Question 1

Question 1(ii)The relationship between pressure and volume at constant temperature is given by :AvogadroCharlesBoyleHenry

Accepted Answer

BoyleReason — According to Boyle's law volume of a given mass of a dry gas is inversely proportional to its pressure at constant temperature.P1V1 = P2V2

Question 2

Question 1(vi)The double covalent molecule is :OxygenAmmoniaSodium chlorideHydrogen sulphide

Accepted Answer

OxygenReason — To attain the stable electronic configuration of the nearest noble gas neon, oxygen needs two electrons.When two oxygen atoms come closer, each contributes two electrons and so they have two shared pair of electrons between them. Both atom attain an octet, resulting in the formation of a double covalent bond [O=O] between them.

Question 3

Question 1(viii)Which of the following will produce water vapour on heating?Potassium chloride crystalsSodium chloride crystalsSodium nitrate crystalsWashing soda crystals

Accepted Answer

Washing soda crystalsReason — Washing soda crystals (Na2CO3.10H2O) contain water of crystallisation. On heating, it is given off as water vapour.

Question 4

Question 1(xii)Which metal gives hydrogen upon reaction with all of the following - Acid, alkali, water?IronZincMagnesiumLead

Accepted Answer

ZincReason — Zn has a unique nature.Zinc is considered to have a unique nature because it can react with acids and can even react with hot conc. alkalis to form hydrogen and a soluble salt.Zn + 2HCl ⟶ ZnCl2 + H2Zn + 2NaOH ⟶ Na2ZnO2 + H2 ↑Zn with water also gives hydrogenZn + H2O ⟶ ZnO + H2

Question 5

Question 2(b)What do you observe when :(i) Ammonium chloride crystals are heated in a test tube ?(ii) Iron nails are added to copper sulphate solution ?(iii) Lead nitrate crystals are heated strongly?(iv) Silver nitrate solution is mixed with sodium chloride solution?(v)	Blue copper sulphate crystals are heated ?

Accepted Answer

(i) On heating ammonium chloride, white crystalline solid sublimates to form a basic gas (NH3) and acidic gas (HCl). The dense white fumes are noticed that form a white mass on the cooler parts of the test tube. No residue is left behind.(ii) When iron nail is kept in a blue coloured copper sulphate solution for sometime, a reddish brown coating is seen on the iron nail and the colour of the solution changes gradually from blue to light green. The reason for this observation is that iron being more reactive than copper, displaces copper from the solution and copper is deposited on the iron nail(iii) When solid lead nitrate is heated strongly, it decomposes to produce light yellow solid lead monoxide, reddish brown nitrogen dioxide gas and colourless oxygen gas.(iv) When silver nitrate solution is mixed with sodium chloride solution, a white ppt of silver chloride is formed along with sodium nitrate.(v) The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate releasing water of crystallization.

Question 6

Question 2(d)(i) Balance each of the following chemical equations :FeCl3 + NH4OH ⟶ NH4Cl + Fe(OH)3Zinc + Nitric acid ⟶ Zinc nitrate + Nitrogen dioxide + Water(ii) Identify the substance which matches the description given below:White crystalline substance which sublimes on heating.The metal that cannot displace hydrogen from dilute hydrochloric acid.

Accepted Answer

(i) 1. FeCl3 + 3NH4OH ⟶ 3NH4Cl + Fe(OH)3Zn + 4HNO3 ⟶ Zn(NO3)2 + 2NO2 + 2H2O(ii) 1. Ammonium chloride [NH4Cl]Copper

Question 7

Question 3(a)The formula of the chloride of the metal 'M' is MCl3. State the formula of its:(i)	Carbonate(ii) Hydroxide

Accepted Answer

(i) Formula of metal carbonateM3+↗CO32−M33 ↘↙ C2O3⇒M22 ↘↙ C3O3\text{M}^{3+} \phantom{\nearrow} \text{CO}_{3}^{2-} \[0.5em] \overset{\phantom{3}{3}}{\text{M}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{C}}\text{O}_{3} \Rightarrow \underset{\phantom{2}{2}}{\text{M}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{C}}\text{O}_{3} \[0.5em]M3+↗CO32−​M33 ↘↙ C2O3​⇒22M​ ↘↙ 3C​O3​So, we get the formula as M2(CO3)3\bold{M}_{\bold{2}}\bold{(CO_{\bold{3}}})_{\bold{3}}M2​(CO3​)3​(ii) Formula of metal hydroxideM3+↗OH1−M33 ↘↙ OH1⇒M21 ↘↙ OH3\text{M}^{3+} \phantom{\nearrow} \text{OH}^{1-} \[0.5em] \overset{\phantom{3}{3}}{\text{M}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{2}{1}}{\text{M}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{OH}} \[0.5em]M3+↗OH1−M33 ↘↙ OH1⇒21M​ ↘↙ 3OH​So, we get the formula as M(OH)3\bold{M}\bold{(OH)}_{\bold{3}}M(OH)3​

Question 8

Question 3(c)Draw the orbital structure of each of the following compounds:(i)	Water(ii) Magnesium chloride[At.No. : H = 1, 0 = 16, Mg = 12, Cl = 17]

Accepted Answer

(i)	Orbital structure of water molecule is shown below :(ii) Orbital structure of Magnesium chloride molecule is shown below :

Question 9

Question 3(d)Write a balanced chemical equation for each of the following reactions:(i)	Magnesium hydrogen carbonate reacts with slaked lime.(ii) Formation of coloured precipitate.(iii) Heating ammonium dichromate.

