1st PUC

Students can Download 1st PUC Physics Model Question Paper 1 with Answers, Karnataka 1st PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Model Question Paper 1 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. All parts are compulsory.
2. Draw relevant figure / diagram wherever necessary.
3. Numerical problems should be solved with relevant formulae.

Part – A

I. Answer the following questions: ( 10 x 1 = 10 )

Question 1.
Write the dimensional formula for Pressure.
Answer:
[pressure] = [M¹ C^-1T^-2]

Question 2.
When does the circular motion become uniform?
Answer:
The circular motion of bodies will be uniform, for uniform speed of particles.

Question 3.
What is the amount of work done by the gravitational force on the body moving on the horizontal plane?
Answer:
Zero

Question 4.
What is elastic collision?
Answer:
A collision in which linear momentum and kinetic energy of the interacting system are conserved, is known as an elastic collision.

Question 5.
Give one example for elastomers.
Answer:
Rubber

Question 6.
What is thermal stress?
Answer:
When thermal expansion is not allowed in a solid, the corresponding stress developed in it is known as thermal stress. The solid acquires a compressive strain due to the external forces, provided by (say) the rigid support at one end of it.

Question 7.
Define mean free path of the molecule in the gas.
Answer:
The average distance traveled by the molecule during two consecutive collisions is called mean free path.

Question 8.
Where does Kinetic energy of the oscillating particle become maximum?
Answer:
The K.E of oscillating particle becomes maximum when it reaches the mean position.

Question 9.
What is resonance?
Answer:
If the frequency of forced oscillation of a system is the same as the natural frequency of the system which induces oscillation in the other, then both the systems are said to be in resonance.

Question 10.
Give the value of phase difference between the particles of adjacent loop of stationary wave.
Answer:
Φ = 180°

Part – B

II. Answer any five of the following questions:  ( 5 x 2 = 10 )

Question 11.
Name any two fundamental forces of nature.
Answer:

• Gravitational force
• Electro – weak force
• Strong nuclear force

Question 12.
Mention any two sources of systematic error.
Answer:

• Instrumental error.
• Personal error.

Question 13.
Draw position time graph for the particle having zero acceleration.
Answer:

Question 14.
Write the equation for maximum horizontal range in projectile motion and explain the terms.
Answer:
R = \(\frac{u^{2} \sin 2 \theta}{g}\)
Where u – initial speed,
θ – angle of projection
g – acceleration due to gravity
R – Horizontal range

Question 15.
Mention any two methods of reducing friction.
Answer:

1. Wear and tear can be minimised by polishing the surfaces and lubricating machinery parts with oil and grease etc.
2. Streamlining the bodies reduce fluid (air or water) resistance.

Question 16.
Mention the conditions for the body in mechanical equilibrium.
Answer:
(1) The vector sum of the forces on the rigid body is zero for translatory equilibrium
\(\sum_{i=1}^{n} \vec{F}_{i}=0\) and (acceleration a = 0 )
(2) The vector sum of the torques on the rigid body is zero for rotatory equilibrium (a=0) i.e.

i.e. the components of X, Y and Z independently vanish to zero for linear equilibrium.

Sum of X components, Y components and Z components of torque on the particles, vanish for rotational equilibrium.

Question 17.
State and explain Hooke’s law.
Answer:
Statement: The ratio of stress to strain is a constant for a material within the elastic limit.
Modulus of elasticity = \(\frac{\text { Stress }}{\text { Strain }}\)

Within the elastic limit, stress v/s strain is a straight line ‘A’ is the elastic limit upto which Hooke’s law is applicable. Beyond ‘B’ the yielding point, the wire extends but does not return to the initial state when the deforming force is removed. ‘F’ is the breaking point. ‘EF’ allows the material to be malleable and ‘DE’, ductile.

Question 18.
What are extensive thermodynamical variables? Give one example.
Answer:
Internal energy, volume, mass are called extensive thermodynamic variables.

Part – C

III. Answer any FIVE of the following questions : ( 5 x 3 = 15 )

Question 19.
Obtain.the equation for maximum height in projectile motion.
Answer:
We know that (v_ocosθ) remains constant.
Instantaneous distance along the horizontal is given by ,
x = (v_ocosθ)t
for x = R (range of the projectile), t = T = time of flight.

Question 20.
State and prove the law of conservation of linear momentum.
Answer:
Statement: In an isolated system of collision of bodies, the total linear momentum before impact is equal to the total linear momentum after impact.

Let m₁ and m₂ be the masses of two bodies moving along \(\vec{v}_{1 i}\) and \(\vec{v}_{2 i}\). Let \(\vec{v}_{1 f}\) and be the \(\vec{v}_{2 f}\) be final velocities after the impact.
At the time of impact the force of action acts on the body B and the force of reaction acts on A.
Applying Newton’s III law of motion
\( |Force of action on \mathrm{B}|=-| Force of reaction on \mathrm{A} |\)

This shows that the total final linear momentum of the isolated system equals its total initial momentum.

