element has a low ionization energy then it is likely to be ............... (metallic/non metallic).

(ii) If an element has seven electrons in its outermost shell then it is likely to have the ............... (largest / smallest) atomic size among all the elements in the same period.

Answer

(i) If an element has a low ionization energy then it is likely to be metallic.

(ii) If an element has seven electrons in it's outermost shell then it is likely to have the smallest atomic size among all the elements in the same period.

The following table shows the electronic configuration of the elements W, X, Y, Z:

(i) What type of bond is formed between:

     1. W and X
     2. Y and Z

(ii) What is the formula of the compound formed between :
     1. X and Z
     2. W and X

Answer

(i) Type of bond formed

(a) W (2,8,1) and X (2,8,7) — Electrovalent bond is formed as W donates 1 electron and X takes up that electron and both attain a stable octet.

(b) Y (2, 5) and Z (1) — Covalent bond as electron sharing takes place between the two elements. The valency of Z element is 1 and that of Y is 3. Z needs one electrons to attain stable duplet structure of nearest noble gas - He [2] and Y needs three electrons to attain stable octet structure of nearest noble gas - Ne [2,8]
One Y atom shares three electron pairs one with each of the three atoms of Z such that Z acquires a duplet configuration and Y attain a stable octet configuration resulting in the formation of three single covalent bonds.

(ii) Formula of the compound formed

(a) X (2,8,7) and Z (1) — XZ as X needs one 1 electron to complete it's octet and Z needs needs one electron to complete it's duplet hence, they both share one electron pair and forms ZX compound.

(b) W (2,8,1) and X (2,8,7) — WX as W has the tendency to donate one electron and gain stability whereas X has the tendency to gain one electron to be stable. Hence, W donates one electron and X takes it and form WX compound.

Write a balanced chemical equation for each of the following:

(i) Burning of ethane in plentiful supply of air.

(ii) Action of water on Calcium carbide.

(iii) Heating of Ethanol at 170°C in the presence of conc. Sulphuric acid.

Answer

(i) Carbon dioxide and water are produced:

$2\text{C}_2\text{H}_6 + 7\text{O}_2[\text{excess}] \longrightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} + \Delta$

(ii) Calcium hydroxide and ethyne are produced

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

(iii) Ethene and water are produced

$\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}\$

Give the structural formulae of each of the following :

(i) 2-methyl propane

(ii) Ethanoic acid

(iii) Butan-2-ol

Answer

(i) 2-methyl propane :

(ii) Ethanoic acid :

(iii) Butan – 2 – ol :

Equation for the reaction when compound A is bubbled through bromine dissolved in carbon tetrachloride is as follows:

(i) Draw the structure of A.

(ii) State your observation during this reaction.

Answer

(i) Compound A is Ethene. Its structure is shown below:

(ii) Brown colour of bromine is discharged.

Fill in the blanks using the appropriate words given below:

(Sulphur dioxide, Nitrogen dioxide, Nitric oxide, Sulphuric acid)

(i) Cold, dilute nitric acid reacts with copper to give ...............

(ii) Hot, concentrated nitric acid reacts with sulphur to form ...............

Answer

(i) Cold, dilute nitric acid reacts with copper to given nitric oxide.

(ii) Hot, concentrated nitric acid reacts with sulphur to form nitrogen dioxide.

Identify the gas evolved and give the chemical test in each of the following cases:

(i) Dilute hydrochloric acid reacts with sodium sulphite.

(ii) Dilute hydrochloric acid reacts with iron (II) sulphide.

Answer

(i) SO₂ is evolved when dilute hydrochloric acid reacts with sodium sulphite.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ [g]

(ii) H₂S gas is evolved when dilute hydrochloric acid reacts with Iron [II] sulphide.

FeS + 2HCl ⟶ FeCl₂ + H₂S

State your observations when ammonium hydroxide solution is added drop by drop and then in excess to each of the following solutions:

(i) copper sulphate solution

(ii) zinc sulphate solution.