Accepted Answer

(i)	Magnesium hydrogen carbonate reacts with slaked lime.Mg(HCO3)2 + Ca(OH)2 ⟶ MgCO3↓ + CaCO3↓ + 2H2O(ii) Formation of coloured precipitate.3NaOH + FeCl3 ⟶ Fe(OH)3 [reddish brown ppt.] + 3NaCl(iii) Heating ammonium dichromate.(NH4)2Cr2O7[Ammonium dichromate (orange)]→ Δ Cr2O3 [Chromic oxide(green)]+4H2O+N2\underset{	ext{[Ammonium dichromate (orange)]}}{(	ext{NH}_4)_2	ext{Cr}_2	ext{O}_7} \xrightarrow{\space \Delta \space} \underset{	ext{ [Chromic oxide(green)]}}{	ext{Cr}_2	ext{O}_3} + 4	ext{H}_2	ext{O} + 	ext{N}_2[Ammonium dichromate (orange)](NH4​)2​Cr2​O7​​ Δ ​ [Chromic oxide(green)]Cr2​O3​​+4H2​O+N2​

Question 10

Question 4(a)Write 2 points of comparison between Halogens and Alkali metals.

Accepted Answer

PropertyHalogensAlkali metalsNatureHighly reactive, highly electronegative, non-metalsHighly reactive, highly electropositive, light soft metalsReducing / Oxidizing natureStrong oxidizing agentsStrong reducing agents

Question 11

Question 4(c)(i) State Boyle's law.(ii) Calculate the volume occupied by 2 g of hydrogen at 270°C and 4 atmosphere pressure if at STP it occupies 22.4 litres.

Accepted Answer

(i) Boyle's law — Temperature remaining constant the volume of a given mass of dry gas is inversely proportional to it's pressure.V ∝ 1P\dfrac{1}{\text{P}}P1​ [T = constant](ii) Initial conditions [S.T.P.] :P1 = Initial pressure of the gas = 1 atmV1 = Initial volume of the gas = 22.4 lT1 = Initial temperature of the gas = 273 KFinal conditions :P2 (Final pressure) = 4 atmT2 (Final temperature) = 270°C = 270 + 273 = 543 KV2 (Final volume) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :1×22.4273=4×V2543V2=543×22.4273×4V2=12163.21092V2=11.13 l\dfrac{1 \times 22.4}{273} = \dfrac{4 \times \text{V}_2}{543} \[1em] \text{V}_2 = \dfrac{543 \times 22.4}{273 \times 4} \[1em] \text{V}_2 = \dfrac{12163.2}{1092} \[1em] \text{V}_2 = 11.13 \text{ l} \[1em]2731×22.4​=5434×V2​​V2​=273×4543×22.4​V2​=109212163.2​V2​=11.13 l∴ Volume occupied = 11.13 l

Question 12

Question 5(a)What is an acid rain ?

Accepted Answer

The term 'acid rain' refers to all precipitations — rain, snow, fog, dew — that are more acidic than normal water (i.e., have pH less than 5.6, the pH value of normal water).

Question 13

Question 5(b)How does acid rain affect soil chemistry and water bodies?

Accepted Answer

Acid rain affects soil chemistry. It removes calcium and potassium, both the basic ingredients of soil, thus making it lose its fertility.Acid rain has serious ecological impacts as it affects water bodies too. The water of lakes and rivers is gradually becoming acidic due to acid rain, which is affecting aquatic life.

Question 14

Question 5(c)Name the first three halogens. Give their physical state and colour.

Accepted Answer

ElementPhysical stateColourFluorine (F)GasPale yellowChlorice (Cl)GasGreenish yellowBromine (Br)LiquidReddish brown

Question 15

Question 6(a)Distinguish between the following pairs of substances :(i)	Sodium sulphide and sodium sulphite(ii) HCl gas and chlorine gas

Accepted Answer

(i) On addition of dil. sulphuric acid to sodium sulphide, hydrogen sulphide (H2S) gas is evolved which can be identified by its rotten egg smell. Hydrogen sulphide gas turns moist lead acetate paper silvery black.Na2S + H2SO4 ⟶ Na2SO4 + H2S ↑On addition of dil. sulphuric acid to sodium sulphite, sulphur dioxide (SO2) gas is evolved. Sulphur dioxide gas has a suffocating odour and turns lime water milky and potassium permanganate solution from pink to clear colourless.Na2SO3 + H2SO4 (dil.) ⟶ Na2SO4 + H2O + SO2↑(ii) HCl and chlorine gasChlorine (Cl2) turns moist starch iodide paper blue black whereas HCl does not.Cl2 + 2KI ⟶ 2KCl + I2Starch + I2 ⟶ Blue black colour

