Oswal 61 Sample Question Papers ICSE Class 10 Chemistry Solutions
Section-A
Answer 1.
(i) (d) (a)-(ii), (b)-(iii), (c)-(i)
Explanation :
Bauxite is an ore of aluminium, while calamine is zinc carbonate and magnetite is triferric tetraoxide.
(ii) (a) More
Explanation :
Ionisation energy is the energy required to remove loosely bonded electron from outermost shell of an isolated gaseous atom. In first ionisation energy, electron from a neutral atom is removed but for second ionisation, electron is removed from a positive atom, where electron are more tightly bounded due to increased attraction force. Therefore, second ionisation energy is high relative to first ionisation potential.
(iii) (d) HSO4–
Explanation :
Reaction of sodium chloride with conc. sulphuric acid produces sodium sulphate and HCl gas. The figure shows the precipitate of NaHSO4. Therefore, anionic part of this precipitate is HSO4–.
(iv) (a) Ionic solid
Explanation :
Ionic solids contain ions in their crystal lattice which are held together by strong electrostatic forces of attraction. Due to this strong forces of attraction, ionic solids are hard and have high melting and boiling points. Most ionic compounds tend to dissociate in polar solvents due to the opposite charges on each ion.
(v) (a) Sodium chloride
Explanation :
Sodium chloride is formed by combination of base NaOH and an acid hydrochloric acid.
(vi) (b) Pd/BaSO4/quinoline
Explanation :
Pd/BaSO4/quinoline reagent is called Linder’s catalyst.
(vii) (c) 44
Explanation :
According to the forumla,
Molecular mass = Vapour density × 2
= 22 × 2 = 44 g
(viii) (a) lowers; pressure
Explanation :
The HCl gas present inside the flask, when mixed with water, lowers the pressure inside the flask as some gas molecules dissolve in water.
(ix) (b) Platinum is costly and can be easily poisoned
Explanation :
Platinum is an efficient catalyst, but it is costly and easily poisoned by arsenic oxide. V2O5 is an efficient catalyst and less expensive than platinum and hence, is a suitable replacement for platinum.
(x) (d) Pink
Explanation :
Phenolphthalein is a colourless indicator which is colourless in acidic medium but turns pink in alkaline medium.
(xi) (a) NO and NO2
Explanation :
Both dilute and concentrated nitric acid reacts with copper to give salt of copper nitrate and water.
However, both of these acids produce different gases.
With conc. HNO3, copper reacts to form copper nitrate and nitrogen dioxide.
$$\text{Cu + 4HNO}_3(\text{conc.})\xrightarrow{}\\\underset{\text{Copper nitrate}}{\text{Cu(NO}_3)_2} +\text{2 NO}_{2} + \text{2H}_{2}\text{O}$$
With dilute nitric acid, copper reacts to form copper nitrate and nitric oxide.
$$\text{3Cu +}\space\underset{\text{Nitric acid}}{\text{8HNO}_{3}}\xrightarrow{}\underset{\text{Copper nitrate}}{\text{3 Cu(NO}_{3})_{2}}\\ + \text{2 NO + 4H}_{2}\text{O}$$
(xii) (b) (i)-(r), (ii)-(p), (iii)-(q)
Explanation :
(xiii) (b) Methoxy methane
Explanation :
The IUPAC name of dimethylether (CH3 – O – CH3) is methoxy methane.
(xiv) (c) Oxidation takes place at cathode
Explanation :
During electrolysis, oxidation i.e., loss of electrons takes at the anode.
(xv) (a) 4
Explanation :
Element with atomic number 14 is silicon, it is placed in period 4.
Answer 2.
(i) (a) Substance B is anhydrous (fused) calcium chloride (CaCl2).
(b) Anhydrous calcium chloride absorbs the moisture present in the apparatus. This helps in keeping the anhydrous ferric chloride dry as it is a highly deliquescent salt.
(c) Iron(III) chloride should be stored in a closed container because of its deliquescent nature. In the presence of moisture, iron(III) chloride absorbs water, which gets dissolved in it and forms a saturated solution.
(ii) 1. (c) Sodium hydrogen sulphate
2. (a) Ferrous ammonium sulphate
3. (e) Contains ions and molecules
4. (b) Contains only ions
5. (d) Contains only molecules
(iii) (a) Increases
(b) Covalent
(c) Acids
(d) Hydrogen
(e) Methanol
(iv) (a) Ca2+
(b) Solution of sodium argentocyanide i.e., Na[Ag(CN)2]
(c) Free molecules
(d) Carboxylic acid
(e) Ammonium chloride
(b) 1. But-1-ene
2. 2-methyl propane
Section-B
Answer 3.
