1st PUC Chemistry Model Question Paper 2 with Answers

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Karnataka 1st PUC Chemistry Model Question Paper 2 with Answers

Time: 3.15 Hours
Max Marks: 100

Instructions:

1. The question paper has four parts A, B, C and D, All parts are compulsory.
2. Write balanced chemical equations and draw labelled diagram wherever required.
3. Use log tables and simple calculators if necessary (Use of scientific calculators is not allowed].

Part -A 

I. Answer ALL of the following (each question carries One mark): ( 10 × 1 = 10 )

Question 1.
What is S.I unit of density?
Answer:
Kgm^-3

Question 2.
States Charles Law.
Answer:
Charle’s law states that volume of fixed mass of gas is directly proportional to absolute temperature at constant pressure. \(\left(\frac{\mathrm{V}_{1}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{T}_{2}}\right)_{\mathrm{P}}\)

Question 3.
Give reason: H^-1 is a Lewis base.
Answer:
Because of electron pair donor.

Question 4.
Name the most electronegative element.
Answer:
Most electronegative element is fluorine.

Question 5.
What is the oxidation state of‘S in HSO^–₄?
Answer:
+6

Question 6.
Which alkali metal is the strongest reducing agent?
Answer:
Lithium is the strongest reducing agent. It is because of low ionisation energy. They can lose electrons easily, that is why they are strong reducing agents.

Question 7.
A mixture of carbon monoxide and hydrogen is known as
Ans.
Water Gas [CO + H₂].

Question 8.
What is dry ice?
Answer:
Solid CO₂ is called dry ice.

Question 9.
Write the IUPAC name of the following bond line formula.
Answer:

Question 10.
Draw the eclipsed conformation of ethane.
Answer:

Part – B

II. Answer any FIVE of the following questions carrying TWO marks: ( 5 × 2 = 10 )

Question 11.
What is heterogeneous mixture? Give an example.
Answer:
These are the mixtures which have different composition in different part. Here component can be observed with naked eyes and also microscope and they are not uniformly distributed. Ex: Sand and iron fillings, sand and water.

Question 12.
Define viscosity. What is the effect of temperature on it?
Answer:
The internal resistance to the flow of the liquids which one layer offers to another layer trying to pass over is called its viscosity. It depends upon the nature of the liquid and temperature.

Question 13.
Define bond order. Write the electronic configuration of Lithium molecule.
Answer:
It is the number of covalent bonds holding the atoms in the molecule.
Atomic number of lithium is 3 : 1 s² 2s¹

Electronic configuration of lithium molecule is : KK σ2s² σ⁺2S°

Question 14.
Give two uses of sodium hydroxide.
Answer:

1. In volumetric analysis.
2. In paper industry.

Question 15.
What are fullereness? Mention the shape of C₆₀ fullerene.
Answer:
Crystalline allotropic forms of carbon are called fullereness. Soccer ball.

Question 16.
Explain Wurtz reaction with an example.
Answer:
When alkyl halides are heated with sodium metal in ether medium higher alkanes are formed. This reaction is known as Wurtz reaction and employed for the synthesis of higher alkanes containing even number of carbon atoms.

Question 17.
How do you convert the following?
(i)Phenol to benzene (ii) Ethene to Ethane
Answer:

(ii) When a mixture of ethene and hydrogen is passed over heated nickel catalyst at 200°C, ethene is obtained.

Question 18.
What is meant by Biochemical Oxygen Demand? Mention its significance.
Answer:
BOD is Biochemical oxygen demand. It is a measure of the dissolved oxygen that would be needed by the microorganisms to oxidise these compounds.
The significance of BOD are: (a) It can measure the total contamination caused by compounds which can be oxidised in the presence of microorganisms, (b) BOD is a realistic measure of water quality.

Part – C

III. Answer any FIVE of the follow ing questions carrying THREE marks: ( 5 × 3 = 15 )

Question 19.
(a) What are isoelectronic ions? Name the species or ion that will be isoelectronic with Mg²⁺ ions.
Answer:
The ions having same number of electrons but different magnitude of nuclear charge are • called isoelectronic ions.
Mg²⁺ < Na⁺ < F^– < O^2- < N^3-
Mg having highest nuclear charge (12 units) has the smallest size whereas N^3- ion having the smallest nuclear charge (7 units) has the largest size.