Question 21.
Prove the work energy theorem in the case of constant force.
Answer:
w.k.t work done W = \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}\)
where F = ma, s = \(\frac{v^{2}-u^{2}}{2 a}\)
W = \(\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}\) becomes
W = \(\frac{m\left(v^{2}-u^{2}\right)}{2}\)
or W = K_f – K_i
or W = Δ K.E
Thus, work come by a constant force is equal to the difference in the K.E of the body.

Question 22.
Obtain the equation for moment of couple.
Answer:
Two equal and opposite forces acting at two different points constitute couple. The moment of couple is measured by taking the product of magnitude of any one force and the arm of the couple.

Hence moment of couple = \(\overrightarrow{\mathrm{F}}|\overrightarrow{\mathrm{OA}}|+\overrightarrow{\mathrm{F}}|\overrightarrow{\mathrm{OB}}|=\overrightarrow{\mathrm{F}}(\mathrm{OA}+\mathrm{OB})=\overrightarrow{\mathrm{FAB}}\)

Question 23.
Derive equation for acceleration clue to gravity on the surface of the earth.
Answer:
Let M be the mass of the Earth and R be its radius.
We know that F = \(-\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
where ‘m’ is a mass body very close to the surface of the Earth.
Force of gravity on the mass ‘m’ is F’ = mg
However F = F’
Therefore mg = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
i.e, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 24.
Show that volume co-efficient of expansion of ideal gas is inversely proportional to absolute temperature.
Answer:
At constant pressure, V ∝ T
i.e., V = KT. For an increase in the temperature ∆T, the corresponding increase in the volume of the gas will be ∆W.
i.e., (V + ∆V) = K(T + ∆T)
i.e., ,V + ∆V = KT+ K∆T
i.e., \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\mathrm{K}=\frac{\mathrm{V}}{\mathrm{T}}\)
or
\(\frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}=\frac{1}{\mathrm{T}}=\gamma\)
where r is volume co-efficient of an ideal gas

Question 25.
Show that C_v = \(\frac{3}{2} \mathbf{R}\) for mono atomic gas.
Answer:
(dQ) = du = C_v dT
ie. C_v dT = \(\frac{f \mathrm{R} d \mathrm{T}}{2}\)
∴ C_v = \(\frac{f R}{2}\) for mono atonic gas, f = 3
C_v = \(\frac{3 \mathrm{R}}{2}\)

Question 26.
Give any three differences between progressive wave and stationary wave.
Answer:

• Progressive waves transfer energy, whereas stationary waves do not.
• No particles of the medium are in a state of rest in a PW but in a SW, the particles at nodes will be at rest.
• The particle velocities will remain the same for a given displacement in a PW, whereas in a SW, particles at nodes will have zero speed and at anti-nodes, particles have maximum speed.
• y = f(x, t) for a PW and y = f(x) g(t) for a SW.
• Distance between two points at which particles have same state of vibration is called , wavelength in a PW whereas distance between two consecutive nodes or anti-nodes is half the wavelength.

Part – D

IV. Answer any TWO of the following questions : ( 2 x 5 = 10 )

Question 27.
Derive equation for centripetal acceleration.
Answer:
Let \(\vec{r}\) and \(\vec{r}^{\prime}\) be the position vectors and \(\vec{v}\) and \(\vec{v}^{\prime}\) veIocities of the object when it is at point P and P’. By definition, velocity at a point is along the tangent at that point in the direction of the motion. Since the path is circular \(\vec{v}\) is perpendicular to \(\vec{r}\) and \(\vec{v}^{\prime}\) is perpendicular to \(\vec{r}^{\prime}\).
Therefore, ∆\(\vec{v}\)is perpendicular to ∆\(\vec{r}\) Average acceleration \(\frac{\Delta \vec{v}}{\Delta t}\) is perpendicular to ∆\(\vec{r}\).
The magnitude of \(\vec{a}\) is, by definition, given by \(|\vec{a}|=\lim _{\Delta t \rightarrow 0} \frac{|\Delta \vec{v}|}{\Delta t}\)
The triangle formed by the position vectors is similary to the triangle formed by the velocity vectors.

lf ∆t is very small. ∆ will also small. The arc PP is approximately equal to \(|\Delta \vec{r}|\)
i.e., \(\lim _{\Delta t \rightarrow 0} \frac{|\Delta \vec{r}|}{\Delta t}\) = v. Thus, centripetal acceleration \(|\vec{a}|=\frac{v}{r} v=\frac{v^{2}}{r}\) and \(\vec{a}=\frac{v}{r} \frac{d \vec{r}}{d t}\).
The centripetal acceleration is always directed towards the centre. The centripetal force = ma.