Answer

(i) Pale blue precipitate of Cu(OH)₂ is formed, which becomes deep blue when excess of ammonium hydroxide is added.

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

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

(ii) Gelatinous white precipitate of zinc hydroxide is formed which becomes colourless solution on adding excess of ammonium hydroxide.

ZnSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Zn(OH)₂ ↓

Zn(OH)₂ +(NH₄)₂SO₄ + 2NH₄OH [in excess] ⟶ 4H₂O + [Zn(NH₃)₄]SO₄

Write equations for the reactions taking place at the two electrodes (mentioning clearly the name of the electrode) during the electrolysis of:

(i) Acidified copper sulphate solution with copper electrodes.

(ii) Molten lead bromide with inert electrodes.

Answer

(i) Electrode — Cathode : Copper, Anode : Copper

Dissociation of aq. CuSO₄ :

CuSO₄ ⇌ Cu²⁺ + SO₄^2-

H₂O ⇌ H¹⁺ + OH^1-

H₂SO₄ ⇌ 2H¹⁺ + SO₄^2-

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]

(ii) Electrode — Cathode : Iron or Graphite, Anode : Graphite

Dissociation of molten PbBr₂ : PbBr₂ ⇌ Pb²⁺ (cathode) + 2Br^1- (anode)

Reaction at cathode:
Pb²⁺ + 2e^- ⟶ Pb [product lead metal]

Reaction at anode:
Br^1- - 1e^-⟶ Br
Br + Br ⟶ Br₂

(i) Name the product formed at the anode during the electrolysis of acidified water using platinum electrodes.

(ii) Name the metallic ions that should be present in the electrolyte when an article made of copper is to be electroplated with silver.

Answer

(i) Oxygen

(ii) Silver ions

A gas cylinder contains 12 x 10²⁴ molecules of oxygen gas.
If Avogadro's number is 6 x 10²³; Calculate:

(i) the mass of oxygen present in the cylinder.

(ii) the volume of oxygen at S.T.P. present in the cylinder. [O=16]

Answer

Gram molecular mass of oxygen = 32 g = 1 mole and contains 6 x 10²³ molecules i.e., 6 x 10²³ molecules weigh 32 g

∴ 12 x 10²⁴ molecules will weigh = $\dfrac{32}{ 6\times 10^{23}}$ x 12 x 10²⁴ = 640 g

(ii) 6 x 10²³ molecules occupy 22.4 lit. of vol.

∴ 12 x 10²⁴ molecules will occupy = $\dfrac{22.4}{ 6\times 10^{23}}$ x 12 x 10²⁴ = 448 lit.

Hence, mass of O₂ present in the cylinder = 640 g and volume of O₂ at S.T.P. present in the cylinder = 448 lit.

A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.

[C=12, H=1]

Answer

Simplest ratio of whole numbers = C : H = 1 : 2.5 = 2 : 5

Hence, empirical formula is C₂H₅

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 29}{29} =2$

∴ Molecular formula = n[E.F.] = 2[C₂H₅] = C₄H₁₀

The equation :

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O,

represents the catalytic oxidation of ammonia. If 100 cm³ of ammonia is used calculate the volume of oxygen required to oxidize the ammonia completely.

Answer

[By Lussac's law]

$\begin{matrix} 4\text{NH}_3 & + & 5\text{O}_2 &\longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \end{matrix}$

To calculate the volume of oxygen required :

$\begin{matrix}\text{NH}_3 & : & \text{O}_2 \\ 4 \text{ vol.} & : & 5 \text{ vol.} \\ 100 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{5}{4} \times 100 = 125 \text{ cm.}^3$

Hence, volume of oxygen required = 125 cm³

By drawing an electron dot diagram show the formation of ammonium ion [N = 7, H = 1]

Answer

Electron dot diagram showing the formation of ammonium ion is given below:

Name the gas evolved when the following mixtures are heated:

(i) Calcium hydroxide and Ammonium Chloride

(ii) Sodium Nitrite and Ammonium Chloride

Answer

(i) Ammonia gas

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

(ii) Nitrogen gas

NH₄Cl + NaNO₂ ⟶ NaCl + NH₄NO₂

NH₄NO₂ ⟶ 2H₂O + N₂

Write balanced chemical equations for each of the following:

(i) When excess of ammonia is treated with chlorine.