(b) Define ionisation enthalpy of an element.
Answer:
The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Ionization energy increases along the period and decreases down the group.

Question 20.
Write any three postulates of Molecular Orbital Theory (MOT).
Answer:

1. The electrons in an atom are found in atomic orbitals, the electrons in a molecule are found in molecular orbitals.
2. The molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper symmetry.
3. The BMO has lower energy and hence greater stability than the corresponding ABMO.
4. The molecular orbitals are filled by electrons in accordance with Aufbau principle, Pauli’s exclusion principle and the Hund’s rule.

Question 21.
Explain SP³ hybridization in methane molecule.
Answer:
CH₄ – Methane Molecule. The Molecular formula of Methane is CH₄
Electronic configuration of C is ground state – 1s²2s²2p²
Electronic configuration of C in excited state – 1s² 2s¹ 2p³
Valance orbital representation –

The valence orbital contains unpaired electron. Elence sp³ hybridized carbon combine with 4 hydrogen atom forms methane molecule.

Question 22.
Write any three differences between Sigma bond and Pi bond.

σ bond :

1. Overlapping is along the axis
2. Overlapping is maximum
3. Due to maximum overlap the bond is strong
4. s-orbital can participate

π bond :

1. Overlapping is on side wise
2. Overlapping is minimum
3. Due to minimum overlap, the band is weak
4. s-orbital cannot participate

Question 23.
Balance the following redox reaction by oxidation number method.

Answer:
Step-1: Write the skeletal equation. The skeletal equation for the given reaction is:
MnO^–₄ (aq) + Br^–(aq) → MnO₂ (s) + BrO₃^– (aq)

Step-2 : Find out the elements which undergo a change in oxidation number (O.N.)

O.N. decreases by 3 per Mn atom ..(i)

Here, O .N. of Br increases from -1 Br^– to + 5 in BrO^–₃ and therefore, Br^– acts as the reductant.
Further, O.N. of Mn decreases from + 7 in MnO^–₄to + 4 in MnO₂, therefore, MnO₄^– acts as the oxidant.

Step-3 : Balance O atoms by adding H₂O molecules.

2MnO^–₄ (s) + Br^– (aq) → 2MnO₂ (s) + BrO^–₃(aq) + H₂O(l)  ….(iv)

Step-4 : Balance H atoms by adding H₂O and OH^– ions since the reaction occurs in basic medium, therefore, add 2₂O to L.H.S. and 2OH^– to R.H.S. of Eq. (iv), we have,

2MnO^–₄ (aq) + Br^– (aq) + H₂O(l) → 2MnO₂ (s) + BrO^–₃(aq) + 2OH^–

Question 24.
(a) What causes the temporary and permanent hardness of water?
Temporary hardness

1. It is caused by bicarbonates of Ca²⁺ and Mg²⁺
2. It is removed by boiling

Permanent hardness

1. It is caused by sulphates and chlorides of Ca²⁺ and Mg²⁺
2. It is removed by adding washing soda

(b) Give the formula of heavy water.
Answer:
D₂O

Question 25.
(a) Compare the hydration of enthalpies and 2nd ionisation enthalpies of alkali and
alkaline earth metals.
Answer:
The hydration enthalpy of alkaline earth metals is larger than alkali metals ions. Therefore . compounds of alkaline earth metals are more extensively hydrated. The second ionisation enthalpies of alkali metals is more than the alkaline earth metals due to inert gas (noble) gas configuration.

(b) Mention the composition of plaster of paris.
Answer:
Plaster of paris is CaSO₄ \(\frac{1}{2}\) H₂O It is prepared by heating gypsum at 373 K.

Question 26.
Write any three difference between graphite and diamond.
Answer

Graphite :

1. Good conductor of electricity
2. Soft and slippery
3. sp² hybridization
4. Planar layer structure

Diamond

1. Bad conductor of electricity
2. Hard andn not slippery
3. sp³ hybridization
4. Tetrahedral structure

Part – D

IV. Answer any FIVE of the following questions carrying FIVE marks: (5 × 5 = 25 )

Question 27.
(a) A compound contains 4.07% hydrogen 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96g mol-1. Find its empirical formula.
Answer:

