Question 1

Question 1(iv)The reddish brown coloured gas formed when nitric acid reacts with carbon isN2O5N2O4NO2N2O3

Accepted Answer

NO2Reason — Nitric acid oxidises carbon to carbon dioxide whereas it itself undergoes reduction to form NO2 which is reddish brown in colour.C + 4HNO3 ⟶ CO2 + 2H2O + 4NO2

Question 2

Question 1(x)The organic compound which undergoes substitution reaction isC2H2C4H8C8H14C5H12

Accepted Answer

C5H12Reason — All alkanes undergo substitution reaction. Pentane [C5H12] is an alkane and hence, undergoes substitution reaction .

Question 3

Question 1(xii)An element X belongs to group II A and another element Y belongs to group V A of the periodic table. Which of the following could be correct?P. Valency of Y will be 5.Q. Element X will have 2 valence electrons.R. Element X belongs to group number 2Only PQRBoth Q and R

Accepted Answer

Both Q and RReason — Q and R are correct however, P is not correct as valency of Y will be equal to = 8 - 5 = 3

Question 4

Question 1(xv)The molecular mass of CO2 and O2 are approximately 44 and 32 respectively. Which of the following statements regarding the mass of oxygen and number of molecules of oxygen is correct?P. The mass of oxygen in 2.24 litres of CO2 is 3.2 gQ. The number of molecules of oxygen present in it is 6.022 x 1023Only POnly QBoth P and QNeither P nor Q

Accepted Answer

Only PReason —P: As 22.4 L of CO2 at S.T.P. has mass = 44 gTherefore, 2.24 L of CO2 at S.T.P. has mass = 4422.4\dfrac{44}{22.4}22.444​ x 2.24 = 4.4 g44 g of CO2 has 32 g of oxygenSo, 4.4 g of CO2 will have = 3244\dfrac{32}{44}4432​ x 4.4 = 3.2 g of oxygen.Hence P is correct.Q: No. of moles of O2 in 3.2 g = 3.232\dfrac{3.2}{32}323.2​ = 0.1 g .1 mole of O2 = 6.023 x 1023 molecules0.1 moles has molecules = 6.023 x 1023 x 0.1 = 6.023 x 1022Hence Q is incorrect.

Question 5

Question 2(ii)Observe the diagram given below used in the electrolysis of aqueous CuSO4 solution. Study the diagram and answer the following.(a) Write the equation for the reactions that is occurring at anode and cathode.(b)	State two appropriate observations for the given electrolysis reaction.(c)	Why does the blue colour of Cu2+ ions in the electrolytic solution fades slowly? Explain.

Accepted Answer

(a) Reaction at cathode:Cu2+ + 2e- ⟶ CuReaction at anode:OH1- - 1e- ⟶ OH x 44OH ⟶ 2H2O + O2(b) Two observations for this electrolysis are:Pink or reddish brown copper is deposited at the cathode.The blue colour of copper sulphate solution slowly fades away and it becomes colourless.(c) As the Cu2+ ions are discharged at the cathode and deposited as copper metal, there is a decrease in Cu2+ ions in the solution. This causes the blue colour to fade slowly and finally the solution becomes colourless as soon as Cu2+ ions are finished.

Question 6

Question 2(iii)Complete the following by choosing the correct answer from the bracket.(a)	Alkenes differ from alkanes due to the presence of ............... (double/single) bonds.(b)	Down the group, ionisation potential ............... (increases /decreases).(c)	By dissolving aluminium oxide in cryolite a ............... (conducting/non-conducting) solution is produced.(d)	The catalyst used for conversion of ethene to ethane is commonly ............... (iron/nickel).(e) The gas released when sodium carbonate is added to a solution of sulphur dioxide is ............... (CO2/SO3)

Accepted Answer

(a)	Alkenes differ from alkanes due to the presence of double bonds.(b)	Down the group, ionisation potential decreases.(c)	By dissolving aluminium oxide in cryolite a conducting solution is produced.(d)	The catalyst used for conversion of ethene to ethane is commonly nickel.(e) The gas released when sodium carbonate is added to a solution of sulphur dioxide is CO2.

Question 7

Question 2(v)(a) Draw the structural formula for the followingMethoxymethane2-methyl propan-2-olEthanol(b) Name the following organic compounds in IUPAC system.

Accepted Answer

(a) Structural formulae are shown below:(1) Methoxymethane(2) 2-methyl propan-2-ol(3) Ethanol(b)	IUPAC names of the organic compounds are:2, 3-dichloropropan-1-oic acidButan-2-ol

Question 8

Question 3(i)Draw the electron dot structure for the following.(a) NH4+(b) CaO

Accepted Answer

(a) NH4+(b) CaO

Question 9

Question 3(ii)Distinguish between the following as directed.(a)	Sodium carbonate and sodium sulphate using H2SO4.(b) Lead (II) oxide and zinc oxide using H2SO4.