(i) (a) Nitrate ion, NO3–
(b) Sulphate ion, SO42–
(ii) (a) Oxygen gas
(b) Carbon dioxide gas
(iii) (a) Li > Na > K
(b) F < Cl < Br < I
(c) Rb > K > Li
(iv) (a) Ammonia
(b) Ferric chloride
(c) Hygroscopic
Answer 4.
(i) (a) Aluminium
(b) Iron
(ii)
| Element | Mass% | Atomic weight | Relative number of atoms | Simple ratio |
| Hydrogen | 2.47 | 1 | 2.47 | 2 |
| Phosphorus | 38.25 | 31 | 1.23 | 1 |
| Oxygen | 59.28 | 16 | 3.70 | 3 |
(a) Empirical formula, H2PO3
(b) Vapour density = 81
Molecular weight = 2 × Vapour density
= 2 × 81 = 162
Empirical weight = (1 × 2 + 31 + 16 × 3) = 81
n = Molecular weight / Empirical weight
$$=\frac{162}{81} = 2$$
Molecular formula = (Empirical formula)n
= (H2PO3)2 = H4P2O6
(iii) (a) Quick lime. (CaO is a hydroscopic salt as it readily absorbs moisture)
(b) By downward displacement of air.
(c) Ammonia gas is not collected over water because it is highly soluble in water.
(iv) (a) When ethanol is heated with conc. H2SO4 at 170°C, it undergoes dehydration and forms ethene.
$$\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow[170\degree\text{C}]{\text{Conc. H}_{2}\text{SO}_{4}}\\\underset{\text{Ethene}}{\text{H}_{2}\text{C} = \text{CH}_{2} + \text{H}_{2} \text{O}}$$
(b) When silver nitrate solution is added to dilute hydrochloric acid, an insoluble white precipitate of silver chloride is formed.
$$\text{AgNO}_{3} + \text{HCl}\xrightarrow{}\\\underset{\text{Silver chloride ppt.}}{\text{AgCl}\darr} + \text{HNO}_{3}$$
(c) When zinc sulphide is heated with dilute hydrochloric acid, a colourless hydrogen sulphide gas is evolved, which has a rotten egg smell.
$$\text{ZnS + 2HCl}\xrightarrow{\Delta}\underset{\text{Hydrogen sulphide gas}}{\text{ZnCl}_{2} + \text{H}_{2}\text{S}\uparrow}$$
Answer 5.
(i) (a) When few drops of purple colour potassium permanganate is added to ethane, its purple colour does not fades but when a few drops of it is added to ethene, the solution decolourises.
(b) Sulphuric acid precipitates the insoluble sulphate from lead nitrate solution.
$$\text{Pb(NO}_3)_2 \text{+ H}_2\text{SO}_4\xrightarrow{}\text{PbCO}_{4} + \text{2HNO}_{3}$$
Lead nitrate reacts with hydrochloric acid to give a white ppt. of lead chloride.
$$\text{Pb(NO}_3)_2 \text{+ 2HCl} \xrightarrow{}\text{PbCl}_2 + \text{2HNO}_3$$
(ii) (a) Copper sulphate and iron(II) sulphate
When little NH4OH is added to CuSO4, pale blue or bluish white precipitate appears which dissolves in excess of NH4OH to give deep blue or inky blue solution.
$$\text{CuSO}_4 + \text{2NH}_4\text{OH}\xrightarrow{}\underset{\text{Pale blue ppt.}}{\text{Cu(OH)}_2}\darr \\+ \text{(NH}_{4})_{2}\text{SO}_{4}$$
$$\text{Cu(OH)}_{2} + \text{4NH}_{4}\text{OH}\xrightarrow{}\underset{\text{Inky blue solution}}{\lbrack\text{Cu(NH}_{3})_{4}(\text{OH})_{2}}\\+ \text{4H}_{2}\text{O} $$
When little NH4OH is added to FeSO4, dirty green precipitate appears which turns reddish brown after sometime. It is insoluble in excess of NH4OH.
$$\text{FeSO}_{4} + \text{2NH}_{4}\text{OH}\xrightarrow{}\underset{\text{Dirty green ppt.}}{\text{Fe}(\text{OH})_{2}}\darr +\\(\text{NH}_{4})_{2}\text{SO}_{4}$$
(b) Zinc nitrate and lead nitrate.
