Oswal Model Specimen Papers ICSE Class 10 Chemistry Solutions (Specimen Paper - 8)

Section-A

1. (i) (a) (1) → (iv), (2) → (i), (3) → (ii), (4) → (iii)

(ii) (b) 3

Explanation :

Atomic number of aluminium is 13 so its electronic configuration is 2, 8, 3. Since it has 3 electrons in its valence shell, it will lose 3 electrons to complete its octet hence, its valency is +3.

(iii) (d) (1)-(ii), (2)-(iii), (3)-(i)

Explanation :

Bauxite is an ore of aluminium, white calamine is zinc carbonate and magnetite is triferric tetraoxide.

(iv) (a) (i), (ii) and (iii)

Explanation :

Aluminium oxide is not reduced by common reducing agents such as carbon, carbon monoxide and hydrogen. It requires much stronger method and hence electrolytic reduction is employed for this purpose.

(v) (d) Weak acid and strong base

Explanation :

Sodium carbonate is a basic salt because it is derived from a weak acid (carbonic acid) and strong base (sodium hydroxide).

(vi) (b) Contains one bond pair and six lone pair

Explanation :

There are one bond pair and six lone pair contains in the given chlorine molecule. Also electronegativity difference is zero (diatomic molecule) and the type of bonding is non-polar covalent single bond.

(vii) (c) C₆H₁₂O₆

Explanation :

The empirical formula of the compound is CH₂O, which has two numbers of atoms of hydrogen, and one atom of oxygen for each carbon. The mass of CH₂O = 12 + 2 × 1 + 16 = 30. The approximate molecular weight of the compound is 180. So, the molecular formula of the given compound is C₆H₁₂O₆.

(viii) (d) 1-(v), 2-(iv), 3-(i), 4-(ii), 5-(vi), 6-(iii)

Explanation :

Protons are responsible for nuclear charge.
Sodium is an alkali metal.
Barium is an alkaline earth metal.
Chlorine is halogen.
Electrons occupy the subshells.
Nobel gases have completely filled shells.

(ix) (c) Exothermic and reversible

Explanation :

The reaction of nitrogen and hydrogen gas to produce ammonia is highly exothermic as it produces an enormous amount of heat. Also, it is reversible in nature. The reason is that some of produce, ammonia converts back to the original reactants, nitrogen and hydrogen under the reaction conditions. Since the reverse reaction occurs under the same conditions as the forward reaction, the reaction is reversible.

(x) (a) Iron chloride

Explanation :

Iron chloride gives HCl on hydrolysis as:

$$\underset{\underset{\text{chloride}}{\text{Ferric}}}{\text{FeCl}_{3}} + 3\text{H}_{2}\text{O}\xrightarrow{}\\\underset{\underset{\text{acid}}{\text{Hydrochloric}}}{3\text{HCl}} +\text{Fe(OH)}_{3}$$

(xi) (b) Aqueous ammonia can also be written as NH₃(aq)

Explanation :

The rate of dissolution of ammonia in water is very high. Also, ammonia has low vapour density than air. Therefore, both the options (a) and (c) are wrong. Hence, only option (b) is correct.

(xii) (b) CH₂

Explanation :

The characteristic property of homologous series is that each member of the series differs from the preceding one by the addition of CH₂ group. Hence, the two neighbours of a homologous series differ by CH₂ whose molecular mass is 14 amu. For example, C₂H₆ differs from CH₄ by CH₂.

(xiii) (a) 148

Explanation :

The general formula of alkyne is C_nH_{2n – 2}.

Here, C = 75 (given)

∴ H = 2 × 75 – 2 = 150 – 2

= 148.

(xiv) (c) A silver-grey deposit at the cathode and reddish-brown fumes at the anode.

Explanation :

During electrolysis, Pb²⁺ ions gain electrons at the cathode, and they become Pb atoms, whereas Br⁻ ions lose electrons and pair up to become Br₂ molecules. Therefore, at anode reddish-brown, bromine gas is formed and a silver-grey deposit of lead is seen at the cathode.

(xv) (a) Nitrogen.

Explanation :

Nitrogen is a non-metal which belongs to group 15.

2. (i) (a) The gas collected in the jar is called ammonia.

$$\text{(b)\space 2NH}_{4}\text{Cl} + \text{Ca(OH)}_{2}\xrightarrow{\text{Heat}}\\\text{CaCl}_{2} + 2\text{H}_{2}\text{O} + 2\text{NH}_{3}\uparrow$$

(d) Ammonia gas is passed through CaO(Y) for removing the water content present in the gas.

(e) A glass rod dipped in hydrochloric acid is brought near the mouth of the gas jar, if the rod produces dense white fumes, this indicates that the glass jar is full of ammonia gas.

