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Model Paper 2

Class 10 - Concise Chemistry Selina

SECTION I [40 Marks]

Question 1(a)

Correct the following statements and rewrite them :

(i) Sodium reacts with chlorine to form sodium chloride.

(ii) The electrolysis of lead bromide liberates lead and bromine.

(iii) Copper sulphate crystals are dehydrated by sulphuric acid.

(iv) Calcium nitrate reacts with sodium sulphate to form calcium sulphate.

(v) Metals of IA group are called alkaline earth metals.


(i) Sodium atom reacts with chlorine atom to form sodium chloride molecule.

(ii) The electrolysis of lead bromide liberates lead at cathode and bromine at anode.

(iii) Copper sulphate crystals are dehydrated by conc. sulphuric acid.

(iv) Calcium nitrate reacts with sodium hydroxide to form calcium hydroxide.

(v) Metals of II A group are called alkaline earth metals.

Question 1(b)

Write ionic reaction at the electrodes during the electrolysis of :

(i) copper II sulphate solution with copper electrodes,

(ii) acidulated water.


Dissociation of aq. copper sulphate

CuSO4Cu2+ + SO42- [ions present — Cu2+, H1+, SO42-, OH1-]

H2O ⇌ H1+ + OH1-

Reaction at cathode

Cu2+ + 2e-Cu [product copper metal]

Reaction at anode

Cu - 2e-Cu2+ [product nil - Cu2+ ions]

(ii) Dissociation of acidified water

H2SO42H1+ + SO42-

H2O ⇌ H1+ + OH1- [ions H1+, SO42-, OH1-]

Reaction at cathode

H1+ + 1e-H x 4

2H + 2H ⟶ 2H2 [product Hydrogen gas]

Reaction at anode

OH1- - 1e-OH x 4

4OH ⟶ 2H2O + O2 [product oxygen gas]

Question 1(c)

(i) What is the atomicity of chlorine.

(ii) Name the compound formed when the above mentioned gas reacts with iron. Give equation.

(iii) Zinc sulphate reacts with sodium hydroxide first a little then in excess. Write a balanced equation for the two reactions with different conditions.


(i) Diatomic

(ii) Iron (III) chloride

2Fe + 3Cl2 ⟶ 2FeCl3

(iii) ZnSO4 + 2NaOH ⟶ Na2SO4 + Zn(OH)2

[Zn(OH)2 + 2NaOH [in excess] ⟶ 2H2O + Na2ZnO2]

Question 1(d)

If a crop of wheat replaces 30 kg of nitrogen per hectare of soil, what mass of the fertilizer calcium nitrate would be required to replace the nitrogen of a 5-hectare field ?

(at wt. N = 14, O = 16, Ca = 40)


Molecular weight of Ca(NO3)2
= 40 + 2[14 + 3(16)]
= 40 + 2[14 + 48]
= 40 + 2(62)
= 40 + 124 = 164 g

Total nitrogen that has to be replaced in 5 hectares of land = 30 x 5 = 150 kg = 150 x 103 g

28 g of N is present in 164 g of calcium nitrate

∴ 150 x 103 g of N is present in = 16428\dfrac{164}{28} x 150 x 103 = 878.57 x 103 g = 878.57 kg

Hence, 878.57 kg of fertilizer will be required

Question 1(e)

(i) Oxygen oxidises ethyne to produce 8.4 litres of carbon dioxide at STP. Calculate the volume of ethyne used in this reaction.

2C2H2 + 5O2 ⟶ 4CO2 + 2H2O

(ii) A2B is the empirical formula of a compound which has vapour density 25. If atomic weight of A and B are 10 and 5 respectively. Find it's molecular formula.


(i) [By Lussac's law]

2C2H2+5O24CO2+2H2O2 vol.:5 vol.4 vol.:2 vol.\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 &\longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} & \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} & : & 2\text{ vol.} \end{matrix}

To calculate the volume of ethyne gas

CO2:C2H24:28.4:x\begin{matrix}\text{CO}_2 & : & \text{C}_2\text{H}_2 \\ 4 & : & 2 \\ 8.4 & : & x \end{matrix}

x=24×8.4=4.2 L\therefore x = \dfrac{2}{4} \times 8.4 = 4.2 \text{ L}

Hence, volume of ethyne gas required = 4.2 L.

(ii) Empirical formula is A2B

Empirical formula weight = 2(10) + 5 = 25

Vapour density (V.D.) = 25

Molecular weight = 2 x V.D. = 2 x 25

n=Molecular weightEmpirical formula weight=2×2525=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 25}{25} = 2

∴ Molecular formula = n[E.F.] = 2[X2Y] = A4B2

Question 1(f)

(i) Name the metalloid(s) of period 3.

