Question 1

Question 1(i)An aqueous solution of copper sulphate turns colourless on electrolysis. Which of the following could be the electrodes?P. anode: copper; cathode: copperQ. anode: platinum; cathode: copperR. anode: copper; cathode: platinumonly Ponly Qonly Rboth Q and R

Accepted Answer

only QReason — If platinum anode is used the blue colour of CuSO4 solution fades since the blue Cu2+ ions which are discharged at the cathode are not replaced at the anode.

Question 2

Question 1(ii)A compound P is heated in a test tube with sodium hydroxide solution. A red litmus paper held at the mouth of the test tube turns blue. Which of the following could compound P be?zinc sulphatecopper sulphateferrous sulphateammonium sulphate

Accepted Answer

ammonium sulphateReason — Ammonium sulphate reacts with sodium hydroxide to produce sodium sulphate, ammonia gas, and water. The red litmus turns blue due to the basic ammonia gas.

Question 3

Question 1(iii)The atomic masses of sulphur (S), oxygen (O), and helium (He) are approximately 32, 16, and 4 respectively.Which of the following statements regarding the number of atoms in 32 g of sulphur, 16 g of oxygen, and 4 g of helium is correct?P. 16 g of oxygen contains four times the number of atoms as 4 g of helium.Q. 16 g of oxygen contains half the number of atoms as 32 g of sulphur.only Ponly Qboth P and Qneither P nor Q

Accepted Answer

neither P nor QReason —Number of atoms = MassMolecular Mass\dfrac{	ext{Mass}}{	ext{Molecular Mass}}Molecular MassMass​ x NAwhere NA is the Avogadro's number = 6.023 x 1023Number of atoms in 32 g of sulphur (s) = 3232\dfrac{32}{32}3232​ x NA = NANumber of atoms in 16 g of oxygen (O) = 1616\dfrac{16}{16}1616​ x NA = NANumber of atoms in 4 g of helium (He) = 44\dfrac{4}{4}44​ x NA = NAHence, all have same number of atoms which is = 6.023 x 1023

Question 4

Question 1(v)Glucose reacts with concentrated sulphuric acid to give a very pure form of carbon called sugar charcoal.The reaction taking place is:oxidationcombustiondehydrationcombination

Accepted Answer

dehydrationReason — Concentrated sulphuric acid is a strong dehydrating agent due to its strong affinity for water, hence, the reaction taking place is a dehydration reaction.C6H12O6 →Conc. H2SO4\xrightarrow {	ext{Conc. H}_2	ext{SO}_4}Conc. H2​SO4​​ 6C + 6H2O

Question 5

Question 1(viii)A ⟶ A+3; B ⟶ B-1Number of electrons present in the outermost shell of atoms A and B respectively are:5, 13, 13, 75, 7

Accepted Answer

3, 7Reason — A has donated three electrons from its outer most shell and has +3 charge. B had 7 electrons and has taken one electrons and has -1 charge.

Question 6

Question 1(ix)A ............... solution is observed after placing Magnesium metal in a solution of Copper sulphate for half an hour.BlueColourlessReddish brownDirty green

Accepted Answer

ColourlessReason — According to the reactivity series of metals, magnesium is more reactive compared to copper. Hence, it replaces copper from CuSO4 and the blue colour fades as colourless magnesium sulphate is formed and brown bits of copper metal form a precipitate:Mg + CuSO4 ⟶ MgSO4 + Cu

Question 7

Question 1(x)An element with atomic no. ............... will form an acidic oxide.3171113

Accepted Answer

17Reason — Element with atomic number 17 has electronic configuration [2, 8, 7] and hence has 7 electrons in the valence shell. It is a non-metal and will form an acidic oxide.

Question 8

Question 1(xiii)The process of electrolysis is an example of:Oxidation reactionReduction reactionRedox reactionDisplacement reaction

Accepted Answer

Redox reactionReason — During electrolysis, the reaction at the cathode involves reduction of cations as they gain electrons to become neutral atoms while that at anode involves oxidation of anions as they lose electrons to become neutral. As redox reactions are reactions where oxidation and reduction takes place simultaneously. Hence, electrolysis is an example of redox reaction.

Question 9

Question 1(xiv)The catalyst used in Ostwald’s process is ............... .Finely divided ironGraphiteVanadium pentoxidePlatinum

Accepted Answer

The catalyst used in Ostwald’s process is platinum.Reason — Ostwald’s process :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 10

Question 1(xv)An element belongs to third period and sixteenth group. It will have ............... electrons in its valence shell.2563

Accepted Answer

6Reason — The element in third period and sixteenth group will have atomic number 16 and electronic configuration 2, 8, 6. Hence, it has 6 valence electrons.