(ii) (a) 2, (b) 3, (c) 4, (d) 5 and (e) 1

(iii) (a) Reducing agents, Donors

(b) High

(d) AgCl

(e) Hydrogenation

(iv) (a) Hygroscopic substance

(b) Electronegativity

(d) Roasting

(e) 1 Mole

(v) (a)

(b) 1. Propyne

2. 1, 2-Dichloro ethane

Section-B

3. (i) (a) Ammonia

(b) Dilute sulphuric acid

(ii) (a) Cryolite

(b) Concentrated sulphuric acid

(iii) (a) 3rd period, 16th group.

(b) Z is a non-metal.

(iv) (a) High

(b) Nitrogen

4. (i) (a) Al₂O₃.2H₂O

(b) Fe₂O₃

(ii) (a)

Element	Mass %	Atomic weight	Relative number of atoms	Simple ratio
Carbon	80	12	6.66	1
Hydrogen	20	1	20	3

Thus, empirical formula of the compound is CH₃

(b) If vapour density of the compound = 15

Molecular weight = 2 × VD = 2 × 15 = 30

Empirical weight = (1 × 12 + 3 × 1) = 15

$$n =\frac{\text{Molecular weight}}{\text{Empirical weight}}\\=\frac{30}{15} = 2$$

Molecular formula = (Empirical formula)_n

= (CH₃)₂ = C₂H₆

(iii) The flow chart shown in the question represents the decomposition of green vitriol to produce sulphuric acid. This is shown in two steps. The first step involves the production of Fe₂O₃, sulphur dioxide, sulphur trioxide and water. In the second step, sulphur trioxide and water react to give sulphuric acid.

$$\text{FeSO}_{4}.7\text{H}_{2}\text{O}\xrightarrow{\Delta}\\\text{FeSO}_{4}(s) + 7\text{H}_{2}\text{O}\\\text{FeSO}_{4} + 7\text{H}_{2}\text{O}\xrightarrow{\Delta}\\\underset{\text{(A)}}{\text{Fe}_{2}\text{O}_{3}(s) } + \underset{\text{B}}{\text{SO}_{2}}\uparrow + \underset{\text{(C)}}{\text{SO}_{2}(g)}$$

Thus, A = Fe₂O₃

B = SO₂

C = SO₃

(iv) (a) A—Sulphur B—Iron pyrites

C—Sulphur dioxide D—Oxygen

E—Concentrated sulphuric acid F—Water

(b) The gas C will turn acidified potassium dichromate paper green.

(c) V₂O₅ or Pt acts as a catalyst and increases the rate of formation of sulphur trioxide from sulphur dioxide and oxygen.

5. (i) (a) HCl gas is 1.28 times heavier than air so, upward displacement method is applied to collect hydrogen chloride gas. It is not collected in water because it is highly soluble in water.

(b) Ammonium nitrate is not used in the preparation of ammonia as it is explosive in nature and it decomposes forming nitrous oxide and water vapours. Thus, it is not used in the preparation of ammonia.

(ii) (a) The two properties of ionic bonds are :

They have high melting and boiling points.
They are hard and brittle.

(b) 1. Coordinate bond

(iv) (a) Manganese dioxide reacts with concentrated hydrochloric acid to give chlorine gas, which is greenish yellow in colour, whereas copper(II) oxide reacts with concentrated hydrochloric acid to give CuCl₂, but no chlorine gas is evolved.

$$\text{MnO}_{2} + 4\text{HCl}\xrightarrow{}\\\text{Cl}_{2} + \text{Cl}_{2}\uparrow{} + 2\text{H}_{2}\text{O} $$

(b) A reddish precipitate of Iron(III) hydroxide is obtained when ferric sulphate reacts with sodium hydroxide solution, whereas no such precipitate is obtained when ferrous sulphate is mixed with sodium hydroxide.

$$\text{Fe}_{2}(\text{SO}_{4})_{3}(aq) + 6\text{NaOH}\space(aq)\xrightarrow{}\\2\text{Fe(OH)}_{3}(s) + 3\text{Na}_{2}\text{SO}_{4}(aq) $$

(c) Lead nitrate reacts with dilute HCl to form the insoluble salt lead chloride, which appears as the white precipitate. The insoluble lead chloride reacts with excess Cl^- ions (of HCl) to form a soluble complex, the tetrachloroplumbate(II) ion,

$$\text{Pb(NO}_{3})_{2} + \text{2HCl}\xrightarrow{}\\\text{PbCl}\downarrow{} + 2\text{HNO}_{3}\\\text{PbCl}_{2} + \underset{\text{Soluble}}{\text{2HCl}}\xrightarrow{}\\\lbrack\text{PbCl}_{4}\rbrack^{2-} + 2\text{H}^{\normalsize+}(aq)$$

Lead nitrate solution reacts with H₂SO₄ to give lead sulphate precipitate, which does not dissolve further in sulphuric acid solution.

$$\text{Pb(NO}_{3})_{2} +\text{H}_{2}\text{SO}_{4}\xrightarrow{}\\\text{PbSO}_{4}\downarrow + \text{HNO}_{3}$$

$$\textbf{6.\space}(i)\space\underset{\text{1 vol.}}{\text{H}_{2}} + \underset{\text{1 vol.}}{\text{Cl}_{2}}\xrightarrow{}\underset{\text{2 vol}}{\text{HCl}}$$

Since, 1 volume of chlorine reacts with 1 volume of hydrogen.