Question 28.
Prove “law of conservation of mechanical energy” for freely falling body.
Answer:
Energy can neither be created nor destroyed but can be transformed into other forms of energy.
Let a particle be at ‘C’. Initially the particle w ill be zero velocity. Let ‘h’ be the height of the particle.

P.E_c = mgh, K.E_c = 0
T.E = mgh
At B, K.E = \(1 / 2 \mathrm{mv}_{\mathrm{B}}=1 / 2 \mathrm{m}(2 \mathrm{g} x)=\mathrm{mg} x\)
P.E_B = mg (h – x)
Hence T.E = (mgh – mgx) + (mgx)
= mgh
At A, v_A²= 2gh
So that K.E = \(\frac{1}{2}\) m2gh = mgh
P.E_A = O.
Hence at all points A, B and C, the total energy is conserved.

Question 29.
Obtain equation for Kinetic energy of rolling motion.
Answer:
K.E of rolling body = Transiatory K.E + Rotational K.E

V. Answer any TWO of the following questions : ( 2 x 5 = 10 )

Question 30.
State and prove Bernaulli’s theorem.
Answer:
Along a steam line, in a steady flow of non viscous fluid, potential energy, kinetic energy and pressure energy remain constant.
i.e., \(\mathrm{mgh}+1 / 2 \mathrm{mv}^{2}+\mathrm{PV}=\mathrm{K}\)
i.e., \(\mathrm{mgh}+1 / 2 \mathrm{mv}^{2}+\mathrm{Pm} / \rho=\mathrm{K} \div \mathrm{m}\)
we get, \(g h+\frac{v^{2}}{2}+\frac{P}{\rho}=K \div g\)
i.e., \(\mathrm{h}+\frac{v^{2}}{2 g}+\frac{\mathrm{P}}{\rho g}=\mathrm{K}\)
where h is called gravitational head, \(\frac{v^{2}}{2 g}\) is velocity head and \(\frac{P}{\rho g}\) is pressure head.
For a horizontal Flow, \(\frac{\rho v^{2}}{2}+\frac{P}{\rho g}\) = K
As the velocity of the fluid increases, pressure of the fluid decreases.

Question 31.
What is isothermal process? Obtain equation for work done in isothermal process.
Answer:
The process in which changes in pressure and volume of a gas system takes place at constant temperature of the system, is known as isothermal process.
For an isothermal process, PV = constant.
At any intermediate stage with pressure P and change in volume from V to V + dV,
work done dw = P dV. However for the entire process W = \(\int_{V_{1}}^{V_{2}} P d V\) .

For V₂ > V₁, W > 0 and for V₂ < V₁, W < 0.
In an isothermal expansion, the gas absorbs heat and does work, while in an isothermal compression, work is done on the gas by the environment and heat is released.

Question 32.
What is closed pipe? Show that modes of vibration in closed pipe are odd harmonics.
Answer:
Let ‘L’ be the length of the closed pipe. A pipe with one end closed is known as a closed pipe system. Le V be the velocity of sound in air. Length of half segment = \(\frac{\lambda_{0}}{4}\)
i.e., L = \(\frac{\lambda_{0}}{4}\) i.e., λ₀ = 4L.

The least mode of vibration is called fundamental mode i.e.,
fundamental frequency f₀ = \(\frac{\mathbf{v}}{\lambda}=\frac{\mathbf{v}}{4 L}\)
In the second mode of vibration.

i.e., f₂ = \(\frac{5 v}{4 L}\)
Hence f₀ : f₁ : f₂ :……..: f_n : : 1 : 3 : 5 :………: (2n – 1)

VI. Answer any THREE of the following questions : ( 3 x 5 = 15 )

Question 33.
A car moving along a straight road with speed of 144 Kmh is brought to a stop within a distance of 200 m. What is the retardation of the car and how long does it take to come to rest?
Answer:
Given u = 144 kmh^-1
ie., u = \(\frac{144 \times 5}{18} \mathrm{ms}^{-1}\)
ie., u = 40 ms^-1
v = 0, s = 200 m, a = ?   t = ?
w.k.t v² = u² + 2aS
i.e., 0 = (40)² + 2a(200)
a = \(\frac{-1600}{400}\)
a = -4ms^-2
By using, v = u + at,
0 = 40 – 4t
i.e, t = \(\frac{40}{4}\) =10 s
Hence time taken by the car to stop
t = 10s

