Oswal Model Specimen Papers ICSE Class 10 Chemistry Solutions (Specimen Paper - 9)

Section-A

1. (i) (b) Mainly ions

Explanation :

Chemical compounds which conduct electricity in the fused or in aqueous solution state and undergo chemical decomposition due to the flow of current through it is known as strong electrolytes.

In strong electrolytes, ions and molecules are the only particles present.

(ii) (c) 1-B, 2-A, 3-C, 4-D

Explanation :

Selenium is a non- metal.
Sodium chloride is electrovalent compound.
Chloride ion is an anion since it contains the negative charge.
Potassium ion is a cation as it contains positive charge.

(iii) (a) The metals can’t be extracted economically from all the minerals.

Explanation :

The naturally occurring compound of metals generally mixed with other matters such as soil, sand, limestone rocks are known as minerals. At the same time, ores are only those minerals in which metals can be extracted commercially and economically. Therefore, metals cannot be extracted economically from all the minerals.

(iv) (a) Polar covalent bond

Explanation :

A polar covalent bond exists when atoms with different electronegativities share electrons in a covalent bond. In a molecule of HCl, each atom requires one more electron to form an inert gas electron configuration.

Therefore, they form the covalent bond in H^δ+ – Cl^δ– which is polar in nature. This is due to the appreciable electronegativity difference of H and Cl.

(v) (a) H₂O molecule and H⁺ion

Explanation :

This ion is formed by the combination of H₂O molecule and H⁺ ion. The oxygen in water has two lone pairs of electrons. H⁺ ion has required 2 more electrons to complete its duplet. During the formation of H₃O⁺, one lone pair (2 electrons) from O atom is donated to the H⁺ ion and oxygen-hydrogen coordinate bond is formed.

(vi) (b) F^–, Ca²⁺ and O^2–

Explanation :

When Na₃AlF₆ dissociates into ions it produces Na⁺, Al³⁺ and F^– ions. Similarly, Y is Ca²⁺ and Z is O^2–.

(vii) (b) Molecular formula

Explanation :

A molecular formula consists of the chemical symbols for the constituent elements followed by numeric subscripts describing the number of atoms of each element present in the molecule.

(viii) (d) NH₃

Explanation :

The reaction of ammonia with HCl gas gives ammonium chloride, which produces white fumes. The ammonia gas has a pungent choking smell. If a glass rod dipped in aqueous ammonia solution is brought near HCl gas, then the white fumes are produced, which act as an identification test for HCl gas.

(ix) (a) To increase the efficiency of finely divided iron

Explanation :

The Haber’s process uses finely divided iron as a catalyst which increases the speed of reaction.

However, to increase the efficiency of this catalyst, a promoter is used, which is usually molybdenum or aluminium oxide.

(x) (a) Zn(OH)₂

Explanation :

Zinc hydroxide is soluble in excess of NaOH due to the formation of soluble sodium.

(xi) (c) Nitric acid

Explanation :

When concentrated sulphuric acid is added to sodium or potassium nitrate, a double displacement reaction occurs in which sodium or potassium bisulphate is formed along with nitric acid.

$$\text{KNO}_{3} + \text{H}_{2}\text{SO}_{4}\xrightarrow{}\\\underset{\text{Potassium bisulphate}}{\text{KHSO}_{4}} +\underset{\text{Nitric acid}}{\text{HNO}_{3}}$$

(xii) (b) n-Pentane

Explanation :

The hydrocarbon 2-methyl butane is an isomer of n-pentane because these two molecules have same molecular formula C₅H₁₂ but their structure is different. Their structure can be represented as :

H₃C — CH₂ — CH₂ — CH₂ — CH₃

n-Pentane

(xiii) (c) Alkyne

Explanation :

C_nH_{2n – 2} → Alkyne

Here, n = 10

Thus, C₁₀H_{2 × 10 – 2} = C₁₀H₁₈

(xiv) (c) Only (iii)

Explanation :

Dilute sulphuric acid is not a strong acid and hence cannot convert common salt to hydrogen chloride.

So it is not employed in the manufacturing of HCl gas.

(xv) (b) Group 2 element

Explanation :

Group 2 elements include beryllium, magnesium, calcium, strontium, barium, and radium. These elements are known as alkaline earth metals for two reasons : Their oxides are heat-stable and persist in the earth’s crust.

2. (i) (a) At cathode reduction takes place and deposition of Ni takes place as Ni²⁺ ions from solution convert to Ni metal.