∴ 4 litre of chlorine will react with only 4 volumes of hydrogen.

∴ (6 – 4) i.e., 2 litre of hydrogen will remain unreacted.

HCl formed will get dissolved in water.

Thus, volume of residual hydrogen gas is 2 litre.

(ii) Given,

Empirical formula = CH

Now, empirical formula mass = 12 × 1 + 1 × 1 = 13

Molecular mass = 2 × Vapour density

= 2 × 13 = 26

Molecular formula mass = n × Empirical formula mass

$$n =\frac{\text{Molecular formula mass}}{\text{Empirical formula mass}}\\=\frac{26}{13} = 2 $$

Molecular formula = n × Empirical formula

= 2 × (CH)

= C₂H₂

Hence, the molecular formula of the compound is C₂H₂.

(iii) (a)

$$\underset{\underset{\text{trioxide}}{\text{Sulphur}}}{\text{SO}_{3}} +\underset{\text{(conc.)}}{\text{H}_{2}\text{SO}_{4}}\xrightarrow{}\underset{\text{Oleum}}{\text{H}_{2}\text{S}_{2}\text{O}_{7}}\\\text{(b)}\\\underset{\text{Methane}}{\text{CH}_{4} + \text{Cl}_{2}}\xrightarrow{\text{hv}}\text{CH}_{3}\text{Cl + HCl}\\\text{CH}_{3}\text{Cl} +\text{Cl}_{2}\xrightarrow{\text{hv}}\text{CH}_{2}\text{Cl}_{2} + \text{HCl}\\\text{CH}_{2}\text{Cl}_{2}\xrightarrow{\text{hv}}\underset{\text{Chloroform}}{\text{CHCl}_{3} + \text{HCl}}\\\text{(c)}\space\underset{\text{Lead oxide}}{3\text{PbO} + 2\text{NH}_{3}}\xrightarrow{}\\\underset{\text{Lead}}{3\text{Pb} + 3\text{H}_{2}\text{O} + \text{N}_{2}}$$

(iv) (a) Concentrated nitric acid is not used for preparation of HCl gas as it is very strong oxidising agent and will end up oxidising HCl gas.

(b) HCl is collected by the upward displacement of air.

7. (i) % of Metal = 100 – 65.5 = 34.5

Chlorine % = 65.5

Relative number of atoms

$$\text{metal} =\frac{34.5}{56} = 0.616\\\text{Chlorine}=\frac{65.5}{35.5} = 1.85$$

Simplest ratio

$$\frac{0.616}{0.616} =1\\\frac{1.85}{0.616} = 3$$

∴ Empirical formula = MCl₃

Molecular = 2 × V.D.

= 2 × 162.5 = 325.0

$$n =\frac{\text{Molecular mass}}{\text{Empirical formula weight}}\\=\frac{325}{162.5} = 2$$

Molecular formula = (Empirical formula)_n

= (MCl₃)₂ = M₂Cl₆

$$\text{(ii)\space (a)}\space\text{Cu}^{2+} + 2e^{\normalsize-}\xrightarrow{}\text{Cu}\\\text{(b)\space}\text{Cu(s)}\xrightarrow{}\text{Cu}^{++} + 2e^{\normalsize-}$$

(iii) (a) When ammonia is passed through excess of chlorine, a highly explosive yellow liquid of nitrogen trichloride is obtained.

$$\underset{\underset{\underset{\text{(yellow liquid)}}{\text{(highly explosive)}}}{\text{Nitrogen trichloride}}}{\text{NH}_{3} + 3\text{Cl}_{2}}\xrightarrow{}\text{NCl}_{3} + \text{3HCl}$$

(b) When a piece of calcium is dropped into water, a vigorous reaction takes place and a colourless, odourless hydrogen gas is evolved, which can be collected in a test tube and water becomes alkaline due to the formation of calcium hydroxide.

$$\text{Ca + 2H}_{2}\text{O}\xrightarrow{}\text{Ca(OH)}_{2} +\text{H}_{2}\uparrow$$

(c) When lead nitrate crystals are heated strongly, a reddish brown coloured gas is released. This is nitrogen dioxide gas. It has an irritating odour. It turns moist blue litmus red.

8. (i) (a) Covalent bonding since L consists of molecules.

(b) L is getting reduced.

(ii) (a) Solution Q is a weak acid as its pH is 5.2

(b) Solution R is a strong alkali as its pH is 12.2

(iii) (a) Conductor has electrons as the conduction particles whereas electrolyte has ions as the conducting species.

(b) Cations are formed when a neutral atom loses electrons whereas anions are formed when a neutral atom gains electron.

(c) An acid dissociates to furnish H⁺ ions and the conjugate base which is negatively charged and a base dissociates to give OH^– ions and a conjugate acid which is positively charged.

(iv) (a) Element C

(b) Element F