Solved Sample Paper 3

Section A

Question 1(i)

Reaction of zinc metal with sodium hydroxide will lead to the formation of

1. sodium zincate
2. sodium chloride
3. zinc hydroxide
4. zinc aluminate

Answer

sodium zincate

Reason — Reaction of zinc metal with sodium hydroxide will lead to the formation of sodium zincate

Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

Question 1(ii)

Ammonia can be obtained by adding water to

1. ammonium chloride
2. ammonium nitride
3. magnesium nitride
4. magnesium nitrate

Answer

magnesium nitride

Reason — Magnesium nitride reacts with warm water to liberate ammonia along with magnesium hydroxide.

Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃ [g]

Question 1(iii)

A few drops of universal indicator is added to four colourless solutions, A, B, C and D having pH 2, 10, 9 and 7, respectively. Which of the following test tube is labelled with incorrect colour.

1. A
2. B
3. C
4. D

Answer

B

Reason — In test tube B, the solution has pH = 10 i.e., basic. Hence, on adding universal indicator, it should show dark blue colour instead of green colour.

Question 1(iv)

The reddish brown coloured gas formed when nitric acid reacts with carbon is

1. N₂O₅
2. N₂O₄
3. NO₂
4. N₂O₃

Answer

NO₂

Reason — 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 1(v)

In the electrolysis of alumina, cryolite is added to

1. lower the melting point of alumina
2. decrease the electrical conductivity
3. minimize the anode effect
4. remove impurities from alumina

Answer

lower the melting point of alumina

Reason — Addition of cryolite lowers the fusion point of the mixture i.e., mixture fuses around 950°C instead of 2050°C.

Question 1(vi)

2 mL of ethanoic acid was taken in test tube I and test tube II each. A red litmus paper was introduced in test tube I and a pH paper was introduced in test tube II. The experiment was performed by four students W, X, Y and Z; and they reported their observation as given in the table.

Which students observation is correct?

1. W
2. X
3. Y
4. Z

Answer

Z

Reason — Red litmus paper remains unchanged in ethanoic acid, while pH paper turns pink in acidic solution.

Question 1(vii)

How many lone pair(s) is/are involved in the formation of coordinate bond(s) in nitric acid ?

1. 1
2. 2
3. 3
4. 4

Answer

1

Reason — One lone pair is involved in the formation of coordinate bond in nitric acid.

Question 1(viii)

Empirical formula and molecular formula are related as

1. Empirical formula = (molecular formula)_n
2. Molecular formula = (empirical formula)_n
3. Empirical formula = (molecular formula)ⁿ
4. None of the above

Answer

Molecular formula = (empirical formula)_n

Reason — Relation between molecular formula and empirical formula is defined as:

Molecular formula = n x Empirical formula

Question 1(ix)

Tetraammine copper (II) sulphate dissolves in ammonia solution to give which colour?

1. Greeen
2. Pink
3. Blue
4. Yellow

Answer

Blue

Reason — Tetraammine copper (II) sulfate dissolves in ammonia solution and gives deep blue color

Question 1(x)

The organic compound which undergoes substitution reaction is

1. C₂H₂
2. C₄H₈
3. C₈H₁₄
4. C₅H₁₂

Answer

C₅H₁₂

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

Question 1(xi)

Magnalium is an alloy of

1. aluminium and magnesium
2. aluminium and copper
3. iron
4. copper

Answer

aluminium and magnesium

Reason — Magnalium is an alloy of aluminium (95%) and magnesium (5%).

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 2

1. Only P
2. Q
3. R
4. Both Q and R

Answer

Both Q and R

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

Question 1(xiii)

An anode used during the extraction of aluminium is

1. platinum rods
2. carbon rods
3. zinc sulphate
4. copper sulphate

Answer

carbon rods

Reason — Thick graphite (carbon) rods are used as anode during the extraction of aluminium.

Question 1(xiv)

Which type of fumes are evolved when copper is added to acidified nitrates?

1. NO₂ fumes
2. N₂O fumes
3. N₂ fumes
4. N₂O₃ fumes

Answer

NO₂ fumes

Reason — Dense reddish brown fumes of nitrogen dioxide are evolved when copper is added to acidified nitrates.

