Oswal Model Specimen Papers ICSE Class 10 Chemistry Solutions (Specimen Paper - 7)

Section-A

1. (i) (c) Oxalic acid.

Explanation :

From the given options, oxalic acid is a weak electrolyte. On the other hand, HCl, nitric acid and sulphuric acid are strong electrolytes.

(ii) (c) Strong reducing agent

Explanation :

Alkali metals are strong reducing agents because they can donate electrons.

(iii) (d) At the Cathode :

$$\text{Cu}^{2 +} + 2e^{\normalsize-}\xrightarrow{}\text{Cu}$$

At the Anode :

$$\text{Cu - 2e}^{\normalsize-}\xrightarrow{}\text{Cu}^{2+}$$

Explanation :

Electrode on the left side is the oxidising electrode because anode is always connected to positive terminal of the battery, where copper atoms lose electrons at this electrode.

(iv) (c) They are insoluble in water.

Explanation :

Ionic compounds are soluble in water but insoluble in organic solvents.

(v) (b) White.

Explanation :

When NaOH is treated with aluminium oxide, a white salt, known as sodium metaaluminate is formed.

(vi) (d) J

Explanation :

J is fluorine and it is the most electronegative element because it has 5 electrons in its 2p shell.

(vii) (c) C₆H₆

Explanation :

Molecular formula = n × Empirical formula

Empirical formula mass = CH = 12 + 1 = 13

Molecular mass = 78

Therefore, n = 6

(viii) (b) Sodium nitrate and sulphuric acid

Explanation :

Nitric acid in the laboratory is prepared by reaction of sodium or potassium nitrate and sulphuric acid. They are distilled to produce the required ammonia.

(ix) (c) Concentrated sulphuric acid

Explanation :

The delivery tube is dipped in concentrated sulphuric acid so that easy movement of HCl gas takes place and no impurity can add to it. It also helps in making the HCl gas moisture-free.

(x) (a) O₂, V₂O₅, H₂SO₄, H₂O

Explanation :

In the contact process, the first step includes oxidation of sulphur to produce SO₂, which on purification, is converted to SO₃ using the catalyst V₂O₅. This SO₃ is then subject to sulphuric acid to produce oleum which on dilution gives sulphuric acid.

(xi) (a) Calcium chloride

Explanation :

The reaction of calcium chloride with water is neutralisation and a double displacement reaction, which produces salt and water. The chlorine atom of HCl displaces oxygen from CaO and hence produces calcium chloride (CaCl₂).

(xii) (a) C_nH_{2n – 2}

Explanation :

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

(xiii) (a) Gangue

Explanation :

An unwanted earthly material associated with tin ore as impurity is known as gangue.

(xiv) (b) Ions

Explanation :

NaCl in molten state conducts electricity as it gets dissociated in to Na⁺ ion and Cl^– ion.

(xv) (a) CH₄

$$\text{CO + 3H}_{2}\xrightarrow{\text{Ni/300\degree C}}\\\text{CH}_{4} + \text{H}_{2}\text{O}$$

2. (i) (a) A is cathode and B is anode.

(b) Molten fluorides of Al, Na and Ba.

(ii)

Column A	Column B
(a) Acid salt	4. Sodium hydrogen carbonate
(b) Copper oxide	1. Black in colour
(c) Zinc hydroxide	5. Soluble in excess sodium hydroxide
(d) Copper metal	2. Reddish brown
(e) Polar compound	3. Hydrogen chloride

(iii) (a) dehydration

(b) sodium bisulphate or sodium hydrogen sulphate

(d) alkanes

(e) sodium

(iv) (a) Cation that does not form a precipitate with ammonium hydroxide but forms one with sodium hydroxide — Ca²⁺

(b) The electrolyte used for electroplating an article with silver is solution of sodium argentocyanide i.e., Na[Ag(CN)₂]

(d) An organic compound containing — COOH functional group is carboxylic acid

(e) A solid formed by reaction of two gases, one of which is acidic and the other basic in nature is ammonium chloride (NH₄Cl) (formed by combining vapours of ammonia with hydrogen chloride gas).