(b) Express 0.00625 in scientific notation.
Answer:
6.25 x 10^-3

Question 28.
(a) Explain significance of four quantum numbers.
Answer:
(i) Principal Quantum number : Determines energy and size of the orbital.
(n)
(ii) Azimuthal Quantum number : Determines the shape of the orbitals (sub-shells).
(l)
(iii) Magnetic Quantum number : Determines the orientation of the orbitals.
(m)
(iv) Spin Quantum number : Determines the spin motion of the orbitals.
(s)

(b) State Pauli’s Exclusion Principle.
Answer:
Pauli’s Exclusion Principle: No two electrons in the same atom can have all the four quantum numbers same.
Illustration: In 1 s orbital there are two electrons, the set of quantum number of one electron differ with another in spin quantum number.

Question 29.
(a) For an element with atomic number Z = 24.
(i) Write its electronic configuration.
Answer:
Electronic configuration of chromium

(ii) Write the value of n and *P value for its electron in valence shell.
Answer:
n = 4, l = 0

(iii) How many unpaired electrons are present in it?
Answer:
six

(b) What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 ms-1?
Answer:
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.625 \times 10^{34} \mathrm{Js}}{0.1 \times 10 \mathrm{ms}}\) = 6.625 × 10³⁴

Question 30.
(a) Write any four postulates of kinetic theory of gases.
Answer:

1. Gases are made up of large number of the minute particles.
2. Pressure is exerted by a gas.
3. There is no loss of kinetic theory.
4. Molecules of gas attract on one another.

(b) State Avogadro’s law.
Answer:
Equal volume of all gases and vapours under the same condition of temperature and pressure contain equal number of molecules. V ∝ n

Question 31.
(a) What is an extensive property? Give an example.
Answer:
It is a property which depends on the amount of the substance present in the system.
E.g: Mass, Volume, Energy.

(b) Write Gibb’s equation.
Answer:
Gibbs energy is defined as the amount of energy available from a system at a given set of
condition, that can be put into useful work. i,e. G = H – TS OR ∆G = ∆H +T∆S

(c) What is entropy?
Answer:
A measure of degree of disorder of a system.

Question 32.
(a) Calculate the enthalpy of formation of methane from the following data:
C(s) + O₂(g) → CO₂(g)   ∆H° = -393.5 kJ
H₂(S) + \(\frac{1}{2}\)O₂(g) → H₂O(l);    ∆H° = -285.8 kJ
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l);   ∆H° = -890 kJ
Answer:
C(graphite) + O₂(g) → CO₂(g);   ∆_fH° = -393.5 kJ …… (1)
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l);   ∆_fH° = -285.8 kJ …… (2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g);   ∆_fH° = +890 kJ …… (3)

Here as we want one mole of C(graphite) as reactant, we write down equation (1) as it is as. We want two moles of H₂(g) as reactant, we multiply equation (2) by 2. As we want one mole of CH₄ as product, we write it down as it is.

C(graphite) + O₂(g) → CO₂(g);   ∆_fH° = -393.5 kJ …… (1)
2H₂(g) + O₂(g) → 2H₂O(l);   ∆_fH° = 2(-285.8 kJ/mol) =-571.6kJ/mol …… (2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g);   ∆_fH° = +890 kJ …… (3)

Adding 1,2 and 3 we obtain:
C(graphite) + 2H₂(g) → CH₄(g);   ∆_fH° = – 74.8 kJ/mol

(b) What are conditions for spontaneity of the reaction at constant pressure and temperature?
Answer:
The value of Kc is directly related to the Gibbs free energy change (AG) of the reaction.
(i) If ∆G = —ve, then the reaction is spontaneous, and proceeds in the forward reaction.
(ii) If ∆G = +ve, then the reaction is non-spontaneous in the forward direction. The reverse reaction will be spontaneous and thus proceed towards the reactants’ side.
(iii) If ∆G = 0, then the reaction is at equilibrium. No net change takes place.

Question 33.
(a) Explain the effect of temperature on,
2NO₂ → N₂O₄; ∆H = -Ve
Answer:
The forward reaction is exothermic and hence the backward reaction is endothermic. If the temperature is increased at equilibrium, the system tries to decrease the temperature by absorbing heat. The backward reaction which is endothermic gets favoured. Equilibrium is shifted to the left side.
If the temperature is decreased, the system tries to increase the temperature by liberating heat. The forward reaction (exothermic) gets favoured. Equilibrium is shifted to the right side.

