Question 1

Question 1(ii)The bond between carbon and hydrogen in a molecule of methane issingle covalent bonddouble covalent bondtriple covalent bondionic bond

Accepted Answer

single covalent bondReason — In a molecule of methane, one carbon atom forms four covalent bonds with four hydrogen atoms by sharing a pair of electrons between itself and each hydrogen (H) atom.

Question 2

Question 1(iii)During the Ostwald's process, the gaseous reactants used arenitrogen and oxygenoxygen and carbon dioxidechlorine and nitrogenammonia and oxygen

Accepted Answer

ammonia and oxygenReason —4NH3+5O2→700−800°CPt4NO+6H2O+21.5K cals4	ext{NH}_3 + 5	ext{O}_2 \xrightarrow[700-800 \degree	ext{C}]{	ext{Pt}} 4	ext{NO} + 6	ext{H}_2	ext{O} + 21.5 	ext{K cals}4NH3​+5O2​Pt700−800°C​4NO+6H2​O+21.5K cals

Question 3

Question 1(viii)The base that neutralizes the acidity in stomach and acts as an antacidAmmonium hydroxideAluminium hydroxideMagnesium hydroxideCalcium oxide

Accepted Answer

Magnesium hydroxideReason — Magnesium hydroxide also known as milk of magnesia is a common component of antacids.

Question 4

Question 1(ix)Reaction of zinc metal with sodium hydroxide will lead to the formation ofsodium zincatesodium chloridezinc hydroxidezinc aluminate

Accepted Answer

sodium zincateReason — Zn + 2NaOH ⟶ H2 + Na2ZnO2 [sodium zincate]

Question 5

Question 1(x)The 'n' of CH3 is 3. Then, the molecular formula isC3H6C3H9CHCH9

Accepted Answer

C3H9Reason — As, molecular formula = n x empirical formulaHence, molecular formula = 3 x CH3 = C3H9Therefore, the molecular formula of the compound is C3H9

Question 6

Question 1(xi)During the electrolysis of aqueous copper sulphate using copper electrodes, copper ions are generated at the anode that go into the copper sulphate solution used as electrolyte.Which of the following is correct regarding the mass of anodeP. Mass of anode remains sameQ. Mass of anode gradually decreasesR. Mass of anode gradually increasesOnly POnly QBoth P and QP, Q and R

Accepted Answer

Only QReason — SO42- and OH1- ions migrate to the anode but neither are discharged since the copper anode loses electrons more easily than SO42- and OH1-. Hence, copper anode shows a loss in mass.

Question 7

Question 1(xiii)During electrolytic reduction of molten alumina, aluminium metal is obtained atcarbon cathodecarbon anodealuminium anodealuminium cathode

Accepted Answer

carbon cathodeReason — At cathode :2Al3+ + 6e- ⟶ 2AlAt anode : 3O2- - 6e- ⟶ 3[O] ⟶ 3O2Al3+ are discharged in preference to Na1+ and Ca2+ ions due to it's lower position in the electrochemical series.

Question 8

Question 1(xv)The atomic masses of nitrogen (N), oxygen (O) and neon (Ne) are approximately 14, 16 and 20.18 respectively.Which of the following statements regarding the number of atoms in 14 g of nitrogen, 16 g of oxygen and 20 g of neon is correct?P. 14 g of nitrogen contains double number of atoms as 20 g of neon.Q. 16 g of oxygen contains two times the number of atoms as 14 g of nitrogen.Only POnly QBoth P and QNeither P nor Q

Accepted Answer

Neither P nor QReason — According to Avogadro's number (NA) = 6.022 x 1023 atoms are present in one mole of element.In case of P,Number of atoms in 14 g of N= Given massAtomic mass\dfrac{	ext{Given mass}}{	ext{Atomic mass}}Atomic massGiven mass​ x NA= 1414\dfrac{14}{14}1414​ X 6.022 x 1023= 6.022 x 1023Number of atoms in 20 g of Ne= Given massAtomic mass\dfrac{	ext{Given mass}}{	ext{Atomic mass}}Atomic massGiven mass​ x NA= 2020.18\dfrac{20}{20.18}20.1820​ X 6.022 x 1023= 0.99 x 6.022 x 1023= 5.96 x 1023 atomsAs number of atoms in 14 g of nitrogen is almost equal to 20 g of Neon. Hence, statement P is incorrect.In case of statement Q,Number of atoms in 16 g of oxygen= Given massAtomic mass mass\dfrac{	ext{Given mass}}{	ext{Atomic mass mass}}Atomic mass massGiven mass​ X NA= 1616\dfrac{16}{16}1616​ X 6.022 x 1023= 6.022 x 1023As number of atoms in 16 g of oxygen is equal to 14 g of nitrogen. Hence, statement Q is also incorrect.

