Q: Question 1(ix)One atomic mass unit is how much part the mass of C-12 atoms?14\dfrac{1}{4}41112\dfrac{1}{12}12118\dfrac{1}{8}81116\dfrac{1}{16}161

112\dfrac{1}{12}121Reason — Atomic mass unit is defined as 112\dfrac{1}{12}121th the mass of an carbon atom C-12.

Question 1

Question 1(ii)Sodium atoms and sodium ionsare chemically samehave the same number of protonshave an identical number of electronsform covalent bonds

Accepted Answer

have the same number of protonsReason — An atom forms its respective ion by accepting or donating electrons. Hence, number of electrons between an atom and its ion will differ but both will have the same number of protons.

Question 2

Question 1(vi)The organic compound having a triple carbon-carbon covalent bond.C3H4C3H6C3H8C4H10

Accepted Answer

C3H4Reason — C3H4 is an Alkyne as it follows the general formula CnH2n-2 and Alkynes contain a triple carbon-carbon covalent bond.

Question 3

Question 1(ix)One atomic mass unit is how much part the mass of C-12 atoms?14\dfrac{1}{4}41​112\dfrac{1}{12}121​18\dfrac{1}{8}81​116\dfrac{1}{16}161​

Accepted Answer

112\dfrac{1}{12}121​Reason — Atomic mass unit is defined as 112\dfrac{1}{12}121​th the mass of an carbon atom C-12.

Question 4

Question 2(i)(a) Draw the structural formula for the following.(1)	Isobutane(2) Propanal(3) But-2-yne(b)	Name the following organic compounds in IUPAC system.

Accepted Answer

(a) Structural formulae are shown below:(1)	Isobutane(2) Propanal(3) But-2-yne(b)	IUPAC names of the organic compounds are:(1) But-1-ene(2) 1-bromo 2, 3-dichloropropane

Question 5

Question 2(ii)Match the following Column I with Column II.Column IColumn II(a) Neon1. CH3OH(b) Vinegar2. Octet(c) Methanol3. 10 electrons(d) Hydrogen chloride gas4. CH3COOH(e) O2-5. Upward displacement

Accepted Answer

Column IColumn II(a) Neon2. Octet(b) Vinegar4. CH3COOH(c) Methanol1. CH3OH(d) Hydrogen chloride gas5. Upward displacement(e) O2-3. 10 electrons

Question 6

Question 3(i)Identify the anion present in each of the following compounds.(a)	Barium chloride on reacting with compound Z will form a white precipitate.(b)	Compound Y on reaction with sulphuric acid will liberate carbon dioxide along with sodium sulphate and water.

Accepted Answer

(a) SO42- (sulphate)Reason — White precipitate formed is of barium sulphate [BaSO4], hence anion present is SO42-.(b) CO32- (carbonate)Reason — Sodium carbonate will react with sulphuric acid liberate carbon dioxide along with sodium sulphate and water.Na2CO3 + H2SO4 ⟶ CO2 + Na2SO4 + H2O

Question 7

Question 3(iii)(a) State whether the following statements are True or False. Justify your answer.(1)	Fluorine is most electronegative element.(2)	The size of potassium atom is greater than the size of sodium atom.(b) Calculate the number of moles in 9 grams of water.

Accepted Answer

(a) (1) TrueReason — Fluorine is placed in period 2 group 17 (VIIA) in the modern periodic table i.e., it occupies the upper right hand corner of the periodic table. Along a period from left to right electronegativity increases and down a group from top to bottom it decreases. Hence, Fluorine being the topmost and rightmost element in its group and period is the most electronegative element of the periodic table.(2) TrueReason — Atomic size increases on moving down the group. As potassium lies below sodium in its group, hence, size of potassium atom is greater than the size of sodium atom.(b) Molar mass of water [H2O] = 2(1) + 16 = 18Number of moles = Given massMolar mass\dfrac{	ext{Given mass}}{	ext{Molar mass}}Molar massGiven mass​ = 918\dfrac{9}{18}189​ = 0.5 molesHence, 0.5 moles are present in 9 gm of water.

Question 8

Question 3(iv)Fill in the blanks selecting the appropriate word from the given choice.(a)	Formation of ............... (NH4+, NH3) involves ammonia molecule and hydrogen ion.(b)	Hydroxyl ion or hydroxide ion is formed when one ............... [H+, OH-] ion is removed from water molecule.(c)	Hydrogen needs one electron to complete	its	............... [singlet, duplet].