Question 11

Question 2(i)The setup shown below is that of the fountain experiment with hydrogen chloride gas in the flask.The fountain starts when a few drops of water from the dropper are introduced into the flask. Instead of the drops of water, Pooja started the fountain by introducing a few drops of Sodium hydroxide into the flask.(a) Explain why the litmus solution gets sucked up when Sodium hydroxide is used.(b) What will be the colour of the fountain when Sodium hydroxide is used? Justify your answer.(c) If instead of HCl gas, ammonia gas is filled in the flask and water is introduced from the dropper, will there be a different observation? Justify your answer.

Accepted Answer

When sodium hydroxide is introduced into the flask, it undergoes a neutralization reaction with HCl to form salt and water:HCl + NaOH ⟶ H2O + NaClAs HCl is used up in the reaction, this lowers the pressure inside the flask. The outside pressure being higher pushes the litmus solution inside, through the jet tube.(b) The colour of the litmus solution will remain unchanged so the fountain will be blue in colour.HCl and sodium hydroxide undergo neutralization reaction to form salt and water which are neutral products. Hence, colour of litmus solution will not change.(c) No the observation will be the same. Ammonia gas and water produce ammonium hydroxide which is basic in nature, hence again a blue colour fountain will be seen.

Question 12

Question 2(ii)Match the following Column A with Column B.Column AColumn B(a) Aluminium1. Covalent compound(b) Sulphuric acid2. Carbonate ore(c) Calcination3. Hall Heroult's process(d) Calcium Chloride4. Contact Process(e) Carbon tetrachloride5. Electrovalent compound

Accepted Answer

Column AColumn B(a) Aluminium3. Hall Heroult's process(b) Sulphuric acid4. Contact Process(c) Calcination2. Carbonate ore(d) Calcium Chloride5. Electrovalent compound(e) Carbon tetrachloride1. Covalent compound

Question 13

Question 2(iii)Complete the following by choosing the correct answers from the bracket:(a) If an element has one electron in the outermost shell then it is likely to have the ................ [smallest/ largest] atomic size amongst all the elements in the same period.(b) ................ [sulphuric acid/ hydrochloric acid] does not form an acid salt.(c) A ................ [reddish brown/ dirty green] coloured precipitate is formed when ammonium hydroxide is added to a solution of ferric chloride.(d) Alkanes undergo ................ [addition/ substitution] reactions.(e) An ................ [alkaline/acidic] solution will turn methyl orange solution pink.

Accepted Answer

(a) If an element has one electron in the outermost shell then it is likely to have the largest atomic size amongst all the elements in the same period.(b) Hydrochloric acid does not form an acid salt.(c) A reddish brown coloured precipitate is formed when ammonium hydroxide is added to a solution of ferric chloride.(d) Alkanes undergo substitution reactions.(e) An acidic solution will turn methyl orange solution pink.

Question 14

Question 2(v)(a) Draw the structural diagram for the following compounds:1-propanal1, 2 dichloro ethaneBut-2-ene(b) Give the IUPAC name of the following organic compounds:

Accepted Answer

(a) Structural diagrams are as follows:1. 1- propanal2. 1, 2 dichloro ethane3. But-2-ene(b) IUPAC name are as follows:1-propanol1-pentene

Question 15

Question 3(iv)(a) State wether the following statements are TRUE or FALSE. Justify your answer.In an electrovalent compound, the cation attains the electronic configuration of the noble gas that comes after it in the periodic table.In the formation of a compound PQ2, atom P gives one electron to each atom of Q. The compound PQ2 is a good conductor of electricity.(b) Calculate the number of moles in 22 grams of carbon dioxide.

Accepted Answer

(a)FalseIn an electrovalent compound, the cation attains the electronic configuration of the noble gas that comes before it in the periodic table.TrueIn the formation of a compound PQ2, atom P gives one electron to each atom of Q. The compound PQ2 is a good conductor of electricity in molten or aqueous solution state.(b) Gram molecular mass of CO2 = 12 + 2(16)= 12 + 32 = 44 gAs,44 g of CO2 = 1 mole∴ 22 g of CO2 = 144\dfrac{1}{44}441​ x 22 = 0.5 molesHence, number of moles in 22 grams of carbon dioxide = 0.5

Question 16

Question 4(ii)A gas cylinder of capacity 40 dm3 is filled with gas X the mass of which is 20 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of hydrogen is 2 g. Find the relative molecular mass of the gas.