Question 9

Question 2(i)The diagram shows an experimental set up for the laboratory preparation of a gas known as marsh gas. It is also present in the air exhaled by animals whose food contain cellulose. Study the diagram and answer the following.(a) Name the gas collected in the jar.(b) Write a balanced chemical equation for the above preparation. How is the gas being collected?(c) Name the type of reaction taking place in this. Why this particular reaction is taking place in this?

Accepted Answer

(a) Methane [CH4](b)CH3COONasodium acetate+NaOHsodalime→ΔCaOCH4methane+Na2CO3\underset{	ext{sodium acetate}}{	ext{CH}_3	ext{COONa}} + \underset{	ext{sodalime}}{	ext{NaOH}} \xrightarrow[\Delta]{	ext{CaO}} \underset{	ext{methane}}{	ext{CH}_4} + 	ext{Na}_2	ext{CO}_3sodium acetateCH3​COONa​+sodalimeNaOH​CaOΔ​methaneCH4​​+Na2​CO3​The gas is collected by downward displacement of water.(c) Decarboxylation reaction takes place. It happens due to elimination of a molecule of CO2 from a carboxylic acid. In this reaction, CO2 is removed as a carbonate.

Question 10

Question 2(ii)Complete the following by choosing the correct answer from the bracket.(a) On moving down the group electron affinity ............... (increases/decreases).(b) The functional group present in acetic acid is ............... (-COOH/-OH/-CHO)(c) ............... (Concentrated sulphuric acid/Dilute sulphuric acid) is a weak acid and a good electrolyte.(d) Compounds which contains one —COOH group are known as ............... (monocarboxylic acid/dicarboxylic acid).(e) Quicklime is not used to dry HCl gas because CaO is ............... (alkaline/acidic) in nature.

Accepted Answer

(a) On moving down the group electron affinity decreases.(b) The functional group present in acetic acid is —COOH.(c) Dilute sulphuric acid is a weak acid and a good electrolyte.(d) Compounds which contains one —COOH group are known as monocarboxylic acid.(e) Quicklime is not used to dry HCl gas because CaO is alkaline in nature.

Question 11

Question 2(v)(a) Draw the structural formula for the following.(1) But-1-ene(2) Butanoic acid(3) Propanol(b) Name the following organic compounds in IUPAC system

Accepted Answer

(a) Structural formulae are shown below:(1) But-1-ene(2) Butanoic acid(3) Propanol(b) IUPAC names of the organic compounds are:(1) Methanoic acid(2) 3-methyl-hexane

Question 12

Question 3(i)Identify the reactant and write the balanced equation for the following.Sodium ethanoate reacts with compound Q to give a strong base Na2CO3 and CH4.

Accepted Answer

The compound Q is sodalime [NaOH]CH3COONasodium ethanoate+NaOHsodalime→ΔCaOCH4methane+Na2CO3\underset{	ext{sodium ethanoate}}{	ext{CH}_3	ext{COONa}} + \underset{	ext{sodalime}}{	ext{NaOH}} \xrightarrow[\Delta]{	ext{CaO}} \underset{	ext{methane}}{	ext{CH}_4} + 	ext{Na}_2	ext{CO}_3sodium ethanoateCH3​COONa​+sodalimeNaOH​CaOΔ​methaneCH4​​+Na2​CO3​

Question 13

Question 3(iv)(a) State whether the following statements are True or False. Justify your answer.(1) According to Mendeleev, periodic properties of elements are periodic functions of their atomic number.(2) Non-metals form oxides that are basic in nature.(b) Write the chemical reaction of Ammonium hydroxide with soluble salt of lead (Pb2+ ion).

Accepted Answer

(a) (1) FalseCorrected Statement — According to Mendeleev, periodic properties of elements are periodic functions of their atomic weight.(a) (2) False.Corrected Statement — Metals form oxide that are basic in nature.(b) Pb(NO3)2 + 2NH4OH ⟶ Pb(OH)2 ↓ + 2NH4NO3

Question 14

Question 4(i)Define:(a) Covalent bond(b) Co-ordinate bond

Accepted Answer

(a) Covalent bond — The chemical bond that is formed between two combining atoms by mutual sharing of one or more pair of electrons is called a covalent bond. Depending on the number of electron pairs shared : bond is single [-], double [=], or triple [≡] covalent.(b) Co-ordinate bond — The bond formed between two atoms by sharing a pair of electrons, provided entirely by one of the combining atoms but shared by both is called a coordinate bond.