Question 34.
A circular racetrack of radius 200 m is banked at the angle of 10°. If the coefficient of friction between the wheels of racecar and the road is 0.15 what is the
(a) optimum speed of the race car to avoid wear and tear on its tyres?
(b) maximum permissible speed to avoid slipping? [Given: Acceleration due to gravity on the earth = 9.8 ms^-2]
Answer:
(i) Give r = 200 m, θ = 10°, μ = 0.15
v = v_max = ?   g = 9.8 ms^-1
w.k.t  v_max = \(\sqrt{\frac{r g(\mu+\tan \theta)}{(1-\mu \tan \theta)}}\)

Question 35.
If the weight of a 4 kg mass on the surface of the earth is 39.2 N, calculate the acceleration due to gravity of earth at
(a) 32 km height from the surface of the earth.
(b) 16 km depth from the surface of the earth. [Given : Radius of earth = 6400 km]
Answer:
Given m = 4kg, F = W = 39.2 N
(i) with attitude
\(g^{\prime}=g\left(1+\frac{h}{R}\right)^{-2}\)
given g = 10ms^-1, h = 32 km, h <<R
hence g¹ = g(1 – 2h/R)
i.e., g¹ = 9.8 \(\left(1-\frac{2 \times 32}{6400}\right)\) = 9.8 x 0.99
g¹ =9.702 ms^-2
(ii) with depth; g¹ = g( 1 – h/R)
i.e., g¹ = 9.8 \(\left(1-\frac{16}{6400}\right)\)
i.e., g¹ = 9.8 x 0.9975
g¹ = 9.775 ms^-2

Question 36.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt?
Answer:
Given m = 2.5 kg θ₁ = 500°C, θ₂ = 0°C
C_s =0.39 x 10³ Jkg^-1k^-1
L = 3.35 x 10⁵ Jkg^-1,  μ_ice = ?
Since m_ice = C_s Δ θ = m_iL

i.e., m_ice = 1.455 kg

Question 37.
A body oscillates with SHM according to the equation (in SI units)
x = 5 cos \(\left(3 \pi+\frac{\pi}{4}\right)\)
Calculate : (a) frequency of oscillation (b) amplitude of oscillation (c) Displacement of oscillation at t = 1s
Answer:
given,  x = 5 cos (3πt + π/4)m,
f = ?   A = ?   x = ?   when t = 1s
Comparing this with x = A cos (wt + Φ) we get,
(i) A = 5m
(ii) w = 3π rad.s^-1
i.e., 2πf= 3π
∴ f = 3/2 = 1.5Hz
(iii) x = 5 cos (3π( 1) + π/4)
i.e x = 5 cos (13π/4)
x = 5 cos (360° + 225°)
i.e x = 5 cos (225°)
i.e x = 5 cos (180°+ 45°)
x = -5 cos 45°
i.e, x= \(-5 / \sqrt{2}\) = -3.535 m

 

Students can Download 1st PUC Chemistry Previous Year Question Paper March 2013 (North), Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

1st PUC Chemistry Previous Year Question Paper March 2013 (North)

Time: 3.15 Hours
Max Marks: 90

Instructions:

1. Write the question number legibly in the margin.
2. Answer for a question should be continuous.

Part – A

I. Answer the EIGHT of the following : ( 8 × 1 = 8 )

Question 1.
Write the IUPAC name of

Answer:
3, 4- dimethyl hexane.

Question 2.
Draw the staggered conformation of ethane.
Answer:

Question 3.
State modern periodic law.
Answer:
Properties of elements are periodic functions of their atomic numbers.

Question 4.
What is meant by octet rule?
Answer:
It is the tendency of atom to attain 8 electrons in valence shell.

Question 5.
Name the process in which hydrogen is obtained as a by product?
Answer:
Electrolysis of brine solution.

Question 6.
Which alkali metal is the strongest reducing agent?
Answer:
Lithium or Li.

Question 7.
How many significant figures in 6.022 × 10²³?
Answer:
4 or Four.

Question 8.
Which orbital is specified by l = 2 and n = 3?
Answer:
3d.

Question 9.
State Boyle’s law.
Answer:
At constant temperature the volume of a given mass of a gas is inversely proportional to the pressure.

Question 10.
Write relationship between ∆H and ∆U?
Answer:
∆H = ∆U + RT∆n.

Question 11.
What is effect of catalyst on the equilibrium of a reversible reaction.
Answer:
A catalyst has no effect on the position of equilibrium but it helps the reaction to attain equilibrium quickly.

Part – B

II. Answer any EIGHT of the following questions. ( 8 × 2 = 16 )

Question 12.
Define functional group.
Answer:
Atom or group of atom which determine the characteristic properties of organic compound.

Question 13.
State Markownikoff’s rule.
Answer:
When an asymmetric reagent adds up to an unsymmetrical alkene, the negative part of the adding molecule goes to the carbon atom with lesser number of hydrogen atoms while the positive part to the other carbon atom.