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

At anode oxidation takes place and Ni metal converts to Ni²⁺ ions.

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

(b) H⁺ is the spectator cation.

$$\text{(c)}\space\text{Ni}\xrightarrow{}\text{Ni}^{2+} + 2e^{\normalsize-}$$

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

(iii) (a) 1

(b) Sodium ethoxide

(d) A downward displacement of air

(e) Nitric oxide.

(iv) (a) Catenation

(b) Electrolyte

(d) Froth flotation process

(e) Coordination bond

(v)

(b)

Ethane
Propanoic acid

Section-B

3. (i) (a) Nitrogen dioxide (NO₂) gas.

$$\text{HNO}_{3}\xrightarrow[\text{Light}]{\text{\text{and heat}}}\\4\text{NO}_{2}\uparrow + \text{O}_{2} +\text{H}_{2}\text{O}$$

(b) Hydrogen sulphide (H₂S) gas.

$$\text{ZnS + HCl}\xrightarrow{}\text{ZnCl}_{2} + \text{H}_{2}\text{S}\uparrow$$

(ii) (a) Ostwald’s process

(b) Quicklime

(iii) (a) The given elements can be arranged in increasing order of reactivity as follows :

Pb < Zn < Ca < K

(b) In the order of preferential discharge at the cathode :

Na⁺ > Mg²⁺ > H⁺ > Cu²⁺

H > Li > Na > K

(iv) (a) Ionisation potential.

(b) Carbon tetrachloride.

4. (i) (a) PbS

(b) ZnO

(ii) Given : C = 75.92%, H = 6.32% and N = 17.76%

Let us assume that the weight of compound is 100 g. So, in that 100 g C is 75.92 g, H is 6.32 g and N is 17.76 g.

Now, lets calculate the moles of each element present in the compound.

75.92g C × (1 mole C/12 g C) = 6.32

6.32g H × (1 mol H/1 g H) = 6.32

Next, dividing all the mole numbers by the smallest among them, which is 1.26. This division yields :

5 mol C, 5 mol H and 1 mol N

So, the compound has the empirical formula : C₅H₅N.

Now, we know that

Molecular mass of a gas = 2 × vapour density of the gas

= 2 × 39.5 = 79

So, molecular mass of the given compound is 79.

Empirical formula mass = (12 × 5) + (1 × 5) + (14 × 1) = 79

As, empirical formula mass = Molecular mass

So, in this case,

Empirical formula = Molecular formula = C₅H₅N.

(iii) (a) Presence of diffused sunlight.

$$\text{(b)\space H}_{2} + \text{Cl}_{2}\xrightarrow{}\text{2HCl}$$

(iv) (a)

$$\text{C + 2H}_{2}\text{SO}_{4}\xrightarrow{}\\\text{CO}_{2}\uparrow + 2\text{SO}_{2}\uparrow + 2\text{H}_{2}\text{O}\\\text{(b)\space}\text{Mg + H}_{2}\text{SO}_{4}\xrightarrow{}\\\text{MgSO}_{4} + \text{H}_{2}\uparrow\\\text{(c)\space}\underset{\text{(Cane sugar)}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}}\xrightarrow{\text{conc. H}_{2}\text{SO}_{4}}\\12\space\text{C + 11 H}_{2}\text{O}$$

5. (i) (a) Concentrated hydrochloric acid cannot be used in place of sulphuric acid because hydrochloric acid is a volatile acid and the produced nitric acid carry away the HCl vapours from reaction mixture.

(b) Hydrogen chloride gas is highly water soluble; when the stopper of a bottle full of hydrogen chloride gas is opened, HCI gas comes in contact with water vapours of air and gives white fumes due to the formation of hydrochloric acid.

(ii) (a) The given equation is :

$$\text{2 NaOH + CuSO}_{4}\xrightarrow{}\\\text{Na}_{2}\text{SO}_{4} +\text{Cu(OH)}_{2}\darr$$

Molecular weight of NaOH, Sodium hydroxide = 23 + 16 + 1 = 40

Molecular weight of Cu(OH)₂,

Copper hydroxide = 64 + 16 + 1 + 16 + 1 = 98

2 × 40 g = 80 g of NaOH is used to precipitate 98 g of Cu(OH)₂

Hence, 200 g of NaOH will be used to precipitate = (98/80) × 200 g of Cu(OH)₂ = 245 g of Cu(OH)₂

So, 490 g of copper hydroxide would be prepared using 200 g of sodium hydroxide.