Accepted Answer

(a)	Formation of NH4+ involves ammonia molecule and hydrogen ion.(b)	Hydroxyl ion or hydroxide ion is formed when one H+ ion is removed from water molecule.(c)	Hydrogen needs one electron to complete	its	duplet.

Question 9

Question 4(i)An organic compound containing C, H and O has 49.3% carbon, 6.84% hydrogen, 43.86% oxygen and its vapour density is 73. What will be the molecular formula of the compound?

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioC49.31249.312\dfrac{49.3}{12}1249.3​ = 4.104.102.74\dfrac{4.10}{2.74}2.744.10​ = 1.5 x 2 = 3H6.8416.841\dfrac{6.84}{1}16.84​ = 6.846.842.74\dfrac{6.84}{2.74}2.746.84​ = 2.5 x 2 = 5O43.861643.8616\dfrac{43.86}{16}1643.86​ = 2.742.742.74\dfrac{2.74}{2.74}2.742.74​ = 1 x 2 = 2Simplest ratio of whole numbers = C : H : O = 3 : 5 : 2Hence, empirical formula is C3H5O2Empirical formula weight = 3(12)+ 5(1) + 2(16) = 36 + 5 + 32 = 73Vapour density (V.D.) = 73Molecular weight = 2 x V.D. = 2 x 73 = 146n=Molecular weightEmpirical formula weight=14673=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{146}{73} = 2n=Empirical formula weightMolecular weight​=73146​=2∴ Molecular formula = n[E.F.] = 2[C3H5O2] = C6H10O4

Question 10

Question 4(iii)(a) Mayank	extracts aluminium	by electrolytic	process. Answer the following questions with respect to the electrolytic process in the extraction of aluminium.(1) A compound which is added to lower the fusion temperature of the electrolytic bath.(2)	Explain, why powdered coke is sprinkled over the electrolytic mixture?(b)	The atomic number of an element is 16. To which group will this element belong to?

Accepted Answer

(a) Cryolite [Na3AlF6](b) The layer of powdered coke is sprinkled over the electrolytic mixture as :It prevents burning of carbon electrodes in air at the emergence point from the bath.It minimizes or prevents heat loss by radiation.(c) The element belongs to group 16 [VI A].Reason — The element Z has atomic number 16 and so the electronic configuration will be 2, 8, 6. The number of valence electrons determine the group of the element. Hence, Z will belong to group 16 [VI A] as it has 6 electrons in the valence shell.

Question 11

Question 4(iv)The following is a list of methods for the preparation of salts:A — direct combination of two elementsB — reaction of a dilute acid with a metalC — reaction of a dilute acid with an insoluble baseD — titration of a dilute acid with a solution of soluble base.E — reaction of two solutions of salts to form a precipitate.Choose from the list A to E, the best method of preparing the following salts by giving a suitable equation in each case:Copper (II) SulphateCalcium CarbonateZinc sulphate

Accepted Answer

(a) C — reaction of a dilute acid with an insoluble baseCu(OH)2 + H2SO4 (dil.) ⟶ CuSO4 + 2H2O(b) E — reaction of two solutions of salts to form a precipitateCaCl2 + Na2CO3 ⟶ CaCO3 ↓ + 2NaCl(c) B — reaction of a dilute acid with a metalZn + H2SO4 (dil.) ⟶ ZnSO4 + H2

Question 12

Question 5(ii)Distinguish between the following as directed.(a) Ammonium chloride and sodium chloride on the basis of heating(b) Ferric salt and ferrous salt by using ammonium hydroxide

Accepted Answer

(a) On strong heating, ammonium chloride sublimes to form dense white fumes which condense to white powdery substance on cooler parts of the test tube whereas no white fumes are seen on heating sodium chloride. Hence, the two can be distinguished.NH4Cl →Δ\xrightarrow{\Delta}Δ​ NH3 + HCl(b) On reaction with ammonium hydroxide, ferrous salt forms a dirty green ppt whereas ferric salt forms a reddish brown ppt. Hence, the two can be distinguished easily.