Accepted Answer

(a) Sodium carbonate reacts with sulphuric acid and gives white coloured sodium sulphate along with water and carbon dioxide.Na2CO3 + H2SO4 ⟶ Na2SO4 + H2O + CO2On the other hand sodium sulphate reacts with sulphuric acid and gives colourless sodium hydrogen sulphate. Hence, the two can be distinguished.Na2SO4 + H2SO4 ⟶ 2NaHSO4(b) Lead oxide reacts with sulphuric acid and gives white coloured lead sulphate along with water.PbO + H2SO4 ⟶ PbSO4 + H2OOn the other hand zinc oxide reacts with sulphuric acid and gives colourless zinc sulphate along with water.ZnO + H2SO4 ⟶ ZnSO4 + H2O

Question 10

Question 3(iii)Name the ions present in(a)	alumina(b) fluorspar(c) sulphuric acid

Accepted Answer

(a)	alumina [Al2O3] — Al3+ and O2-(b) fluorspar [CaF2]— Ca2+ and F-(c) sulphuric acid [H2SO4] — H+ and SO42-

Question 11

Question 4(i)Define:(a)	Alkanes(b) Alkenes

Accepted Answer

(a)	Alkanes	— Saturated aliphatic hydrocarbons containing a carbon carbon single bond are called alkanes. The general formula is CnH2n+2(b) Alkenes — Unsaturated aliphatic hydrocarbons containing a carbon carbon double bond are called alkenes. The general formula is CnH2n

Question 12

Question 4(ii)An organic compound has an empirical formula (CH2O). Its vapour density is 45. Then what will be the molecular formula of the compound?

Accepted Answer

Given,Empirical formula = CH2OHence, empirical formula mass = 12 + 2(1) + 16 = 12 + 2 + 16 = 30 gMolecular weight = 2 x Vapour density = 2 x 45 = 90Molecular formula = n x empirical formulawhere,n = molecular massempirical formula mass\dfrac{	ext{molecular mass}}{	ext{empirical formula mass}}empirical formula massmolecular mass​∴ n = 9030\dfrac{90}{30}3090​ = 3Hence, molecular formula = 3 x CH2O = C3H6O3Therefore, the molecular formula of the compound is C3H6O3

Question 13

Question 4(iii)State the conditions required for the following reactions.(a)	Conversion of ammonia to nitric oxide in presence of air.(b)	Conversion of hydrogen chloride to chlorine.(c)	Conversion of sodium bisulphate to sodium sulphate.

Accepted Answer

(a) Conditions required for the conversion of ammonia to nitric oxide in presence of air 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)(b) Conditions required for the conversion of hydrogen chloride to chlorine are:Temperature should be above 500°C.Reaction:2HCl →>500°C\xrightarrow{\gt 500 \degree	ext{C}}>500°C​ H2 + Cl2(c) Conditions required for the conversion of sodium bisulphate to sodium sulphateTemperature should be above 200°C.Reaction:NaNO3 + NaHSO4 →>200°C\xrightarrow{\gt 200 \degree	ext{C}}>200°C​ Na2SO4 + HNO3

Question 14

Question 4(iv)(a) State whether the following statements are True or False. Justify your answer.Weak acids have low electrical conductivity.The concentration of hydrogen ions is equal in strong acids and weak acids.(b)	Calculate the percentage of sodium (Na) in Na2B4O7.10H2O.

Accepted Answer

TrueReason — Weak acids are partially dissociated in fused or aqueous solution hence allow small amount of electricity to flow through them and thus, have low electrical conductivity.FalseReason — Acids with a higher concentration of H+ (aq) ions are strong acids and acids with a lower concentration of H+ (aq) ions are weak acids.(b) Molecular weight of Na2B4O7.10H2O= 2(23) + 4(11) + 7(16) + 20(1) + 10(16)= 46 + 44 + 112 + 20 + 160= 382 g382 g of Na2B4O7.10H2O contains 46 g of Na∴ 100 g of Na2B4O7.10H2O will contain = 46382\dfrac{46}{382}38246​ x 100 = 12.04%Hence, 12.04% of sodium is present in Na2B4O7.10H2O

Question 15

Question 5(i)An organic compound, whose vapour density is 45, has the following percentage compositionH = 2.22%, O = 71.19% and remaining carbon.Calculate,(a) it's empirical formula, and(b) it's molecular formula

Accepted Answer

https://www.knowledgeboat.com/question/an-organic-compound-whose-vapour-density-is-45-has-the--586471767444377600Element% compositionAt. wt.Relative no. of atomsSimplest ratioHydrogen2.2212.221\dfrac{2.22}{1}12.22​ = 2.222.222.21\dfrac{2.22 }{2.21}2.212.22​ = 1Oxygen71.191671.1916\dfrac{71.19}{16}1671.19​ = 4.444.442.21\dfrac{4.44}{2.21 }2.214.44​ = 2Carbon26.591226.5912\dfrac{26.59}{12}1226.59​ = 2.212.212.21\dfrac{ 2.21 }{2.21}2.212.21​ = 1Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1Hence, empirical formula is CHO2Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45V.D. = 45Molecular weight = 2 x V.D. = 2 x 45 = 90n=Molecular weightEmpirical formula weight=9045=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{90}{45} = 2n=Empirical formula weightMolecular weight​=4590​=2So, molecular formula = 2(CHO2) = C2H2O4

Question 16

Question 5(iii)State your observations during the process of electrolysis of given compounds.(a)	Fused lead bromide using graphite electrodes.(b)	Copper sulphate solution using copper electrodes.(c)	Acidified water using platinum electrodes.