Question 14.
What is electron gain enthalpy?
Answer:
The enthalpy change occurs when an electron is added to isolated gaseous atom to convert into anion.

Question 15.
Write electronic configuration of C₂ molecule ? What is its magnetic property?
Answer:
\(\sigma_{1 \mathrm{s}^{2}} \sigma_{1 \mathrm{s}^{2}}^{*} \sigma_{2 \mathrm{s}^{2}} \sigma_{2 \mathrm{s}^{2}}^{*}\left(\pi 2 \mathrm{p}_{x}^{2}\right)\left(\pi 2 \mathrm{p}_{y}^{2}\right)\) Diamagnetic.

Question 16.
What is diagonal relationship? Give an example.
Answer:
Lithium shows similarities to magnesium and beryllium to aluminium in many of their properties. This type of diagonal similarity is commonly referred to as diagonal relationship in periodic table.

Question 17.
Diamond is a bad conductor of electricity but graphite is a good conductor. Justify the statement.
Answer:
Due to sp³ hybridisation in diamond, no free electrons are present.
In graphite due to sp² hybridisation there are free electrons to conduct electricity.

Question 18.
What mass of calcium carbonate is to be decomposed to obtain 4.4 g of CO₂ in the following reaction.

(molecular mass of CaCO₃ = 100g) .
Answer:

Question 19.
State and illustrate Pauli’s exclusion principle.
Answer:
No two electrons in an atom can have all same set of four quantum numbers alike.

Question 20.
Write Vander Waals equation for one mole of a gas and name any two terms.
Answer:
\(\left(p+\frac{a}{V^{2}}\right)(V-b)=R T\)
P = pressure, V = volume, R = universal gas constant, T = absolute temperature

Question 21.
What type of system, the following systems are
i) a cup containing hot tea
ii) hot coffee placed in a thermos flask?
Answer:
i) open system
ii) isolated system.

Question 22.
State Le Chatelier’s principle.
Answer:
If a system under equilibrium be subjected to a change in temperature, pressure or concentratio, then the equilibrium shifts itself in such a way so as to neutralise the effect of the change.

Question 23.
Chemical equation is dynamic. Give reason.
Answer:
Both forward and backward reactions are occuring at the same rate.
The concentration of the reactants and products remains constant.

Part – C

III. Answer any FOUR of the following questions : ( 4 × 4 = 16 )

Question 24.
What is resonance effect? Name one group each showing + R and – R effect.
Answer:
The resonance effect is defined as the polarity produced in the molecule by the interaction of two π bonds or between a π -bond and lone pair of electrons present in conjugated molecular.
+R_effect : -NH₂, – NHCOCH₃, – CH₃
-R_effect : -NO₂, -CN, -CH0, -COOH

Question 25.
Define chain and functional isomerism. Give examples for each isomerism.
Answer:
Compounds having same molecular formula but differing in arrangement of atoms in chain is called isomerism.
Ex – Butane C₄H₁₀

Functional Isomerism
Compounds having same molecular formula but differing in functional group is called functional isomerism.
Ex :

Question 26.
(a) Explain the mechanism of nitration of benzene?
Answer:
Nitration 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 NO
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) Mention the catalyst used in Friedel Crafts reaction.
Answer:
anhydrous AlCl₃.

Question 27.
(a) Explain the mechanism of chlorination of methane.
Answer:
Mechanism of chlorination of methane involves three types.
Step 1 : Initiation : Chlorine absorbs energy and undergoes homolysis to give chlorine free radicals.

Step 2 : Propagation : Chlorine free radical reacts with methane to give methyl free radical.
Cl^•+ CH₄ → CH₃^•+HCl

The methyl free radical reacts with chlorine to form methyl chloride and chlorine free radical.
CH₃^•+ Cl₂ → CH₃Cl + Cl^•

Step 3 : Termination : Free radiais combine to form stable products.
Cl^• + Cl^• → Cl (Chlorine)
CH₃^•+CH₃^•→ C₂H₆ (Ethane)
CH₃^•+ Cl^• → CH₃Cl (Methyl Chloride)

(b) Draw the structure of trans 1, 2- dibromo ethene.
Answer:

Question 28.
(a) Define geometrical isomerism. Give an example.
Answer:

restricted or hindered rotation about C = C bond is called geometrical isomerism.

(b) Name the product when calcium carbonte is treated with water.
Answer:
Acetylene.