Question 13

Question 5(iii)Name the ions present in each of the following(a) NaCl(b) CH3COOH(c) H2O

Accepted Answer

(a) NaCl — Na+, Cl-(b) CH3COOH — CH3COO-, H+(c) H2O — H+, OH-

Question 14

Question 5(iv)With reference to the first three periods of the modern periodic table, answer the questions given below.(a) What features of the atomic structure accounts for the similarities in the chemical properties of the elements in group VII A of the periodic table?(b) Name the element which has the highest ionisation potential.(c)	What is the atomic number and name of the element present in 3rd period and 17th group of the periodic table?

Accepted Answer

(a) All elements in group VII A have same number of valence electrons (i.e., 7), hence they have similar chemical properties.(b) Helium (He) has highest ionisation potential, as it has fully-filled electronic configuration due to which it requires maximum energy to remove an electron from its valence shell.(c) The element is chlorine (Cl) with atomic number 17.

Question 15

Question 6(i)The following questions are pertaining to various laboratory preparations(a) Which property of hydrogen chloride is demonstrated by fountain experiment?(b)	Why hydrogen chloride gas is not collected over water?(c)	What is the source of nitrogen gas used in Haber's process?

Accepted Answer

(a) The high solubility of HCl gas in water is demonstrated by the Fountain experiment.(b) Hydrogen chloride gas is not collected over water since it is highly soluble in water.(c) Nitrogen is obtained by fractional distillation of liquid air.

Question 16

Question 6(ii)Explain the following:(a) Zinc oxide can be reduced to zinc by using carbon monoxide, but aluminium oxide cannot be reduced by reducing a agent.(b) During electrolysis of molten lead bromide, a graphite anode is preferred to other electrodes.(c)	Electrolysis of molten lead bromide is considered to be a redox reaction.

Accepted Answer

(a) Aluminium is placed above zinc in reactivity series. Hence, aluminium is more reactive than zinc and its oxide is stable and cannot be reduced by a reducing agent. Zinc being less reactive metal, its oxide is less stable and so it can be easily reduced by carbon monoxide.(b) As graphite is unaffected by the reactive bromine vapours released at the anode hence, a graphite anode is preferred during the electrolysis of molten lead bromide(c) Electrolysis of molten lead bromide involves oxidation and reduction reactions and hence is a redox reaction.Reduction reaction at cathode:Pb2+ + 2e- ⟶ PbOxidation reaction at the anode:Br1- - 1e2- ⟶ BrBr + Br ⟶ Br2

Question 17

Question 6(iii)4.2 grams of magnesium carbonate is decomposed by heating according to the question.MgCO3 ⟶ MgO + CO2Calculate the following(a)	Volume of carbon dioxide obtained at STP.(b) Mass of MgO formed[Atomic weight : Mg = 24, C = 12, O = 16]

Accepted Answer

MgCO3⟶MgO+CO224+12+3(16) g=24+16[12+2(16)]=84=4012+3244 g\begin{matrix} 	ext{MgCO}_3 & \longrightarrow & 	ext{MgO} & + & 	ext{CO}_2 \ 24 + 12 + 3(16) 	ext{ g} & & = 24 + 16 & & [ 12 + 2(16)] \ = 84 & & = 40 & & 12 + 32 \ & & & & 44 	ext{ g} \ \end{matrix}MgCO3​24+12+3(16) g=84​⟶​MgO=24+16=40​+​CO2​[12+2(16)]12+3244 g​(a) Number of moles of MgCO3 = Given weightmolecular weight\dfrac{	ext{Given weight}}{	ext{molecular weight}}molecular weightGiven weight​ = 4.284\dfrac{4.2}{84}844.2​ = 0.05 moles1 mole of MgCO3 gives 1 mole of CO2Hence, 0.005 mole of MgCO3 gives 0.05 mole of CO2(a) Volume of carbon dioxide at STP = no. of moles of carbon dioxide x volume in litres= 0.05 x 22.4 = 1.12 LHence, volume of carbon dioxide obtained at STP = 1.12 L(b) 84 grams of MgCO3 produce 40 grams of MgO∴ 4.2 g will produce = 4084\dfrac{40}{84}8440​ x 4.2 = 2 gHence, mass of MgO formed = 2g

Question 18

Question 7(iii)Give balanced chemical equation for the following.(a)	Preparation of ethyne from (ethylene dibromide) 1, 2-di bromoethane.(b)	Ethanoic acid from ethanol.(c)	Sodium acetate from acetic acid.