Accepted Answer

Vapour density of gas =Wt. of certain volume of gas Wt. of same volume of H2=202=10 g\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \[0.5em] = \dfrac{20}{2} \[0.5em] = 10 \text{ g}Wt. of same volume of H2​Wt. of certain volume of gas ​=220​=10 gMolecular weight = 2 x Vapour density= 2 x 10 = 20 gHence, relative molecular mass of gas is 20 g

Question 17

Question 4(iii)Give balanced equations for each of the following:(a) Action of warm water on Aluminium nitride.(b) Oxidation of carbon with conc. Nitric acid.(c) Dehydration of ethanol by conc. Sulphuric acid at a temperature of 170°C.

Accepted Answer

(a) AlN + 3H2O ⟶ Al(OH)3 + NH3 [g](b) C + 4HNO3 ⟶ CO2 + 2H2O + 4NO2(c) C2H5OH →170°CConc. H2SO4\xrightarrow[170\degree	ext{C}]{	ext{Conc. H}_2	ext{SO}_4}Conc. H2​SO4​170°C​ C2H4↑ + H2O

Question 18

Question 5(iii)Seema takes a blue crystalline salt P in a test tube. On heating it produces a white anhydrous powder. P is dissolved in water. Zinc is added to one part of the solution and to another part of the solution Barium chloride is added.(a) Name the compound P.(b) Mention one observation when zinc is added to the solution of P.(c) State the colour of the precipitate formed when barium chloride is added to the solution of P.

Accepted Answer

(a) P is blue vitriol [CuSO4.5H2O](b) Blue colour slowly fades as ZnSO4 is formed.(c) White ppt. of BaSO4 is obtained.CuSO4 + BaCl2 ⟶ BaSO4 ↓ + CuCl2

Question 19

Question 5(iv)Give reasons:(a) Ethene undergoes addition reaction.(b) Hydrocarbons can be used as fuels.(c) Hydrogen chloride gas cannot be collected over water.

Accepted Answer

(a) In case of ethene, the valencies of atleast 2 carbon atoms are not fully satisfied by hydrogen atoms. The availability of electrons in the double bond makes them more reactive and hence they undergo addition reactions only.(b) Hydrocarbons are highly combustible and they produce good amount of heat hence they are used as a fuel.(c) Hydrogen chloride gas is not collected over water since it is highly soluble in water.

Question 20

Question 7(iii)8.2 grams of calcium nitrate is decomposed by heating according to the equation2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2Calculate the following:(a) Volume of nitrogen dioxide obtained at STP(b) Mass of CaO formed[Atomic weights: Ca –40 , N—14, O—16]

Accepted Answer

2Ca(NO3)2⟶2CaO+4NO2+O22[(40)+2(404[22.4]2(14+3(16))]+16) lit.=328 g=112g\begin{matrix} 2	ext{Ca}(	ext{NO}_3)_2 & \longrightarrow & 2	ext{CaO} & + & 4	ext{NO}_2 & + & 	ext{O}_2 \ 2[(40) + & & 2(40 & & 4[22.4] \ 2(14 +3(16))] & & + 16) & & 	ext{ lit.} \ = 328 	ext{ g} & & = 112 	ext{g} \ \end{matrix}2Ca(NO3​)2​2[(40)+2(14+3(16))]=328 g​⟶​2CaO2(40+16)=112g​+​4NO2​4[22.4] lit.​+O2​(i) 328 g of Ca(NO3)2 produces 4(22.4) lit of nitrogen dioxide.∴ 8.2 g of Ca(NO3)2 will produce 4×22.4328\dfrac{4 	imes 22.4}{328}3284×22.4​ x 8.2 = 2.24 lit of nitrogen dioxide.Hence, vol of nitrogen dioxide evolved = 2.24 lit.(ii) 328 g of Ca(NO3)2 produces 112 g of calcium oxide.∴ 8.2 g of Ca(NO3)2 will produce 112328\dfrac{112}{328}328112​ x 8.2 = 2.8 g of calcium oxide.Hence, 2.8 g of calcium oxide is produced.

Solved 2024 Specimen Paper ICSE Class 10 Chemistry

Section A 40 Marks

Section B 40 Marks