(ii) An equation to illustrate the reducing nature of ammonia.

Answer

(i) 8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

(ii) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.

A: Typical acid property

B: Non-volatile acid

C: Oxidizing agent

D: Dehydrating agent

Choose the property [A, B, C or D] depending on which is relevant to each of the following:

(i) Preparation of hydrogen chloride gas.

(ii) Preparation of copper sulphate from copper oxide.

(iii) Action of conc. sulphuric acid on sulphur.

Answer

(i) B: Non-volatile acid

$\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}$

(ii) A: Typical acid property

CuO + H₂SO₄ (conc.) ⟶ CuSO₄ + H₂O

(iii) C: Oxidizing agent

S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

Give reasons why:

(i) Sodium Chloride will conduct electricity only in fused or aqueous solution state.

(ii) In the electroplating of an article with silver, the electrolyte sodium argento-cyanide solution is preferred over silver nitrate solution.

(iii) Although copper is a good conductor of electricity, it is a non-electrolyte.

Answer

(i) The ions Na⁺ and Cl^– are not free but held together by strong electrostatic force of attraction. In fused or molten state the ions break free and move. Hence, NaCl will conduct electricity only in fused state or aq. soln. state.

(ii) Migration of Ag¹⁺ ions from sodium argentocyanide solution is slow compared to that from silver nitrate. Hence, an even deposition of metal silver is obtained on the article. Therefore, the electrolyte sodium argentocyanide soln. is preferred over silver nitrate solution.

(iii) 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.

(i) Name the solution used to react with Bauxite as a first step in obtaining pure aluminium oxide, in the Baeyer's process.

(ii) Write the equation for the reaction where the aluminium oxide for the electrolytic extraction of aluminium is obtained by heating aluminium hydroxide.

(iii) Name the compound added to pure alumina to lower the fusion temperature during the electrolytic reduction of alumina.

(iv) Write the equation for the reaction that occurs at the cathode during the extraction of aluminium by electrolysis.

(v) Explain why it is preferable to use a number of graphite electrodes as anode instead of a single electrode, during the above electrolysis.

Answer

(i) Sodium hydroxide

(ii) Conversion of Al(OH)₃ - to pure alumina - heat on Al(OH)₃

$\underset{\text{[aluminium hydroxide]}}{\text{2Al(OH)}_3} \xrightarrow{1100\degree\text{C}} \underset{\text{[pure alumina]}}{\text{Al}_2\text{O}_3} + \text{3H}_2\text{O}$

(iii) Cryolite (Na₃AlF₆)

(iv) 2Al³⁺ + 6e^- ⟶ 2Al

(v) The oxygen evolved at the anode escapes as a gas or reacts with the carbon anode. 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] The carbon anode is hence consumed and renewed periodically after a certain period of usage. It is therefore preferable to use a number of graphite electrodes as anode instead of a single electrode.

State what would you observe when:

(i) Washing Soda Crystals are exposed to the atmosphere.

(ii) The salt ferric chloride is exposed to the atmosphere.

Answer

(i) When crystals of washing soda are exposed to air, it loses its water of crystallisation due to efflorescence.

(ii) When ferric chloride is exposed to the atmosphere, it absorbs moisture from the atmospheric air to become moist and ultimately dissolves in the absorbed water, forming a saturated solution.

Identify the cations in each of the following case:

(i) NaOH solution when added to the solution 'A' gives a reddish brown precipitate.

(ii) NH₄OH solution when added to the solution 'B' gives a white ppt which does not dissolve in excess.

(iii) NaOH solution when added to the solution 'C' gives a white ppt which is insoluble in excess.

Answer

(i) Ferric (Fe³⁺) ion

(ii) Lead (Pb²⁺) ion

(iii) Calcium (Ca²⁺) ion