(v)

(b) 1. 2-Methyl butanoic acid

2. 1,2- Dibromoethane

Section-B

3. (i) (a) Vanadium pentoxide (V₂O₅)

(b) Ammonium chloride (NH₄Cl)

(ii) (a) The blue coloured hydrated copper sulphate loses its water of crystallisation and white anhydrous copper(II) sulphate is formed.:

$$\underset{\underset{\text{(Blue crystals)}}{\text{Hydrated copper sulphate}}}{\text{CuSO}_{4}.5 \text{H}_{2}\text{O}} + \underset{\underset{\text{Sulphuric acid}}{\text{Conc.}}}{\lbrack \text{H}_{2}\text{SO}_{4}\rbrack}\xrightarrow{}\\\underset{\underset{\text{(White powder)}}{\underset{\text{sulphate}}{\text{Anhydrous copper}}}}{\text{CuSO}_{4}} + \lbrack \text{H}_{2}\text{SO}_{4}.5\text{H}_{2}\text{O}\rbrack$$

(b) The reddish brown nitrogen dioxide gas which has pungent smell is evolved along with the formation of cupric nitrate and water.

$$\text{Cu + 4HNO}_{3}\xrightarrow{}\\\underset{\text{(Cupric nitrate)}}{\text{Cu(NO}_{3})_{2}} +\underset{\text{Nitrogen dioxide gas}}{2\text{H}_{2}\text{O} + 2\text{NO}_{2}\uparrow}$$

(iii) (a) Element A is in group 1 and period 3 while element B is in group 17 and period 3 of the periodic table.

(b) A⁺ + 1e^– → A, B + 1e^– → B^–1

Physical state : Solid state

Solubility : Completely soluble in water

(iv) (a) He < Ne < Ar

(b) K < Na < Li

4. (i) (a) Bauxite

(b) Calamine

(ii) (a) We know that, the molecular weight of the gas is twice its vapour density.

Molecular weight = 2 × Molecular weight of carbon dioxide

CO₂ = 12 + 2 (16) = 12 + 32 = 44 gram.

The vapour density of carbon dioxide

V.D. (CO₂) = Molecular weight of CO₂/2

$$\text{V.D. (CO}_2) =\frac{44}{2}$$

V.D. (CO₂) = 22

The vapour density of CO₂ is 22. This indicates that the mass of x litres of hydrogen gas or the molecular mass of CO₂ is 44 g.

Thus, the vapour density of carbon dioxide (CO₂) is 22.

(b)

Relative number of atoms	Simplest ratio
$$\text{X} =\frac{24}{12} = 2\\\text{O} =\frac{64}{16} =4$$	$$\text{X} =\frac{2}{2} =1\\\text{O} =\frac{4}{2} = 2$$

Therefore, simplest ratio between X and O is X : O = 1 : 2

Thus, empirical formula of the compound is XO₂.

(iii) (a)

$$\underset{\text{(Conc.)}}{\text{NaCl + H}_{2}\text{SO}_{4}}\xrightarrow{\Delta\lt 200\degree\text{C}}\\\underset{\text{(Hydrogen chloride)}}{\text{NaSO}_{4} + \text{HCl}\uparrow}$$

(b)

$$\text{2 HCl + Fe}\xrightarrow{}\underset{\text{(Iron(II) chloride)}}{\text{FeCl}_{2} +\text{H}_{2}\uparrow}$$

$$\text{(c)\space}\text{HCl}_\text{(g)} + \text{NH}_{3(g)}\xrightarrow{}\\\underset{\text{Ammonium chloride}}{\text{NH}_{4}\text{Cl}}$$

(iv) Chemical compounds that have identical chemical formulae but differ in properties and the arrangement of atoms in the molecule are called isomers. Therefore, the compounds that exhibit isomerism are known as isomers.

Structural formula of the isomers of butane are:

Chain isomerism : When two or more compounds have a same molecular formula, but they differ
in length of the main chain.
Position isomerism : When two or more compounds have same molecular formula but differ in the position of substituent or functional group on the carbon atom.

$$\textbf{5.\space}(i) (a)\text{KNO}_{3} + \text{conc. H}_{2}\text{SO}_{4}\xrightarrow{\text{(below 200°C)}}\\\text{KHSO}_{4} + \text{HNO}_{3}$$

(b)

The reaction mixture should not be heated beyond 200°C because nitric acid decompose at higher temperatures.
The apparatus must be made of all glass as the vapours of nitric acid are corrosive therefore, it damages the rubber and cork.