Question 15

Question 4(ii)SolveA compound has the following percentage composition by mass, carbon 14.4%. hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. (H=1, C=12 and Cl=35.5)

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon14.41214.412\dfrac{14.4}{12}1214.4​ = 1.21.21.2\dfrac{1.2}{1.2}1.21.2​ = 1Hydrogen1.211.21\dfrac{1.2}{1}11.2​ = 1.21.21.2\dfrac{1.2}{1.2}1.21.2​ = 1chlorine84.535.584.535.5\dfrac{84.5}{35.5}35.584.5​ = 2.382.381.2\dfrac{2.38}{1.2}1.22.38​ = 1.98 = 2Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2Hence, empirical formula is CHCl2

Question 16

Question 4(iii)State the conditions required for the following reactions(a) Conversion of potassium nitrate to potassium bisulphate.(b) Conversion of nitrogen and hydrogen to ammonia.(c) Conversion of ammonia to nitric oxide.

Accepted Answer

(a) Conditions required for the conversion of potassium nitrate to potassium bisulphate are:Temperature of the reaction should not exceed 200°C.All glass apparatus should be used.Reaction:KNO3 + H2SO4 →<200°C\xrightarrow{\lt 200 \degree	ext{C}}<200°C​ KHSO4 + HNO3(b) Conditions required for the conversion of nitrogen and hydrogen to ammonia are:Temperature should be between 450-500°C.Pressure should be above 200 atm.Catalyst used is finely divided iron.Promoter used is traces of molybdenum or Al2O3.Reaction:N2 + 3H2 ⇌450−500°C, above 200 atmFe and Mo\xrightleftharpoons[450-500	ext{°C},	ext { above }200 	ext{ atm}]{	ext{Fe and Mo}}Fe and Mo450−500°C, above 200 atm​ 2NH3 + Δ(c) Conditions required for the conversion of ammonia to nitric oxide are:Temperature should be around 800°C.Catalyst used is Platinum.Reaction:4NH3 + 5O2 →800°CPt.\xrightarrow[800 \degree 	ext{C}]{	ext{Pt}.}Pt.800°C​ 4NO↑ + 6H2O + Δ2NO + O2 ⟶ 2NO2 (brown gas)

Question 17

Question 5(iv)X [Oxygen, Sulphur, Selenium] and Y [Neon, Argon, Xenon] are two sets of elements. Using these elements complete the following.(a) Set ............... is known as noble gases.(b) In set X ............... has the highest electronegativity.(c) The ............... set contains the non-metals.

Accepted Answer

(a) Set Y is known as noble gases.(b) In set X oxygen has the highest electronegativity.(c) The X set contains the non-metals.

Question 18

Question 6(ii)Calculate:(a) The mass of 12.044 x 1023 oxygen atoms.(b) What is the vapour density of ethylene?

Accepted Answer

(a) No. of moles of oxygen atoms = 12.044×10236.022×1023\dfrac{12.044 	imes 10^{23}}{6.022 	imes 10^{23}}6.022×102312.044×1023​ = 2 molesMass of oxygen atoms = no. of moles x atomic mass= 2 x 16 = 32 g.∴ The mass of 12.044 x 1023 oxygen atoms is 32 grams.(b) Gram molecular mass of ethylene (C2H4) = 2(12) + 4(1) = 24 + 4 = 28 gVapour density = Molecular weight2\dfrac{	ext{Molecular weight}}{2}2Molecular weight​ = 282\dfrac{28}{2}228​ = 14Hence, vapour density is 14

Question 19

Question 7(ii)Consider the elements 8B, 20A. What type of bond will be formed when element B and A react with each other. Also find out in which group, A and B elements are present in the periodic table.

Accepted Answer

Electronic configuration of 8B is [2, 6]. It needs two electrons to complete its octet.Electronic configuration of 20A is [2, 8, 8, 2]. It wants to donate two electrons in order to obtain a stable octet configuration.Hence, ionic bond is formed as 20A loses two electrons and 8B gains two electrons and both attain a stable octet.8B [2, 6] is present in group VIA and 20A [2, 8, 8, 2] is present in group in IIA as the group of an elements is predicted from the number of electrons in the valence shell.

Question 20

Question 7(iii)(a) List any two uses of ammonia.(b) Give the balanced equations for the following conversionsAmmonia solution to an amphoteric hydroxideAmmonia to brown gas

Accepted Answer

(a) Two uses of ammonia are:Liquid ammonia is used as a refrigerant in ice plants.Ammonia is used in the manufacture of nitrogenous fertilisers like ammonium sulphate, ammonium nitrate, urea, etc.(b) Balanced equations for the conversions are given below:AlCl3 (aq) + 3NH4OH (aq) ⟶ Al(OH)3 ↓ + 3NH4Cl (aq)4NH3 + 5O2 →800°CPt.\xrightarrow[800 \degree 	ext{C}]{	ext{Pt}.}Pt.800°C​ 4NO↑ + 6H2O + Δ2NO + O2 ⟶ 2NO2 (brown gas)

Solved Sample Paper 1

Section A

Section B