(b) The precipitate of copper hydroxide is bluish green solid or pale blue solid.

(iii) (a) When acetylene is passed through an ammonical solution of cuprous chloride, at room temperature, red colour precipitate of cuprous acetylide is formed.

$$\underset{\text{Acetylene}}{\text{HC ≡ CH}} +\underset{\underset{\text{Chloride}}{\text{cuprous}}}{\text{Cu}_{2}\text{Cl}_{2}}+ 2\text{NH}_{4}\text{OH}\xrightarrow{}\\\underset{\underset{\text{(red ppt)}}{\text{Cuprous acetylide}}}{\text{Cu}-\text{C}≡\text{C}-\text{Cu}\darr + 2\text{NH}_{4}\text{Cl} + 2\text{H}_{2}\text{O}}$$

(b) When bromide in carbon tetrachloride is added to ethane, there will be no change in the orange colour of bromine solution. However, when bromide in carbon tetrachloride is added to ethene, the orange colour of the bromine disappears due to the formation of the colourless ethylene bromide.

$$\underset{\text{Ethene}}{\text{CH}_{2}=\text{CH}_{2}} +\underset{\underset{\underset{\text{Color}}{\text{orange}}}{\text{(Raddish)}}}{\text{Br - Br}}\xrightarrow{\text{CCl}_{4}}$$

(c) When acetic acid or ethanoic acid reacts with sodium hydrogen carbonate, a colourless and odourless CO₂ gas is released with brisk effervescence. The presence of CO₂ can be checked by passing the gas through lime water which turns milky on reacting with CO₂, thus indicating its presence.

(iv) (a) L is copper nitrate, M is nitrogen dioxide gas and N is hydrogen sulphide gas.

$$\text{(b)\space 2Cu(NO}_{3})_{2}\xrightarrow{\Delta}\\\text{2CuO + 4 NO}_{2} +\text{O}_{2}$$

$$\text{(c)\space Cu(NO}_{3})_{2} + \text{H}_{2}\text{S}\xrightarrow{}\\\text{CuS + HNO}_{3}$$

6. (i)

Element	Percentage	Molecules	Simple Ratio	Simple whole ratio
C	82.76	$$\frac{82.76}{12} = 6.89$$	1	2
H	17.24	$$\frac{17.24}{1} = 17.24$$	2.5	5

∴ Empirical formula = C₂H₅

Empirical formula mass = (12 × 2) + (1 × 5) = 24 + 5 = 29

Vapour density → 29 (Given)

Molecular mass = Vapour density × 2 = 29 × 2

= 58 g

Molecular formula mass = n × Empirical formula mass

$$=\frac{\text{Molecular formula mass}}{\text{Empirical formula mass}}\\=\frac{58}{29}= 2$$

Molecular formula = n × Empirical formula

= 2 × C₂H₅

= C₄H₁₀

(ii) Nitrogen : 82.35% and Hydrogen : 17.64%

Element	N	H
Percentage	82.35	17.64
Relative Ratio	$$\frac{82.35}{14} = 5.88$$	$$\frac{17.64}{1} = 17.64$$
Simple Ratio	$$\frac{5.88}{5.88} = 1$$	$$\frac{17.64}{5.88} = 3$$

So, the empirical formula of the gas would be NH₃.

$$\text{(iii)\space(a)}\underset{\underset{\text{chloride}}{\text{Ammonium}}}{2\text{NH}_{4}\text{Cl}(s)} +\underset{\underset{\text{lime}}{\text{Slaked}}}{\text{Ca(OH)}_{2}}\xrightarrow{\Delta}\\\underset{\underset{\text{chloride}}{\text{Calcium}}}{\text{CaCl}_{2}(s)} + \underset{\text{Water}}{2\text{H}_{2}\text{O}} + \underset{\text{Ammonia gas}}{2\text{NH}_{3}}\uparrow$$

$$\text{(b)\space}\text{NH}_{3} + 3\text{Cl}_{2}\xrightarrow{}\underset{\underset{acid }{\text{Hydrochloric}}}{3\text{HCl}} +\underset{\underset{\text{chloride}}{\text{Nitrogen}}}{\text{NCl}_{3}}$$

$$\text{(c)\space 2NH}_{3} + \text{H}_{2}\text{SO}_{4}\xrightarrow{}\underset{\underset{\text{sulphate}}{\text{Ammonium}}}{(\text{NH}_{4})_{2}\text{SO}_{4}}$$