$$\text{(ii)\space(a)}\underset{\underset{\text{(Blue solution)}}{\text{Copper sulphate}}}{\text{CuSO}_{4}} + \underset{\underset{\text{hydroxide}}{\text{Potassium}}}{2\text{KOH}}\xrightarrow{}\\\underset{\underset{\text{(Blue ppt.)}}{\text{Copper hydroxide}}}{\text{Cu(OH)}_{2}\darr} + \underset{\underset{\text{sulphate}}{\text{Potassium}}}{\text{K}_{2}\text{SO}_{4}}$$

$$\text{Cu(OH)}_{2} + \underset{(Excess)}{\text{KOH}}\xrightarrow{}\text{No reaction}$$

$$\text{(b)}\space\underset{\text{sulphate}}{\underset{\text{Calcium}}{\text{CaSO}_{4}}} +\underset{\underset{\text{hydroxide}}{\text{Potassium}}}{2\text{KOH}}\xrightarrow{}\\\underset{\underset{\text{(A white ppt.)}}{\text{Calcium hydroxide}}}{\text{Ca(OH)}_{2}\darr} + \underset{\underset{\text{sulphate}}{\text{Potassium}}}{\text{K}_{2}\text{SO}_{4}}$$

$$\text{Ca(OH)}_{2} + \underset{\text{(Excess)}}{\text{KOH}}\xrightarrow{}\text{No reaction}$$

(iii) (a) Isomer of n-butane

(b) Propyne is the three membered unsaturated hydrocarbon with a triple bond.

$$\text{H}_{3}\text{C} - \text{C}≡\text{CH}$$

(iv) (a) Fe³⁺

(b) Pb²⁺

6. (i) The chemical equation for the complete combustion of methane is :

$$\underset{\underset{1 \text{mol}}{1\text{mol}}}{\text{CH}_{4}} + \underset{\underset{2×22.4 \space\text{lit. at S.T.P}}{2\text{mol}}}{2\text{O}_{2}}\xrightarrow{}\\\text{CO}_{2} + 2\text{H}_{2}\text{O}$$

Now,

Molecular weight of CH₄ = 12 + 1 × 4 = 16 g

16 g of CH₄ requires = 2 × 22.4 lit. oxygen

∴ 8.8 g of methane would requires = (2 × 22.4 × 8.8)/16 = 24.64 lit. oxygen

$$\text{(ii)}\space \underset{2 \text{vol.}}{2\text{C}_{4}\text{H}_{10}} + \underset{13 vol.}{13\text{O}_{2}}\xrightarrow{}\\8\text{CO}_{2} +10\text{H}_{2}\text{O} $$

∴ According to Gay Lussac’s law, 2 vol. of butane require 13 vol. of oxygen

$$\therefore\space \text{90 \text{dm}}^{3} \text{of butane require =}\frac{13×90}{2}$$

= 585 dm³

Thus, 585 dm³ of oxygen is required to burn 90 dm³ of butane.

(iii) (a) Preparation of NH₃gas using alkali can be done by reacting ammonium sulphate with sodium hydroxide.

$$\underset{\underset{\text{sulphate}}{\text{Ammonium}}}{(\text{NH}_{4})_{2}\text{SO}_{4}} +\underset{\underset{\text{hydroxide}}{\text{Sodium}}}{2\text{NaOH}}\xrightarrow{}\\\underset{\text{Ammonia}}{2\text{NH}_{3}} + 2\text{H}_{2}\text{O} + \text{Na}_{2}\text{SO}_{4}$$

(b) Concentrated sulphuric acid is not used for drying ammonia gas because concentrated sulphuric acid (H₂SO₄) being acidic in nature reacts with basic ammonia gas to give ammonium sulphate [(NH₄)₂SO₄].

(c) Ammonia gas can not be collected over water because it has a high solubility in water and it dissolves in water to give a basic solution. Thus, it is collected by downward displacement of air as being lighter than air.

$$\text{NH}_{3}(g)\xrightarrow{}\text{H}_{2}\text{O(I)}\xrightarrow{}\\\text{NH}_{4}^{+}(aq) + \text{OH}^{\normalsize-}(aq)$$

(iv) (a) Dilute nitric acid is generally considered a typical acid but not so in its reaction with metals because it does not liberate hydrogen with all metals except Mg and Mn. It is a powerful oxidising agent and the nascent oxygen formed oxidises the hydrogen to water.