When little NH4OH is added to Zn(NO3)2, a white precipitate appears which is soluble in excess of NH4OH.
$$\text{Zn(NO}_3)_2 \text{+ 2NH}_4\text{OH}\xrightarrow{}\underset{\text{White ppt.}}{\text{Zn(OH)}_{2}\darr}\\+\text{2NH}_{4}\text{NO}_{3}\\\text{Zn(OH)}_2 + \text{4NH}_{4}\text{OH}\xrightarrow{}\underset{\text{Soluble salt}}{\lbrack\text{Zn(NH}_{3})_{4}\rbrack}(\text{OH})_{2}+\\\text{4H}_{2}\text{O}$$
When little NH4OH is added to Pb(NO3)2 a white precipitate appears which is insoluble in excess of NH4OH.
$$\text{Pb(NO}_3)_2 + \text{2NH}_4\text{OH}\xrightarrow{}\underset{\text{White ppt.}}{\text{Pb}(\text{OH})_{2}}\darr +\\\text{2NH}_{4}\text{NO}_{3}$$
(iii) (a) Hydration
(b) Pyrolysis
(c) Polymerisation
(iv) (a) Those metals which react with both alkalies as well as acids are called amphoteric metals e.g. Zn, Sn, Al etc. They react with caustic alkalies like NaOH, KOH on heating and liberates H2 gas. e.g.
$$\underset{\text{Zinc}}{\text{Zn}} + \underset{\text{(Conc.)}}{\text{2NaOH}(\text{aq})}\xrightarrow{}\underset{\text{Sodium zincate}}{\text{Na}_{2}\text{ZnO}_{2} +\text{H}_{2}}\uparrow\\\underset{\text{Zinc}}{\text{Zn}} + \underset{\text{(Conc.)}}{\text{2 KOH}}(\text{aq})\xrightarrow{}\underset{\text{Potassium zincate}}{\text{K}_{2}\text{ZnO}_{2} + H_{2}}\uparrow\\\underset{\text{Tin}}{\text{Sn}} + \underset{\text{(Conc.)}}{\text{2NaOH}(\text{aq})} + \text{H}_{2}\text{O}\xrightarrow{}\\\underset{\text{Sodium stannate}}{\text{Na}_{2}\text{SnO}_{3} + 2 H_{2}}\uparrow{}$$
$$\underset{\text{Tin}}{\text{Sn} + }\space\underset{\text{(Conc.)}}{\text{Sn} + \text{2NaOH(aq)}} +\text{H}_{2}\text{O}\xrightarrow{}\\\underset{\text{Sodium stannate}}{\text{Na}_{2}\text{SnO}_{3} + 2\text{H}_{2}}\uparrow{}\\\underset{\text{Tin}}{\text{Sin}} + \underset{\text{(Conc.)}}{\text{2KOH}(\text{aq})} +\text{H}_{2}\text{O}\xrightarrow{}\\\underset{\text{Potassium stannate}}{\text{K}_{2}\text{SnO}_{3} + 2\text{H}_{2}\uparrow{}}\\\underset{\text{Aluminium}}{\text{2Al + 2NaOH + 2H}_2\text{O}}\xrightarrow{}\\\underset{\text{Sodium aluminate}}{\text{2NaAlO}_{2}} + \text{3H}_{2}\uparrow{}\\\underset{\text{Aluminium}}{\text{2AI + 2KOH + 2H}_{2}\text{O}}\xrightarrow{}\\\underset{\text{Potassium aluminate}}{\text{2KAIO}_{2} + 3\text{H}_{2}\uparrow}$$
(b) Those oxides which react with both acids as well as bases are called amphoteric oxides. Oxides of amphoteric metals like Zn, Sn, Al, etc., react with strong alkalies like NaOH to form complex salt and water, e.g.
$$\underset{\text{Zinc oxide}}{\text{ZnO}} + \underset{\text{(Conc.)}}{\text{2NaOH}}(\text{aq})\xrightarrow{}\underset{\text{Sodium zincate}}{\text{Na}_{2}\text{ZnO}_{2} + \text{H}_{2}\text{O}}$$
$$\underset{\text{Tin oxide}}{\text{SnO}_{2}} +\underset{\text{(Conc.)}}{2\text{NaOH}}(\text{aq})\xrightarrow{}\\\underset{\text{Sodium stannate}}{\text{Na}_{2}\text{SnO}_{3} + \text{H}_{2}\text{O}}\\\underset{\text{Aluminium oxide}}{\text{Al}_{2}\text{O}_{3}} +\underset{\text{(Conc.)}}{\text{2 NaOH}(\text{aq})}\xrightarrow{}\\\underset{\text{Sodium aluminate}}{\text{2 Na AIO}_{2} + \text{H}_{2}\text{O}}$$
(c) Concentrated sulphuric acid act as drying agent.