Accepted Answer

(a) Ethyne from (ethylene dibromide) 1, 2-di bromoethane.C2H4Br2 ethylene dibromide+2KOH [aq.] alcoholic KOH→boilC2H2 Ethene [ethylene]+2KBr+2H2O\underset{	ext{ ethylene dibromide}}{	ext{C}_2	ext{H}_4	ext{Br}_2} + \underset{	ext{ alcoholic KOH}}{	ext{2KOH [aq.]}} \xrightarrow{	ext{boil}} \underset{	ext{ Ethene [ethylene]} }{	ext{C}_2	ext{H}_2} + 	ext{2KBr} +	ext{2H}_2	ext{O} ethylene dibromideC2​H4​Br2​​+ alcoholic KOH2KOH [aq.]​boil​ Ethene [ethylene]C2​H2​​+2KBr+2H2​O(b) Ethanoic acid from ethanol.C2H5OHethanol→K2Cr2O7[acidified][O]CH3CHOethanal→K2Cr2O7[acidified][O]CH3COOHethanoic acid\underset{ 	ext{ethanol}}{	ext{C}_2	ext{H}_5	ext{OH}} \xrightarrow[\underset{	ext{[acidified]}}{	ext{K}_2	ext{Cr}_2	ext{O}_7}]{	ext{[O]}} \underset{	ext{ethanal}}{	ext{CH}_3	ext{CHO}} \xrightarrow[\underset{	ext{[acidified]}}{	ext{K}_2	ext{Cr}_2	ext{O}_7}]{	ext{[O]}} \underset{ 	ext{ethanoic acid} }{	ext{CH}_3	ext{COOH}}ethanolC2​H5​OH​[O][acidified]K2​Cr2​O7​​​ethanalCH3​CHO​[O][acidified]K2​Cr2​O7​​​ethanoic acidCH3​COOH​(c) Sodium acetate from acetic acid.CH3COOH + NaOH ⟶ CH3COONa + H2O

Question 19

Question 7(iv)Give one relevant observation for each of the following reactions.(a)	Zinc carbonate is heated.(b)	Excess of ammonia is passed through an aqueous solution of lead nitrate.(c)	Sodium hydrogen carbonate is added to nitric acid.

Accepted Answer

(a) On heating, zinc carbonate, which is a light amorphous white solid, changes to pale yellow and gives off a colourless and odourless gas that turns lime water milky. The milkiness disappears on passing excess of gas. This proves that the gas evolved is carbon dioxide.ZnCO3 →Δ\xrightarrow{\Delta}Δ​ ZnO + CO2 ↑(b) Chalky white precipitate of lead hydroxide is formed which is insoluble in excess of ammonium.Pb(NO3)2 + 2NH4OH ⟶ 2NH4NO3 + Pb(OH)2 ↓(c) When sodium hydrogen carbonate is added to nitric acid, a colourless and odourless gas is evolved that turns lime water milky. The milkiness disappears on passing excess of gas. Hence, the gas evolved is carbon dioxide.NaHCO3 + HNO3 ⟶ NaNO3 + H2O + CO2 ↑

Question 20

Question 8(i)The equation,4NH3 + 5O2 ⟶ 4NO + 6H2Orepresents the catalytic oxidation of ammonia. If 100 cm3 of ammonia is used, calculate the volume of oxygen required to oxidise the ammonia completely.

Accepted Answer

[By Lussac's law]4NH3+5O2⟶4NO+6H2O4 vol.:5 vol.⟶4 vol.\begin{matrix} 4	ext{NH}_3 & + & 5	ext{O}_2 &\longrightarrow & 4	ext{NO} & + & 6	ext{H}_2	ext{O} \ 4 	ext{ vol.} & : & 5 	ext{ vol.} & \longrightarrow & 4	ext{ vol.} \end{matrix}4NH3​4 vol.​+:​5O2​5 vol.​⟶⟶​4NO4 vol.​+6H2​OTo calculate the volume of oxygen required :NH3:O24 vol.:5 vol.100 cm.3:x\begin{matrix}	ext{NH}_3 & : & 	ext{O}_2 \ 4 	ext{ vol.} & : & 5 	ext{ vol.} \ 100 	ext{ cm.}^3 & : & 	ext{x} \end{matrix}NH3​4 vol.100 cm.3​:::​O2​5 vol.x​∴x=54×100=125 cm3	herefore x = \dfrac{5}{4} 	imes 100 = 125 	ext{ cm}^3∴x=45​×100=125 cm3Hence, volume of oxygen required = 125 cm3

Solved Sample Paper 4

Section A

Section B