(b) State Le-Chatelier’s principle.
Answer:
“When a constraint is applied to a system at equilibrium in a reversible reaction, the equilibrium shifts so as to nullify the constraint” [Constraint is either change in temperature or pressure or addition of reactant].

(c) Give an example of homogeneous equilibrium.
Answer:
N₂(g) + O₂(g) ⇌ 2NO(g)

Question 34.
(a) Define acid and base by Bronsted Lowry concept.
Answer:
A substance or molecule which donates a proton is called acid [Protogenic].
A substance or molecule which accepts a proton is called a base [Protophilic],
Example:

(b) What is buffer solution? Give an example. –
Answer:
It is a solution which resists the pH or pOH when an acid or base is added to it. Buffer solution is a mixture of a weak electrolyte and its salt.

(c) Give an example for solid vapour equilibrium.
Answer:
I_2(s) ⇌ I_2(v)

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 × 5 = 10 )

Question 35.
(a) For the compound CH s C – CH = CH – CH₃
(i) Write the bond line formula.
Answer:

(ii) Mention whether the given compound is saturated or unsaturated.
Answer:
unsaturated

(iii) Identify the number of sigma-bond and pi-bond.
Answer:
10σ2π

(b) What is homologus series? Give an example.
Answer:
A series of structurally related organic compounds containing same functional group and have similar chemical characteristics and the two consecutive members differ in their molecular
formula by -CH₂ -.

Question 36.
(a) Explain the position isomerism with an example.
Answer:
Two or more compounds having same molecular formula but differ in the position of the functional group are called position isomers. This phenomenon is called position isomerism.
Ex: C₃H₈O

(b) How is halogen estimated by Carius method?
Answer:
When an organic compound containing halogen (Cl, Br or 1) is heated in a sealed tube with fuming nitric acid and excess of silver chloride, silver halide is formed from the mass of silver halide obtained, the percentage of the halogen can be calculated.

Procedure: In a hard glass tube (Carius tube), 5 ml of fuming HN03 and 2 to 2.5 g AgN03 are taken. A small narrow weighing tube, containing a small amount (nearly 0.15 -0.2g) of accurately weighed organic compound, is introduced in the Carius tube in such a way that nitric acid does not enter the weighing tube. The Carius tube is now sealed and heated in a furnace at 300°C for about six hours.

The tube is then cooled and its narrow end is cut off and the contents are completely transferred to a beaker by washing with water. The precipitate of silver halide formed is filtered through a weighed sintered glass crucible. It is washed, dried and weighed.

Observation and calculation:

(i) Mass of organic compound taken = w₁g
(ii) Mass of silver halide obtained = w₂g
. (a) For chlorine: \(\underset{143.5 \mathrm{g}}{\mathrm{AgCl}} \equiv \underset{35.5 \mathrm{g}}{\mathrm{Cl}}\)
143.5g of AgCl contains 35.5 g of chlorine
w2g of AgCl will contain \(\frac{35 \cdot 5 \times w_{2}}{143 \cdot 5}\) 0f chlorine
∴ % Cl₂ = \(\frac{35 \cdot 5 \times w_{2}}{143 \cdot 5} \times \frac{100}{w_{1}}\)

(b) For bromine: \(\underset{188 \mathrm{g}}{\mathrm{AgBr}} \equiv \underset{\mathrm{BO}_{\mathrm{g}}}{\mathrm{Br}}\)
188g of AgBr contains 80g of bromine
w₂ g of AgBr will contain \(\frac{980 \times w_{2}}{188} g\) of bromine.

Question 37.
(a) Explain the mechanism of nitration of Benzene.
Answer:
Nitration of benzene reacts with a mixture of concentrated nitric acid and concentrated sulphuric acid at 50°C to form nitrobenzene.

Mechanism: This involves the following steps:

Step 1: Generation of electrophile nitronium ion NO₂⁺
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄^–

Step 2: The electrophile NO₂⁺ attacks the benzene ring to form a carbocation which is resonance stabilized.

Step 3: Loss of a proton to give nitrobenzene. The proton is removed by HSO₄^–.

(b) How do you convert ethyne to benzene?
Answer:
When ethyne is passed over red hot iron tube polymerise to give benzene. This process is called cyclic polymerisation.