Question 29.
What is pollutant? Mention the three gaseous air pollutants.
Answer:
Chemicals which bring about undesirable changes in the environment are called pollutants.
Ex : Nitric oxide (NO), Nitrogen dioxide (NO₂), sulphur dioxide (SO₂)

Part – D

Answer any FIVE of the following questions : ( 5 × 5 = 25 )

Question 30.
(a) Define ionisation enthalphy. How does it vary along the period and down the group in the periodic table.
Answer:
The minimum amount of energy which is needed to remove the most loosely bounded electron from a neutral isolated gaseous atom in its ground state to form a cation also in a gaseous state.

The ionisation enthalpy decreases down the group because the atomic size increases down the group and effective nuclear charge decreases. Hence removal of electrons become easier.

Ionisation energy increases along the period because the atomic size decreases from left to right in a period. Hence effective nuclear charge increases and removal of electrons becomes more and more difficult.

(b) How many periods are there in the periodic table?
Answer:
7 – periods.

Question 31.
(a) Draw the energy level diagram of O₂ molecule and calculate the bond order, why O₂ is paramagnetic?
Answer:

Energy level diagram of oxygen molecule
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{8-4}{2}=\frac{4}{2}=2\)

(b) Give two differences between bonding and antibonding molecular orbitals.
Answer:

Bonding molecular orbitals	Anti-bonding molecular orbitals
1. Formed by the addition wave functions atomic orbitals.	1. Formed by the subtraction of wave functions atomic orbitals.
2. Have less energy than the atomic orbitals which combined	2. Have more energy than the atomic orbitals which combined.

Question 32.
(a) What is hydrogen bond? Give an example each for inter and intra molecular hydrogen bond.
Answer:
A hydrogen bond is the attractive force which binds hydrogen atom of one molecule with electronegative atom CF₁ O or N present in another molecule.
Intermolecular hydrogen bond – Ex -HF
Intramolecular hydrogen bond – Ex – O – nitrophenol.

(b) According to VSEPR theory, predict the shape of H₂O and NH₃ molecules.
Answer:
H₂O – bent V shaped.
NH₃ – pyramidal.

Question 33.
(a) Balance the following redox reaction using oxidation number method.
Cr₂O₇^-2+ S^2- + H⁺ → Cr³⁺ + S + H₂O.
Answer:

Multiply oxidation equation by the extent of reduction & reduction equation by the extent of oxidation.
Eqn (1) × 6; 6S^2- → 6S
Eqn (2) × 2; 2Cr₂O₇² → 4Cr³⁺
Add both the equations
2CrO-+ 6S^2- → 4Cr³⁺ + 6S
Balance oxygen
2CrO ^2-+ 6S^2- → 4Cr³⁺+ 6S + 14 H₂O
Balance hydrogen
2Cr₂O₇^2-+ 6S^2- + 28H⁺ → 4Cr³⁺+ 6S + 14 H₂O

(b) Identify the type of redox reactions.
i) Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu(s).
ii)
Answer:
i) Displacement type
ii) Decomposition type

Question 34.
(a) What happens when metal Mg is treated with dilute hydrochloric acid and write the equation.
Answer:
Mg + 2HCl → MgCl₂ + H₂ ↑
When Mg is treated with HCl, MgCl₂ and H displaces hydrogen gas.

(b) Why are metallic hydrides also called interstitial hydrides.
Answer:
In these hydrides, hydrogen occupies interstices in the metal lattice producing distortion without any change in its type so they are also called interstitial hydrides.

(c) Give two uses of hydrogen.
Answer:

1. Dihydrogen is used in the manufacture of vanaspathi fat by the hydrogenation of polyunsaturated vegetable oil.
2. Widely used in the manufacture of metal hydrides.

Question 35.
(a) Name the raw materials used in manufacture of Na₂CO₃ by Solvay process.
Answer:
Ammonia, CO₂, sodium chloride, ammonium hydrogen carbonate.

(b) Mention biological role of Na and Ca ions.
Answer:
Na – are involved in transmission of nerve signals.
Ca – 99% of body calcium is present in bones and teeth.

(c) Which alkali metal gives golden yellow colour to the flame?
Answer:
Sodium.

Question 36.
(a) Define catenation and name the element showing maximum property of catenation.
Answer:
Self linking of carbon atom to form a long /branched chain/cyclic ring is called catenation.
Carbon has maximum catenation property.

(b) What is the repeating unit in Organo Silicon polymer? Name the starting material used in the manufacture of Organo Silicon Polymer.
Answer:
(-R₂SiO-) is the repeating unit.
Methyl chloride and silicon in presence of copper as catalyst at temperature 573 K are the starting material.

OR (Internal Choice)

Question 36.
(a) Name the compound of group – 13 element which exist as dimer and draw the structure of the compound.
Answer:
AlCl₃ Structure

(b) What is inert pair effect? Illustrate it with suitable example.
Answer:
The occurrence of oxidation states two unit less than that of group oxidation states are attributed as inert pair effect.
Ex : Ge, Sn and Pb prefer to exhibit + 2 state because of inert pair effect i.e. ns² electrons remain inert and do not involve in bonding.

