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) CnH2n – 2

Explanation :    

The general formula of alkyne is CnH2n – 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.

(c) Graphite rods dipped in molten pure aluminium.

(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

(c) nitrogen trichloride

(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]

(c) The particles present in a liquid such as kerosene, is a non-electrolyte are free molecules

(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)

7ds_Chemistry_m2(V_1)
7ds_Chemistry_m2(V_2)
7ds_Chemistry_m2(V_3)

(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

(c) Ionic bond exists between A and B.

Physical state : Solid state

Solubility : Completely soluble in water

(iv) (a) He < Ne < Ar

(b) K < Na < Li

(c) Br < Cl < F

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:

7ds_Chemistry_m4(iv)
  1. Chain isomerism : When two or more compounds have a same molecular formula, but they differ
    in length of the main chain.
  2. 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)

  1. The reaction mixture should not be heated beyond 200°C because nitric acid decompose at higher temperatures.
  2. 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.

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

(c) The compound n-pentane represents the straight chain structure of 2-methyl butane.

(iv) (a) Fe3+

(b) Pb2+

(c) Ca2+

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 NHgas 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

(c) Non-metal

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