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 g

Q. The number of molecules of oxygen present in it is 6.022 x 10²³

1. Only P
2. Only Q
3. Both P and Q
4. Neither P nor Q

Answer

Only P

Reason —

P: As 22.4 L of CO₂ at S.T.P. has mass = 44 g

Therefore, 2.24 L of CO₂ at S.T.P. has mass = $\dfrac{44}{22.4}$ x 2.24 = 4.4 g

44 g of CO₂ has 32 g of oxygen

So, 4.4 g of CO₂ will have = $\dfrac{32}{44}$ x 4.4 = 3.2 g of oxygen.

Hence P is correct.

Q: No. of moles of O₂ in 3.2 g = $\dfrac{3.2}{32}$ = 0.1 g .

1 mole of O₂ = 6.023 x 10²³ molecules

0.1 moles has molecules = 6.023 x 10²³ x 0.1 = 6.023 x 10²²

Hence Q is incorrect.

Question 2(i)

Match the following Column I with column II

Answer

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.

Answer

(a) Reaction at cathode:
Cu²⁺ + 2e^- ⟶ Cu

Reaction at anode:
OH^1- - 1e^- ⟶ OH x 4
4OH ⟶ 2H₂O + O₂

(b) Two observations for this electrolysis are:

1. Pink or reddish brown copper is deposited at the cathode.
2. 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 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)

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 2(iv)

Identify the following.

(a) The acid that is present in vinegar.

(b) A metal which catches fire, when kept in open air.

(c) A sweet smelling compound formed when ethanol reacts with ethanoic acid.

(d) A catalyst used in Haber's process.

(e) An organic compound used for ripening and preservation of fruits.

Answer

(a) Ethanoic acid

(b) Sodium (Na)

(c) Ethyl acetate (Ester)

(d) Finely divided iron (Fe)

(e) Ethene (C₂H₄)

Question 2(v)

(a) Draw the structural formula for the following

1. Methoxymethane
2. 2-methyl propan-2-ol
3. Ethanol

(b) Name the following organic compounds in IUPAC system.

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:

1. 2, 3-dichloropropan-1-oic acid
2. Butan-2-ol

Section B

Question 3(i)

Draw the electron dot structure for the following.

(a) NH₄⁺

(b) CaO

Answer

(a) NH₄⁺

(b) CaO

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.

Answer

(a) Sodium carbonate reacts with sulphuric acid and gives white coloured sodium sulphate along with water and carbon dioxide.

Na₂CO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + CO₂

On 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 + H₂O

On the other hand zinc oxide reacts with sulphuric acid and gives colourless zinc sulphate along with water.

ZnO + H₂SO₄ ⟶ ZnSO₄ + H₂O

Question 3(iii)

Name the ions present in

(a) alumina

(b) fluorspar

(c) sulphuric acid

Answer

(a) alumina [Al₂O₃] — Al³⁺ and O^2-

(b) fluorspar [CaF₂]— Ca²⁺ and F^-

(c) sulphuric acid [H₂SO₄] — H⁺ and SO₄^2-

Question 3(iv)

An element 'Y' is in II period and group 16 of the periodic table.

Answer the following questions:

(a) Is it a metal or a non-metal?

(b) Write the valence electrons present in the element Y.

(c) What is its valency?

Answer

(a) Electronic configuration of Y is [2, 6]. Hence, it is a non-metal.

(b) 6

(c) Valency of Y = 8 - 6 = 2 as it needs 2 more electrons to complete its octet and attain a stable configuration.

Question 4(i)

Define:

(a) Alkanes

(b) Alkenes

Answer

(a) Alkanes — Saturated aliphatic hydrocarbons containing a carbon carbon single bond are called alkanes. The general formula is C_nH_2n+2

(b) Alkenes — Unsaturated aliphatic hydrocarbons containing a carbon carbon double bond are called alkenes. The general formula is C_nH_2n

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?