(ii) Define atomic size.

(iii) What would be seen on mixing a solution of calcium chloride with a solution of sodium carbonate? Write an equation also.

(iv) Write an equation for the preparation of a hydroxide of a metal which is

  1. Green

  2. Light blue


(i) Silicon [Si]

(ii) Atomic size is the distance between the centre of the nucleus of an atom and it's outermost shell.

(iii) White precipitate of calcium carbonate appears.

CaCl2 + Na2CO3 ⟶ 2NaCl + CaCO3


  1. Green : Iron [II] hydroxide
    FeSO4 + 2NaOH ⟶ Na2SO4 + Fe(OH)2

  2. Light blue : Copper [II] hydroxide
    CuSO4 + 2NaOH ⟶ Na2SO4 + Cu(OH)2

Question 1(g)

Give a chemical test in each case to distinguish between the following pairs of chemicals.

(i) Sodium chloride solution and sodium nitrate solution.

(ii) Sodium sulphate solution and sodium chloride solution.

(iii) Calcium nitrate solution and zinc nitrate solution.


(i) Add silver nitrate soln. to the given solns., sodium chloride reacts to form a white ppt. which is soluble in NH4OH and insoluble in dil. HNO3. The other soln. is sodium nitrate.

NaCl + AgNO3 ⟶ AgCl + NaNO3

NaNO3 + AgNO3 ⟶ no white ppt.

(ii) Sodium sulphate solution reacts with barium chloride to form white ppt. of barium sulphate and sodium chloride, whereas, no reaction takes place in case of sodium chloride solution because both of them have the same anion.

Na2SO4 + BaCl2 ⟶ BaSO4 ↓ [white ppt.] + 2NaCl

NaCl + BaCl2 ⟶ no reaction

(iii) When NaOH is added to the given soln., Zn(NO3)2 reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Ca(NO3)2 forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.

Question 1(h)


(i) The volume of 2.5 moles of a gas X at stp.

(ii) The weight of 2.8 dm3 of a basic gas you have learnt, at STP.

(iii) The molecular weight of a gas Y whose 5.6 litres of volume at STP weighs 3.5 gms.

(at wt of C = 12, X = 14, H = 1, O = 16)


(i) Vol. occupied by 1 mole = 22.4 lit

∴ Vol. occupied by 2.5 moles = 22.4 x 2.5 = 56 lit.

Hence, vol. occupied is 56 lit.

(ii) Let the basic gas be NH3

Molecular mass of NH3 = 14 +3(1) = 17 g

At STP, 1 mole of ammonia weighs 17 g and occupy 22.4 dm3 volume.

∴ 1 dm3 volume has weight = 1722.4\dfrac{17}{22.4} = 0.76 g

∴ 2.8 dm3 of NH3 has weight = 2.8 x 0.76 = 2.128 g

Hence, weight of 2.8 dm3 of ammonia = 2.128 g

(iii) 5.6 L of volume at STP weighs 3.5 gms.

∴ 22.4 L of volume weighs = 3.55.6\dfrac{3.5}{5.6} x 22.4 = 14 g

Hence, molecular weight = 14g

SECTION II [40 Marks]

Question 2(a)

Concentrated nitric acid oxidises phosphorus to phosphoric acid as

P + HNO3 ⟶H3PO4 + H2O + NO2

Balance the equation and answer the following questions based on it.

(i) What mass of phosphoric acid can be prepared from 6.2 g of phosphorus?

(ii) What mass of nitric acid will be consumed at the same time ?

(iii) What would be the volume of steam produced at the same time if measured at 760 mm. Hg pressure and 273°C.

(Atomic wt of H = 1, N = 14, O = 16, P = 31) ?


Balanced equation is :

P + 5HNO3 ⟶ H3PO4 + H2O + 5NO2

(i) Gram molecular mass of phosphorus = 31 g

gram molecular mass of H3PO4 = 3(1) + 31 + 4(16) = 3 + 31 + 64 = 98 g

31 g of P makes 98 g of phosphoric acid

∴ 6.2 g of P will make 9831\dfrac{98}{31} x 6.2 = 19.6 g

Hence, mass of phosphoric acid prepared = 19.6 g

(ii) Gram molecular mass of nitric acid (HNO3) = 1 + 14 + 3(16) = 63 g

31 g of P needs 63 g of nitric acid

∴ 6.2 g of P will need 6331\dfrac{63}{31} x 6.2 = 12.6 g of nitric acid

Hence, mass of nitric acid needed = 12.6 g

(iii) 31 g of P gives 22.4 lit of steam

∴ 6.2 g of P will give 22.431\dfrac{22.4}{31} x 6.2 = 4.48 L

Hence, volume of steam produced = 4.48 L

Question 2(b)

What is an 'alloy'. Give composition of

(i) duralumin

(ii) brass

(iii) fuse metal.