(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.

$$\underset{\text{Black}}{\text{MnO}_{2}} + 4\text{HCl}\xrightarrow{}\\\underset{\text{Colourless}}{\text{MnCl}_{2} + \text{Cl}_{2}}\uparrow +\underset{\text{Greenish Yellow gas}}{2\text{H}_{2}\text{O}}$$

(b) A reddish precipitate of iron(III) hydroxide is obtained when ferric sulphate reacts with sodium hydroxide solution, whereas dirty green precipitate is obtained when ferrous sulphate is mixed with sodium hydroxide.

$$\text{FeSO}_{4} + 2\text{NaOH}\xrightarrow{}\\\underset{\text{Dirty green}}{\text{Fe(OH)}_{2}\darr + \text{Na}_{2}\text{SO}_{4}}\\\text{Fe}_{2}(\text{SO}_{4})_{3} + 6\text{NaOH}\xrightarrow{}\\\underset{\text{Reddish brown}}{2\text{Fe(OH)}_{3} + 3\text{Na}_{2}\text{SO}_{4}}$$

(c) Lead nitrate reacts with dilute HCl to form the insoluble salt lead chloride, which appears as the white precipitate. While lead chloride is insoluble in cold water but soluble in warm water.

$$\text{Pb(NO}_{3})_{2} + 2\text{HCl}\xrightarrow{}\underset{\text{(White ppt.)}}{\text{PbCl}_{2}\darr + 2\text{HNO}_{3}}$$

(Insoluble in cold water but dissolves in warm water)

Whereas lead nitrate solution reacts with H₂SO₄ to give white precipitate of lead sulphate which is insoluble both in cold and warm water.

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

(Insoluble both in cold and warm water)

7. (i)

Element	Percentage ratio	Atomic mass	Relative number of atoms	Simplest ratio
C	12.67	12	$$\frac{12.67}{12} = 1.055$$	$$\frac{1.055}{1.055} = 1$$
H	2.13	1	$$\frac{2.13}{1} = 2.13$$	$$\frac{2.13}{1.055} ≈ 2$$
Br	85.11	80	$$\frac{85.11}{80} = 1.063$$	$$\frac{1.063}{1.063} ≈ 1$$

∴ Empirical formula of the compound is CH₂Br

Molecular formula = (Empirical formula) × n

$$n =\frac{\text{M.W.}}{\text{Empirical formula weight}}\\=\frac{2 × \text{V.D}}{\text{Empirical formula weight}}\\=\frac{2×94}{(12 + 2 +80)}\\=\frac{2×94}{94} = 2$$

∴ Molecular formula = (CH₂Br) × 2 =

C₂H₄Br₂

(ii) (a) C₂H₄ contains a double bond between two carbon atoms.

(b) C₂H₂ contains a triple bond between two carbon atoms.

(iii) (a) Na₂O, MgO

(b) SO₂

The type of bonding in hydronium ion is coordinate bonding.

(ii) (a) When crystals of washing soda are exposed to air, they lose 9 water molecules of crystallisation and becomes monohydrate forming a white powder. Thus, shows the phenomenon of efflorescence.

$$\text{Na}_{2}\text{CO}_{3}.10\text{H}_{2}\text{O}\xrightarrow{\text{Dry air}}\\\text{Na}_{2}\text{CO}_{3}.\text{H}_{2}\text{O} + 9\text{H}_{2}\text{O}$$

(b) The salt ferric chloride, when exposed to the atmosphere, absorbs water molecules to become moist and show the phenomenon of deliquescence.

(iii) (a) NaCl is an ionic compound. Sodium chloride ions in the solid state are held by the electrostatic force of attraction, thus are not free to move and not conduct electricity but in the fused state, the crystal lattice breaks down and the charged particles (ions) are free to move and thus are able to conduct electricity.

(b) Silver nitrate solution undergoes rapid dissociation that can cause non-uniform coating, therefore, it is not preferred. Whereas, sodium argentocyanide to cyanide solution is a complex salt and undergoes slow decomposition and ensure smooth and uniform coating.

(c) Copper is a metals so it conducts electricity by flow of electrons but it does not form ions so it is a non-electrolyte.

(iv) (a) Atomic number of S is 17.

(b) S belongs to group 17 and period 3.