Part – E

Answer any FIVE of the following questions :  ( 5 × 5 = 25 )

Question 37.
(a) carbohydrate containing 40% carbon, 6.73% hydrogen and 53.3% oxygen. The molecular mass of compound is 180. Determine its molecular formula.
Answer:

∴ Empirical formula = CH₂O.
Molecular mass of compound =180
Empirical mass = CH₂O. = 12 + 2 × 1.008 + 16 = 30
\(\mathrm{n}=\frac{\mathrm{M} \cdot \mathrm{M}}{\mathrm{E} \cdot \mathrm{M}}=\frac{180}{30}=6\)
Molecular formula = C₆H₁₂O₆ = Glucose

(b) Define empirical formula.
Answer:
It is the simplest formula of compound which gives the relative number of atoms of different elements present in a molecule of a compound.

Question 38.
(a) Calculate the wavelength of spectral line when an electron in hydrogen atom undergoes transition from 3rd energy level to 2nd energy level (R = 109678 cm^-1).
Answer:
\(\frac{1}{\lambda}\) = R \(\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\) = 109678 cm^-1 \(\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]\)
\(\frac{1}{\lambda}\) = 15245.242 cm^-1
l = \(\frac{1}{15245.242 \mathrm{cm}^{-1}}\) = 0.00006559 cm = 6.559 × 1O^-5 cm.

(b) Mention the shape of d-orbitals.
Answer:
double dumbell shaped.

Question 39.
(a) Write four postulates of kinetic theory of gases.
Answer:

1. All gases are made up of a large number of very small particles called molecules.
2. The gas molecules move randomly in all possible directions in straight lines. The direction of motion changes when they collide with other molecules.
3. The molecules are so small that their individual values is negligible as compared to the total volume of gas.
4. There is no force of attraction or repulsion between the molecules of a gas.

(b) What is the value of gas constant.
Answer:
R = 8.314 JK^-1 mol^-1.

Question 40.
(a) Calculate the enthalpy of formation of liquid benzene (C₆H₆) given the enthalpies of combustion of carbon, hydrogen and benzene as -393.5 KJ, -285.83 KJ and -3267.0 KJ respectively.
Answer:
Required equation
6C(s) + 3H₂(g) → C₆H₆ (l) Δ_fH = ?
Given Data : C(s) + O₂(g) → CO₂(g)  Δ_cH = -393.5KJ
H₂(g) + \(\frac{1}{2}\) O₂(g) → H₂O(l)  ∆_cH = -285.83K
C₆H₆(l) + \(\frac{15}{2}\) O₂(g) → 6CO₂ + 3H₂O  Δ_cH = -3267. 0KJ

(b) What is aspontaneous process?
Answer:
A spontaneous process is one which occurs by itself without any external aid.

OR (Internal choice)

(a) Calculate the enthalpy of combustion of methanol (CH₃OH) given enthalpies of formation of CH₃OH ₍₁₎ CO_2(g) and H₂O₍₁₎ as -239 KJ, -393.5 KJ and -286 KJ respectively.
Answer:
Required Eqn. CH₃OH(l) + \(\frac{3}{2}\)O₂(g) → f CO₂(g) + 2H₂O(l)  ∆_cH = -?
Given data
(1) C(s) + 2H₂(g) + \(\frac{1}{2}\) O₂(g) → CH₃OH(l)  ∆_fH = = -239 kJ
(2) C(s) + O₂(g) → CO₂(g)   ∆_fH = -393.5 Id
(3) H₂(g) + \(\frac{1}{2}\)O(g) → H₂O(l)  ∆_fH = -286 kJ
Reverse eqn( 1) and multiply 3 by 2 and add all three

(b) What happens to the entropy, when liquid is converted into vapours?
Answer:
Entropy increases.

Question 41.
(a) Calculate K. for the following equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). Given equilibrium constant of SO₂,O₂ and SO₃ are 0.3 M, 0.41 M, and 1.45 M respectively. What happens to the above equilibirum if (i) SO₂ is added (ii) O₂ is removed.
Answer:
\(K_{c}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}\)
= \(\frac{(1.45)^{2}}{(0.3)^{2}(0.41)}\) = 56.97
i) Addition of SO₂ favours forward reaction.
ii) Removal of O₂ favours backward reaction.

(b) Identify the two conjugate acid base pairs in the following NH₃ + HCl ⇌ NH₄⁺+ Cl^–
Answer:
NH₃ / NH₄⁺ , HCl / Cl^–

Question 42.
(a) What is common ion effect? Give an example.
Answer:
Suspension of degree of dissociation of weak electrolyte by the addition of strong electrolyte having a common ion is called common ion effect.