Alloy — An alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements.

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

(ii) Brass — Cu [60-80%], Zn [40-20%]

(iii) Fuse metal — Pb [50%], Sn [50%]

Question 3(a)


(i) Acid

(ii) molar volume

(iii) neutralisation.


(i) Acid — An acid is a compound which when dissolved in water yields hydronium ions [H3O+] as the only positively charged ion.

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

(iii) Neutralization — It is the process due to which [H+] ions of an acid react completely or combine with [OH-] ions of a base to give salt and water only.

Question 3(b)

(i) Define pH. If the hydrogen ion concentration of a solution is 10-12 moles per litre, calculate its pH value.

(ii) What fluid is injected when yellow wasps or white ants bite? How can the pain be relieved?

(iii) A solution of Iron (III) chloride has a pH less than 7. Is the solution acidic or alkaline. Write the equation of its hydrolysis.

(iv) If iron is reacted with dilute sulphuric acid, what will be the product formed ?


(i) pH is defined as the negative logarithm [to the base 10] of the hydrogen ion concentration expressed in moles/litre.

Thus, pH = -log10H+. It represents the strength of acids and alkalis, expressed in terms of hydrogen ion concentration [H+ aq.]

pH = -log10H+ = -log10[10-12] = 12

Hence, pH value = 12

(ii) When yellow wasps or white ants bite, they inject formic acid into the skin. So by applying base it neutralises the acid and gives relief.

(iii) Solution will be acidic as pH is less than 7.

Hydrolysis of Iron (III) chloride:

FeCl3 ⟶ Fe3+ + 3Cl-

H2O ⟶ H+ + OH-

Further, the ferric ions partially combine with hydroxide ions to form ferric hydroxide.

Fe3+ + 3OH- ⟶ Fe(OH)3

This precipitation removes the hydroxyl ions [OH-] from the solution leaving a relative excess of hydrogen ions and this excess of hydrogen ions makes the solution acidic.

(iv) Iron sulphate will be formed

Fe + H2SO4 ⟶ FeSO4 + H2

Question 4(a)

What do you observe when —

(i) Ammonia is passed over heated copper oxide.

(ii) Conc. H2SO4 is added to blue vitriol.

(iii) Acetylene is passed into bromine water.

(iv) NaCl is added to silver nitrate solution.

(v) Conc. HCl acid is exposed to atmosphere.


(i) Ammonia gas reduces black copper [II] oxide to brown Copper.

2NH3 + 3CuO ⟶ 3Cu + 3H2O + N2

(ii) Blue hydrated copper sulphate becomes white anhydrous copper sulphate

CuSO4.5H2OConc. H2SO4CuSO4+5H2O\text{CuSO}_4.5\text{H}_2\text{O} \xrightarrow{\text{Conc. H}_2 \text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O}

(iii) Reddish brown bromine dissolved in water turns colourless when acetylene (ethyne) is passed through it.

C2H2 + Br2 ⟶ C2H2Br2 + Br2 ⟶ C2H2Br4

(iv) NaCl reacts with silver nitrate solution to form a white ppt.

NaCl + AgNO3 ⟶ AgCl ↓ [white ppt.] + NaNO3

(v) Due to strong affinity for water, concentrated hydrochloric acid pulls moisture from the air towards itself. This moisture forms droplets of water and hence, a cloud can be seen.

Question 4(b)

(i) Give 2 tests to distinguish between HCl solution and Ammonia solution.

(ii) Distinguish between MnO2 and CuO


(i) Two tests to distinguish between HCl solution and Ammonia solution:

  1. HCl turns moist blue litmus paper red, whereas, ammonia gas turns moist red litmus blue.
  2. With HCl phenolphthalein solution remains colourless whereas, ammonia gas turns it pink.

(ii) Distinguish between MnO2 and CuO:

(a) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper oxide and filtrate is bluish in colour.