(b) Concentrated nitric acid appears yellow when it is left standing in a glass bottle because when nitric acid is left standing in a glass bottle, it decomposes to give reddish brown NO₂ gas which dissolves in undecomposed nitric acid to give a yellow colour.

(c) An all glass apparatus is used in the laboratory preparation of nitric acid because nitric acid vapours are corrosive in nature and destroy materials like rubber, cork or metal.

7. (i) Determination of empirical formula

Element	%	Atomic mass	Relative number of atoms	Simplest Ratio
K	47.9	39	$$\frac{47.9}{39} = 1.2$$	$$\frac{1.2}{0.6} = 2$$
Be	5.5	9	$$\frac{5.5}{9} = 0.6$$	$$\frac{0.6}{0.6} = 1$$
F	46.6	19	$$\frac{46.6}{19} = 2.4$$	$$\frac{2.4}{0.6} = 4$$

Empirical formula = K₂BeF₄

(ii) (a) Cryolite

Molten cryolite acts as solvent for alumina and also lowers the fusion temperature from 2050°C to 950°C and enhances conductivity and thereby saves electrical energy.

$$\text{(b)\space} 2\text{AI}^{3 +} + 6e^{\normalsize-}\xrightarrow{}\text{AI}.$$

(c) During electrolysis, oxygen gas is formed at anode which oxidises graphite or carbon anode to carbon dioxide, so it is necessary to replace anode periodically.

(iii) (a) Copper sulphate by neutralisation

$$\underset{\underset{\text{hydroxide}}{\text{Copper}}}{\text{Cu(OH})_{2}} +\underset{\underset{\text{acid}}{\text{Sulphuric }}}{\text{Cu(OH})_{2}}\xrightarrow{}\\\underset{\underset{\text{sulphate}}{\text{Copper}}}{\text{CuSO}_{4}}\darr + 2\text{H}_{2}\text{O}$$

(b) Zinc carbonate by precipitation

$$\underset{\text{Zinc nitrate}}{\text{Zn(NO}_{3})_{2}}(aq) + \underset{\text{Sodium carbonate}}{2 \text{NaCO}_{3}(aq)}\xrightarrow{}\\\underset{\text{Zinc carbonate}}{Zn(\text{CO}_{3})_{2}} + 2\text{NaNO}_{3}(aq)$$

8. (i) (a) Covalent bonding

(b) Ionic or electrovalent bonding

(ii) (a) When NH₄OH solution is added drop by drop to copper sulphate solution a pale blue or bluish white precipitate is formed which is soluble in excess of NH₄OH and a deep blue or inky blue solution is formed with excess of ammonium hydroxide.

$$\text{CuSO}_{4} + 2\text{NH}_{4}\text{OH}\xrightarrow{}\\\text{Cu(OH)}_{2} + (\text{NH}_{4})_{2}\text{SO}_{4}\\\text{Cu(OH)}_{2} + 2\text{NH}_{4}\text{OH}\xrightarrow{}\\\text{Cu(NH}_{3})_{4}(\text{OH})_{2} + 4\text{H}_{2}\text{O}$$

Copper solution forms a blue precipitate with ammonium hydroxide solution.

(b) On adding lead nitrate to HCl (hydrochloric acid) we will get a white precipitate. On heating the solution the one whose precipitate will redissolve will be dil. HCl and the one with insoluble precipitate will be dil. H₂SO₄.

Actually, on adding lead nitrate to HCl, PbCl₂ precipitates out and on heating the solution it redissolves. But in case of H₂SO₄, PbSO₄ is formed which is insoluble even on heating dissolves it.

$$\text{Pb(NO}_{3})_{2} + 2\underset{\text{(dil.)}}{\text{HCl}}\xrightarrow{}\\\text{PbCl}_{2} + 2\text{HNO}_{3}\\\text{Pb}(\text{NO}_{3})_{2} +\underset{\text{(dil.)}}{\text{H}_{2}\text{SO}_{4}}\xrightarrow{}\\\text{PbSO}_{4} + 2\text{HNO}_{3}$$

(iii) (a) Cathode

(b) Conc. H₂SO₄

(c) Platinum-rhodium catalyst gauzes are used in the Ostwald process for the manufacture of nitrogen monoxide by the oxidation of ammonia.

(iv) (a) Third period

(b) Six electrons