Question 16

Question 6(b)Al2O3 is the formula of Aluminium oxide. Write the formulae of :(i) Aluminium nitrate(ii) Aluminium phosphate

Accepted Answer

(i) Formula of aluminium nitrateAl3+↗NO31−Al33 ↘↙ N1O3⇒Al21 ↘↙ N3O3\text{Al}^{3+} \phantom{\nearrow} \text{NO}_{3}^{1-} \[0.5em] \overset{\phantom{3}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{N}}\text{O}_{3} \Rightarrow \underset{\phantom{2}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{N}}\text{O}_{3} \[0.5em]Al3+↗NO31−​Al33 ↘↙ N1O3​⇒21Al​ ↘↙ 3N​O3​So, we get the formula as Al(NO3)3\bold{Al}\bold{(NO}_{\bold{3}}\bold{)}_\bold{3}Al(NO3​)3​(ii) Formula of aluminium phosphateAl3+↗PO43−Al33 ↘↙ P3O4⇒Al21 ↘↙ P1O4\text{Al}^{3+} \phantom{\nearrow} \text{PO}_{4}^{3-} \[0.5em] \overset{\phantom{3}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{3}{\text{P}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{P}}\text{O}_{4} \[0.5em]Al3+↗PO43−​Al33 ↘↙ P3O4​⇒21Al​ ↘↙ 1P​O4​So, we get the formula as AlPO4\bold{Al}\bold{PO}_{\bold{4}}AlPO4​

Question 17

Question 6(c)The electronic configuration of three elements A, B and C are as follows :A — 2, 8, 8, 2;B — 2, 8, 7C — 2, 4(i) Which of these can form cations ?(ii) Which of these can form anions ?(iii) Write a formula of a covalent compound formed using these elements

Accepted Answer

(i) A will form cations as it will lose electrons to attain a stable octet.(ii) B will form anions as it will gain electrons to attain a stable octet.(iii) Formula of a covalent compound formedC4↗B1−C34 ↘↙ B1⇒C21 ↘↙ B4\text{C}^{4} \phantom{\nearrow} \text{B}^{1-} \[0.5em] \overset{\phantom{3}{4}}{\text{C}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{B}} \Rightarrow \underset{\phantom{2}{1}}{\text{C}} \space {\searrow}\mathllap{\swarrow} \space \underset{4}{\text{B}} \[0.5em]C4↗B1−C34 ↘↙ B1⇒21C​ ↘↙ 4B​So, we get the formula as CB4\bold{CB}_{\bold{4}}CB4​

Question 18

Question 6(d)(i) What do you understand by the term 'isotopes'?(ii) Why do isotopes of an element possess identical chemical properties ?(iii) Name the isotope of hydrogen which do not have a neutron in it.

Accepted Answer

(a) The atoms of the same element, having same atomic number (Z), but different mass number (A), are called isotopes.​(ii) Chemical properties of isotopes of the same element are similar because chemical properties are dependent on the electronic configuration of an atom. As Isotopes have same atomic number so they have same number of electrons and hence same electronic configuration.(iii) Protium (11H).

Question 19

Question 7(b)What temperature in degree centigrade would	be necessary to double the volume of a gas (initially at STP), if the pressure is reduced to half ?

Accepted Answer

Initial conditions [S.T.P.]Final conditionsP1 = Initial pressure of the gas = 1 atmP2 = Final pressure of the gas = 0.5 atmV1 = Initial volume of the gas = VV2 = Final volume of the gas = 2VT1 = Initial temperature of the gas = 273 KT2 = Final temperature of the gas = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :1×V273=0.5×2VT2T2=0.5×2×273T2=273K\dfrac{1\times \text{V}}{273} = \dfrac{0.5\times\text{2\text{V}}}{\text{T}_2} \[0.5em] \text{T}_2 = 0.5\times 2 \times 273\[0.5em] \text{T}_2 = 273\text{K}2731×V​=T2​0.5×2V​T2​=0.5×2×273T2​=273K∴ Final temperature of the gas = 273 K = 0 °C

Question 20

Question 7(d)Mohr's salt has the formula (NH4)2SO4.FeSO4.6H2O(i)	Calculate its molecular mass.(ii) What is the percentage of Nitrogen in Mohr's salt ? (Atomic mass: N =14, H = 1, S = 32, Fe = 56, O = 16)

Accepted Answer

(a) Molar mass of Mohr's salt [(NH4)2SO4.FeSO4.6H2O]= 2[14 + 4(1)] + 32 + 4(16) + 56 + 32 + 4(16) + 6[2(1) + 16]= 2[18] + 32 + 64 + 56 + 32 + 64 + 6[18]= 36 + 32 + 64 + 56 + 32 + 64 + 108= 392 g(ii) Molar mass of Mohr's salt [(NH4)2SO4.FeSO4.6H2O] = 392 gMass of nitrogen in one mole of Mohr's salt = 2 x 14 = 28 g392 g Mohr's salt has mass of nitrogen = 28 g∴ 100 g Mohr's salt will have mass= 28×100392\dfrac{28 	imes 100 }{392}39228×100​= 7.14%

Model Question Paper — 1

Section A

Section B