Answer 6.
(i) (a)
| Element | Mass% | Atomic weight | Relative number of atoms | Simple ratio |
| C | 85.7% | 12 | 7.14 | 7.14/7.14 = 1 |
| H | 14.3% | 1 | 14.3 | 14.3/7.14 = 2 |
(b) Empirical formula CH2.
Molecular formula = n × Empirical formula
$$n = \frac{\text{Molecular wt.}}{\text{Empirical formula wt.}} =\frac{28}{14} = 2$$
Molecular formula = 2 × (CH2) = C2H4.
(ii) Empirical formula weight of X2Y = 2 × 10 + 5 = 25
Molecular weight = 2 × Vapour density
= 2 × 25 = 50
$$n = \frac{\text{Molecular weight}}{\text{Empirical weight}}\\=\frac{50}{25} = 2$$
Molecular formula = (X2Y)2 = X4Y2
$$\text{(iii)\space (a)\space}\underset{\text{Sulphur}}{\text{S(s)}} +\underset{\text{Nitric acid}}{\text{6HNO}_3\text{(aq)}}\xrightarrow{}\\\underset{\text{Water}}{\text{2H}_{2}\text{O}(1)} + \underset{\text{Sulphuric acid}}{\text{H}_{2}\text{SO}_{4}(aq)} +\underset{\text{Nitrogen dioxide}}{\text{6NO}_{2}(g)}$$
(b) The equation for catalytic oxidation of ammonia is :
$$\underset{\text{Ammonia}}{\text{4NH}_{3} + 5\text{O}_{2}}\xrightarrow[\text{700-800\degree C}]{\text{Pt}}\\\underset{\text{Nitric oxide}}{\text{4NO + 6 H}_{2}\text{O}} +\text{Heat}$$
Catalyst is a wire mesh consisting of platinum and rhodium.
$$\text{(c)}\space\underset{\text{Potassium nitrate}}{\text{KNO}_{3}(s)} + \underset{\text{Conc. sulphuric acid}}{\text{H}_{2}\text{SO}_{4}(l)}\xrightarrow{(\text{T}\lt 200\degree C)}\\\underset{\text{Nitric acid}}{\text{HNO}_{3}(l)} + \underset{\text{Potassium hydrogen sulphate}}{\text{KHSO}_{4}(s)}$$
(iv) (a) The equation for the laboratory preparation of hydrogen chloride gas :
$$\text{NaCl + H}_{2}\text{SO}_{4}\xrightarrow{\lt200\degree \text{C}}{\text{NaHSO}_{4} + \text{HCl}}\uparrow$$
Although it is a reversible reaction, it goes to completion as hydrogen chloride continously escapes as a gas.
The reaction can occur up to the stage of the formation of sodium sulphate on heating above 200°C.
$$\text{NaHSO}_{4} + \text{NaCl}\xrightarrow{\text{above 200\degree C}}\text{Na}_{2}\text{SO}_{4} +\text{HCl}\uparrow$$
(b) The drying agent used in the laboratory preparation of hydrochloride acid is concentrated sulphuric acid. The other drying agents such as phosphorus pentoxide (P2O5) and quick lime (CaO) cannot be used because they react with hydrogen chloride.
$$\text{2P}_{2}\text{O}_{5} + \text{3HCl}\xrightarrow{}\text{POCl}_{3} + 3\text{HPO}_{3}\\\text{CaO + 2HCl}\xrightarrow{}\text{POCl}_{3} + \text{3HPO}_{3}$$
(c) A safety precaution which should be taken during the preparation of hydrochloric acid : Always wear chemical splash goggles, chemical-resistant gloves and a chemical resistant apron in the laboratory during the preparation of hydrochloric acid.
Answer 7.