MnO2 + 4HCl ⟶ MnCl2 + 2H2O + Cl2

CuO + 2HCl ⟶ CuCl2 + H2O

(b) On adding ammonium hydroxide to the above filtrate, no ppt. is formed in case of MnO2, whereas, in case of copper oxide, pale blue ppt. is formed. The ppt. dissolves in excess of ammonium hydroxide forming azure blue colour.

Question 5(a)

Write an ionic reaction at the electrodes during electroplating of silver over brass spoon after giving the ionisation of the electrolyte.


Dissociation of sodium silver cyanide

Na[Ag(CN)2] ⇌ Na1+ + Ag1+ + 2CN1-

H2O ⇌ H1+ + OH1-

Reaction at cathode [article to be plated]

Ag1+ + 1e1- ⟶ Ag [Ag deposited on article]

Reaction at anode [block of active-silver]

Ag - 1e1- ⟶ Ag1+ [product nil - Ag1+ ions]

Question 5(b)

Draw an electron dot diagram of :

(i) a non-polar hydrocarbon molecule.

(ii) a polar covalent molecule.


Electron dot diagram of :

(i) a non-polar hydrocarbon molecule : Methane

Draw the electron dot structure of covalent compound methane. Chemical Bonding, Concise Chemistry Solutions ICSE Class 10

(ii) a polar covalent molecule : HCl

Draw the electron dot structure of covalent compound HCl. Chemical Bonding, Concise Chemistry Solutions ICSE Class 10

Question 5(c)

State three harmful effects of acid rain.


Harmful effects of acid rain are:

  1. Decolourises leaf pigments, retards growth of crops and reduces rate of photosynthesis.
  2. Acid rain gets absorbed by plants, animals and directly or indirectly toxicity enters food chain affecting humans.
  3. Corrodes metallic surfaces, disintegrates paper and leather, weakens building material such as statues, sculptures, marbles, limestone [CaCO3] etc.

Question 6(a)

State the elements of group 17. What are they called and why ?


The elements of group 17 are Fluorine, Chlorine, Bromine, Iodine, and Astatine.

Group 17 elements are called halogens. The name halogen (Greek, halo = sea or salt + gen = producing) means 'salt former'. Group 17 elements form salts, hence, they are called Halogens.

Question 6(b)


(i) Ionisation potential

(ii) Electron affinity

How do they vary in VII A group or IIIrd period.


(i) Ionization Potential is the amount of energy required to remove an electron from the outer most shell of an isolated gaseous atom.

(ii) Electron affinity is the amount of energy released when an atom in the gaseous state accepts an electron to form an anion.

Both ionization potential and electron affinity decreases down the group VII A.

Both ionization potential and electron affinity increases as one moves from left to right in IIIrd period.

Question 6(c)

Name the common refrigerant. How does it deplete ozone layer ?


The main refrigerants used are Freon chlorofluorocarbons (CFC). They deplete ozone layer and contribute to global warming. The chlorofluorocarbons are decomposed by ultraviolet rays to highly reactive chlorine which is produced in the atomic form.

CF2Cl2UltravioletCF2Cl+Cl[free radical]\text{CF}_2\text{Cl}_2 \xrightarrow{\text{Ultraviolet}} \text{CF}_2\text{Cl} + \text{Cl} [\text{free radical}]

The free radical [Cl] reacts with ozone and chlorine monoxide is formed.

Cl + O3 [ozone] ⟶ ClO + O2

This causes depletion of ozone. Chlorine monoxide further reacts with atomic oxygen and produces more free radicals.

ClO + O ⟶ Cl + O2

Again this free radical [Cl] destroys ozone, and the process continues thereby giving rise to ozone depletion.

Question 7(a)

Draw the structures of,

(i) an alcohol with 2 carbon atoms

(ii) an alkyne with 3 carbon atoms

Give their IUPAC names.


(i) Ethanol [C2H5OH]

Draw the structure of an alcohol with 2 carbon atoms. Concise Chemistry Solutions ICSE Class 10.

(ii) Propyne [C3H4]

Draw the structure of an alkyne with 3 carbon atoms. Concise Chemistry Solutions ICSE Class 10.

Question 7(b)

Write the equations for the following and name the products formed.

(i) Ethyl alcohol reacts with sulphuric acid at 170°C.

(ii) Acetic acid reacts with conc. sodium hydroxide solution.

(iii) Acetic acid reacts with potassium hydrogen carbonate.


(i) Products formed — ethene and water.

C2H5OH ethyl alcohol170°CConc. H2SO4[excess]C2H4ethene+H2\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}\

(ii) Products formed — sodium acetate and water


(iii) Products formed are potassium acetate, water and carbon dioxide gas.