Accepted Answer

(a) The observations at the anode & at the cathode are:At cathode — silvery grey deposit of lead metal.Reaction at cathode:Pb2+ + 2e- ⟶ PbAt anode — reddish brown fumes of bromine vapours.Reaction at anode:Br1- - 1e- ⟶ BrBr + Br ⟶ Br2(b) The observations at the anode & at the cathode are:At cathode — Brownish pink copper metal is deposited at cathode during electrolysis of copper sulphate.Reaction at cathode:Cu2+ + 2e- ⟶ CuAt anode — Copper ions are formed. Copper anode diminishes in mass.Reaction at anode:Cu - 2e- ⟶ Cu2+The blue colour of copper sulphate solution does not fade.(c) The observations at the anode & at the cathode are:At cathode — Hydrogen gas is liberated at cathode.Reaction at cathode:H+ + e- ⟶ HH + H ⟶ H2At anode — Oxygen gas is liberated at anode.Reaction at anode:OH- ⟶ OH + e-OH + OH ⟶ H2O + OO + O ⟶ O2

Question 17

Question 6(i)For each of the formula of ores given below, what is their common names and role/use.(a)	Al2O3.2H2O(b)	Na3AlF6

Accepted Answer

The common name and use of the ores are given below:S.No.OreCommon NameUseaAl2O3.2H2OBauxiteIt is the main ore from which aluminium is extracted.bNa3AlF6CryoliteLowers the fusion temperature from 2050°C to 950°C and enhances conductivity.

Question 18

Question 6(ii)Calculate.(a)	The vapour density of gas is 11. What would be the volume occupied by 19 g of the gas at STP?(b)	Give the empirical formula of butane.

Accepted Answer

(a) Given,Vapour density = 11Molecular weight = 2 x vapour density = 2 x 11 = 22 gVolume of 22 g of gas = 22.4 L∴ Volume of 19 g of gas = 22.422\dfrac{22.4}{22}2222.4​ x 19 = 19.34 LHence, volume occupied by 19 g of the gas at STP = 19.34 L(b)	Molecular formula of butane is C4H10∴ Ratio of C and H is 4 : 10Simple ratio is 2 : 5Hence, empirical formula = C2H5

Question 19

Question 6(iv)Give reasons(a)	Brass is preferred over copper for door settings.(b) Metals like iron, aluminium, cobalt and nickel become passive (inert) with concentrated nitric acid.(c)	Alloying of gold is done with silver or copper.

Accepted Answer

(a) Brass is preferred over copper for door settings as:Hardness and tensile strength — Brass is stronger than its components, Copper and Zinc.Corrosion resistance — Brass has good corrosion resistance due to the presence of zinc in its composition.(b) When these metals react with concentrated nitric acid, the nitric acid oxidises the metals, forming metal oxides on their surface. The metal oxides, in turn, form a protective layer preventing further reaction with nitric acid.(c) As pure gold is very soft so in order to increase its strength such that jewellery could be made, alloying of gold is done with silver or copper.

Question 20

Question 7(iii)Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:P + 5HNO3 ⟶ H3PO4 + 5NO2 + H2O(a)	What mass of phosphoric acid can be prepared from 3.1 g of phosphorus?(b)	What mass of nitric acid will be consumed at the same time?

Accepted Answer

P+5HNO3⟶H3PO4+H2O+5NO231 g5[(1)+14+3(16)]3(1)+31=5[63]+4(16)=315g=98 g\begin{matrix} 	ext{P} & + & 5	ext{HNO}_3 & \longrightarrow & 	ext{H}_3	ext{PO}_4 & + & 	ext{H}_2	ext{O} & + & 5	ext{NO}_2 \ 31 	ext{ g} & & 5[(1)+14+3(16)] & & 3(1) + 31 \ & &= 5[63] & & + 4(16) \ & & = 315 g & & = 98 	ext{ g} \ \end{matrix}P31 g​+​5HNO3​5[(1)+14+3(16)]=5[63]=315g​⟶​H3​PO4​3(1)+31+4(16)=98 g​+H2​O+5NO2​(a) 31 g of P forms 98 g of phosphoric acid∴ 3.1 g will form 9831\dfrac{ 98}{31}3198​ x 3.1 = 9.8 g.Hence, 9.8 g of phosphoric acid is formed(b) 31 g of P consumes 315 g of nitric acid.∴ 3.1 g will consume = 31531\dfrac{315}{31}31315​ x 3.1 = 31.5 g

Solved Sample Paper 3

Section A

Section B