Answer

Given,

Empirical formula = CH₂O

Hence, empirical formula mass = 12 + 2(1) + 16 = 12 + 2 + 16 = 30 g

Molecular weight = 2 x Vapour density = 2 x 45 = 90

Molecular formula = n x empirical formula

where,

n = $\dfrac{\text{molecular mass}}{\text{empirical formula mass}}$

∴ n = $\dfrac{90}{30}$ = 3

Hence, molecular formula = 3 x CH₂O = C₃H₆O₃

Therefore, the molecular formula of the compound is C₃H₆O₃

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.

Answer

(a) Conditions required for the conversion of ammonia to nitric oxide in presence of air are:

1. Temperature should be around 800°C.
2. Catalyst used is Platinum.
3. Reaction:
   4NH₃ + 5O₂ $\xrightarrow[800 \degree \text{C}]{\text{Pt}.}$ 4NO↑ + 6H₂O + Δ
   2NO + O₂ ⟶ 2NO₂ (brown gas)

(b) Conditions required for the conversion of hydrogen chloride to chlorine are:

1. Temperature should be above 500°C.
2. Reaction:
   2HCl $\xrightarrow{\gt 500 \degree\text{C}}$ H₂ + Cl₂

(c) Conditions required for the conversion of sodium bisulphate to sodium sulphate

1. Temperature should be above 200°C.
2. Reaction:
   NaNO₃ + NaHSO₄ $\xrightarrow{\gt 200 \degree\text{C}}$ Na₂SO₄ + HNO₃

Question 4(iv)

(a) State whether the following statements are True or False. Justify your answer.

1. Weak acids have low electrical conductivity.
2. The concentration of hydrogen ions is equal in strong acids and weak acids.

(b) Calculate the percentage of sodium (Na) in Na₂B₄O₇.10H₂O.

Answer

1. True
   Reason — 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.
2. False
   Reason — 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 g

382 g of Na₂B₄O₇.10H₂O contains 46 g of Na

∴ 100 g of Na₂B₄O₇.10H₂O will contain = $\dfrac{46}{382}$ x 100 = 12.04%

Hence, 12.04% of sodium is present in Na₂B₄O₇.10H₂O

Question 5(i)

An organic compound, whose vapour density is 45, has the following percentage composition

H = 2.22%, O = 71.19% and remaining carbon.

Calculate,

(a) it's empirical formula, and

(b) it's molecular formula

Answer

https://www.knowledgeboat.com/question/an-organic-compound-whose-vapour-density-is-45-has-the--586471767444377600

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, empirical formula is CHO₂

Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{90}{45} = 2$

So, molecular formula = 2(CHO₂) = C₂H₂O₄

Question 5(ii)

Identify the functional group in the following organic compounds.

(a) CH₃CH₂CH₂COCH₃

(b) CH₃CH₂CH₂COOH

Answer

(a) Ketonic group [C=O]

(b) Carboxyl group [COOH]

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.

Answer

(a) The observations at the anode & at the cathode are:

At cathode — silvery grey deposit of lead metal.

Reaction at cathode:
Pb²⁺ + 2e^- ⟶ Pb

At anode — reddish brown fumes of bromine vapours.

Reaction at anode:
Br^1- - 1e^- ⟶ Br
Br + 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^- ⟶ Cu

At 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^- ⟶ H
H + H ⟶ H₂

At anode — Oxygen gas is liberated at anode.

Reaction at anode:
OH^- ⟶ OH + e^-
OH + OH ⟶ H₂O + O
O + O ⟶ O₂

Question 5(iv)

Choose the most appropriate answer from the following list of substances which fits best in the description.

[Silver chloride, ammonium chloride, aqua-regia, explosive, N₂ , NCl₃, volatile]

(a) Silver nitrate reacts with hydrochloric acid to give a white precipitate.

(b) The solution formed by the concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3:1.

(c) The property which does not allow conc. HNO₃ used in the preparation of HCl.

Answer

(a) Silver chloride

(b) Aqua regia

(c) Volatile

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

Answer

The common name and use of the ores are given below:

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.

Answer

(a) Given,

Vapour density = 11

Molecular weight = 2 x vapour density = 2 x 11 = 22 g

Volume of 22 g of gas = 22.4 L

∴ Volume of 19 g of gas = $\dfrac{22.4}{22}$ x 19 = 19.34 L

Hence, 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 : 10

Simple ratio is 2 : 5

Hence, empirical formula = C₂H₅

Question 6(iii)

Raju is going to the laboratory for the preparation of sulphuric acid.