(i)
| Element | Percentage | Molecules | Simple ratio | Simple whole ratio |
| C | 82.76 | $$\frac{82.76}{12} = 6.89$$ | $$\frac{6.89}{6 89} = 1$$ | 2 |
| H | 17.24 | $$\frac{17.24}{1} = 17.24$$ | $$\frac{17.24}{6.89} = 2.5$$ | 5 |
Empirical formula mass = (12 × 2) + (1 × 5)
= 24 + 5 = 29
Vapour density → 29 (Given)
Molecular mass = V.D. × 2 = 29 × 2 = 58 g
Molecular formula mass = n × Empirical formula mass
$$\Rarr\space n = \frac{\text{Molecular formula mass}}{\text{Empirical formula mass}}\\=\frac{58}{29}= 2$$
Molecular formula = n × Empirical formula
= 2 × C2H5
= C4H10
(ii) (a) Electrode on the left side is the oxidising electrode because copper atoms lose electrons at this electrode, therefore, it is called as anode.
(b) At anode : Cu – 2e– → Cu2+
At cathode : Cu2+ + 2e– → Cu
(c) 1. Reddish brown copper metal is deposited at cathode so anode becomes thin and cathode grows thick gradually.
2. Blue colour of aqueous copper(II) sulphate solution remains unchanged because the effective concentration of copper ions in solution remains the same.
(iii) (a) Calcium nitrate and zinc nitrate solutions can be distinguish by reacting with ammonium hydroxide solution :
On adding ammonium hydroxide gelatinous white precipitates of zinc hydroxide are formed.
$$\text{Zn(NO}_{3})_{2} + \text{2NH}_{4}\text{OH}\xrightarrow{}\\\underset{\text{Zinc hydroxide}\text{(Gelatinous white precipitate)}}{\text{Zn(OH)}_{2}}\darr +\text{2NH}_{4}\text{NO}_{3}$$
On adding excess of ammonium hydroxide, the precipitates dissolve forming a soluble complex.
$$\text{Zn(NO)}_2 + \text{2NH}_4\text{NO}_3\xrightarrow{}\\\text{2NH}_{4}\text{OH}\underset{\text{Soluble complex}}{\lbrack Zn(\text{NH}_{3})_{4}\rbrack}(\text{NO}_{3})_{2} + \text{4H}_{2}\text{O}$$
No visible reaction occurs when we add calcium nitrate to ammonium hydroxide.
$$\text{CaNO}_3 \text{+ NH}_4\text{OH}\xrightarrow{}\text{No reaction}$$
(b) Ammonium sulphate crystals give pungent colourless gas ammonia (NH3) when heated with any alkali or base. When NH3 gas comes in contact with a glass rod dipped in HCl white fumes of this released NH4Cl are produced. While there is no reaction with sodium sulphate or no pungent gas is released.
(c) Magnesium chloride reacts with silver nitrate solution and form a white precipitate of silver
chloride, whereas magnesium nitrate does not react with silver nitrate solution.
$$\text{MgCl}_2 \text{(aq) + 2AgNO}_3 \text{(aq)}\xrightarrow{}\\\underset{\text{Silver chloride(white precipitate)}}{\text{Mg(NO}_{3})_{2}(\text{aq}) +\text{2AgCl} (s)\darr}\\\text{MgNO}_{3} +\text{AgNO}_{3}\xrightarrow{}\text{No reaction} $$
Answer 8.
(i) (a) Compound XY2 has ionic bonds in nature because it is formed by transfer of electrons,
Four properties of compound XY2 are :
- It is hard and brittle.
- It is soluble in water.
- It has high melting and boiling point.
- It does not conduct electric current in the solid state but conducts electric current in the molten or dissolved state.
(b) In the first step, the donor atom (say A) transfers one electron of its lone pair to the acceptor atom (say B). Due to this the atom A develops a unit positive charge and atom B develops a unit negative change. This charge is known as formal charge and is similar to the formation of an ionic bond.
In the second step, the two electrons, one each with A+ and B– are shared by both the ions. This is similar to the formation of covalent bond.
$$\text{A}^{\normalsize+} + \text{B}^{\normalsize-}\xrightarrow{}\text{A : B or A}\xrightarrow{}\text{B}$$
Thus, a coordinate bond is equivalent to a combination of an electrovalent bond and a covalent bond. Hence, it is also called a semi polar bond or dative bond.
(ii) (a) The pH of :
- Pure water is 7
- Milk is 6.6
- Human blood is 7.3
(b) pH value = 7 indicates a neutral solution.
pH value > 7 indicates an alkaline solution.
pH value < 7 indicates an acidic solution.
(iii) (a) Reduction
(b) Oxidation
(c) Reduction
(iv) (a) Thallium
(b) Boron
(c) Three
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