(a) Write the equations for sulphuric acid combining directly with metals and non-metals.

(b) Which concentrated acid will oxidise sulphur directly to sulphuric acid ? Write the equation for the same.

(c) What is the name of the process by which sulphuric acid is manufactured ? Name the catalyst used in the process.

Answer

(a) Reaction with metal :
Fe + H₂SO₄ (dil.) ⟶ FeSO₄ + H₂ [g]

Reaction with non-metal :
C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2SO₂ + 2H₂O

(b) Nitric acid [HNO₃]

S + 6HNO₃ [conc.] ⟶ H₂SO₄ + 2H₂O + 6NO₂

(c) Contact process.

The catalyst used in this process is vanadium pentaoxide (V₂O₅)

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.

Answer

(a) Brass is preferred over copper for door settings as:

1. Hardness and tensile strength — Brass is stronger than its components, Copper and Zinc.
2. 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 7(i)

(a) Give balanced equation for the following.

(1) Sodium hydroxide is added to hydrogen chloride.

(2) Sodium aluminate is treated with water.

(b) If the valency of an element is 2, then to which group will this element belong to?

Answer

(a) Balanced equations:

1. NaOH + HCl ⟶ NaCl + H₂O
2. NaAlO₂ + 2H₂O ⟶ NaOH + Al(OH)₃

(b) Group 2

Reason — The number of valence electrons determine the group of the element. Hence, the element belongs to group 2 as it has 2 electrons in the valence shell.s

Question 7(ii)

Nitric acid cannot be concentrated beyond 68% by the distillation of a dilute solution of HNO₃. Explain why?

Answer

An aqueous solution of nitric acid (68% concentration) forms a constant boiling mixture at 121°C. A constant boiling mixture is one which boils without change in composition.

Hence on boiling further, the mixture evolves out the vapours of both acid and water in the same proportion as in the liquid.

Therefore, dilute HNO₃ cannot be concentrated beyond 68% by boiling.

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?

Answer

$\begin{matrix} \text{P} & + & 5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & \text{H}_2\text{O} & + & 5\text{NO}_2 \\ 31 \text{ g} & & 5[(1)+14+3(16)] & & 3(1) + 31 \\ & &= 5[63] & & + 4(16) \\ & & = 315 g & & = 98 \text{ g} \\ \end{matrix}$

(a) 31 g of P forms 98 g of phosphoric acid

∴ 3.1 g will form $\dfrac{ 98}{31}$ 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 = $\dfrac{315}{31}$ x 3.1 = 31.5 g

Question 8(i)

Identify the name of the salt on the basis of the following description.

(a) A salt prepared by the action of strong acid on metal.

(b) A nitrate which decomposes into its respective metal on heating.

Answer

(a) Magnesium sulphate (MgSO₄)

(b) Silver nitrate (AgNO₃)

Question 8(ii)

Complete and balance the equations.

(a) SO₂ + NaOH ⟶

(b) CuSO₄ + NH₄OH ⟶

Answer

(a) SO₂ + 2NaOH ⟶ Na₂SO₃ + H₂O

(b) CuSO₄ + 2NH₄OH ⟶ Cu(OH)₂ ↓ + (NH₄)₂SO₄

Question 8(iii)

Arrange the following as per the instruction given in the brackets.

(a) C, Pb, Sn, Ge, Si (increasing order of non-metallic character).

(b) Na, Al, Cl (increasing order of ionisation potential).

(c) Zn²⁺, Na⁺, Cu²⁺(order of preference of discharge at the cathode).

Answer

(a) Pb < Sn < Ge < Si < C

(b) Na < Al < Cl

(c) Na⁺ < Zn²⁺ < Cu²⁺

Question 8(iv)

The following questions refer to the periodic table:

(a) What are the elements of group 17 called?

(b) Name the most electronegative element of group 17.

(c) Name the weakest oxidising agent of group 17.

Answer

(a) Halogens

(b) Fluorine

(c) Iodine