(b) Calculate the solubility of lead chloride (PbCl₂) at 298K, if its solubility product is 1.6 × 10^-5.
Answer:
\(S=\sqrt{\frac{K_{s p}}{4}}=\sqrt{\frac{1.6 \times 10^{-5}}{4}}=1.58 \times 10^{-2} \mathrm{mol} / \mathrm{dm}^{3}\)

Question 43.
(a) State postulates of Bohr’s theory.
Answer:

1. Electrons revolve around the nucleus in an circular path of fixed radius and energy. These paths are called orbits.
2. The energy of an electron in the orbit does not change with time.
3. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E.
   \(\gamma=\frac{\Delta \mathrm{E}}{\mathrm{h}}=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{h}}\)
4. The angular momentum of an electron in a given stationary state can be expressed as
   mvr = n. \(\frac{h}{2 \pi}\) n = 1, 2, 3……

(b) Write de Broglies equation.
Answer:
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) or \(\frac{\mathrm{h}}{\mathrm{mc}}\)

Question 44.
(a) Define the terms molarity and mole fraction.
Answer:
Molarity (M) = \(\frac{\text { No. of moles of solute }}{\text { Volume of solution in litres }}\)
Mole fraction of A = \(\frac{\text { No. of moles of } \mathrm{A}}{\text { No. of moles of solution }}\)
= \(\frac{\mathbf{n}_{A}}{n_{A}+n_{B}}\)

(b) What is limiting reagent.
Answer:
The reactant which gets consumed, limits the amount of product formed is called limiting reagent.

Students can Download 1st PUC Chemistry Previous Year Question Paper March 2013 (South), Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

1st PUC Chemistry Previous Year Question Paper March 2013 (South)

Time: 3.15 Hours
Max Marks: 90

Instructions:

1. Write the question number legibly in the margin.
2. Answer for a question should be continuous.

Part – A

I. Answer the EIGHT of the following: ( 8 × 1 = 8 )

Question 1.
What is functional group?
Answer:
Properties of certain organic compounds determined by the presence of an atom or group of atom is called functional group.

Question 2.
Give one example which has K_c = K_p
Answer:
H₂(g) + I₂(g) ⇌ 2NO(s)
or
N₂ + O₂ ⇌ 2NO(g).

Question 3.
State modern periodic law.
Answer:
Properties of elements are the periodic function of their atomic number.

Question 4.
Give any one example for solid and vapour equilibrium.
Answer:
I₂ (solid) ⇌ I₂ (vapour)
OR
camphor (s) ⇌ camphor (vapour)

Question 5.
What is oxidation number of oxygen in hydrogen peroxide?
Answer:
Oxygen in H₂O₂ = -1

Question 6.
Mention one use of hydrogen.
Answer:
It is used as rocket fuel in space research.

Question 7.
What is Avogadro’s number?
Answer:
Avogadro’s number = 6.022 × 10²³

Question 8.
A wavelength of 100 nm of a radiation lies in which region?
Answer:
100 nm = UV

Question 9.
State Boyle’s law.
Answer:
At constant temperature the volume of given mass of a gas is inversely proportional to pressure.

Question 10.
Whether internal energy is extensive property or intensive property?
Answer:
Extensive property.

Question 11.
Write electronic configuration of alkali metals.
Answer:
[Nobel gas] ns¹.

Part – B

II. Answer any EIGHT of the following questions. ( 8 × 2 = 16 )

Question 12.
Calculate percentage composition of carbon and hydrogen in ethanol.
Answer:
Molecular mass of ethanol = 46.06
% carbon = \(\frac{24}{46}\) x 100 = 52
% of hydrogen = \(\frac{6}{46}\) x 100 = 13

Question 13.
Explain hydrogen bonding.
Answer:
Hydrogen bonding is defined as “Attractive force which binds hydrogen atom of one molecule with electronegative atom (F, O, or N) of another molecule.

Question 14.
Define Hund’s rule of maximum multiplicity.
Answer:
Hund’s rule: “Pairing of electron in the orbitals having same subshell doesn’t take place until each orbital has one electron each”.

Question 15.
Represent the behaviour of real gases from ideal gases graphically.
Answer:

Question 16.
Represent the chemical equation when butane (Cooking gas – LPG) burnt completely in air?
Answer:
C₄H₁₀(g) + \(\frac{13}{2}\) O₂ (g) → 4CO₂ (g) + 5H₂O(g)   ∆_cH = -2658.0 KJ

Question 17.
Explain lewis acid and base concept.
Answer:
Acid → Accept a pair of electron
Base → Donate a pair of electron

Question 18.
Define ionisation enthalphy with an example.
Answer:
Ionisation enthalpy “The amount of energy to remove outermost electron from an isolated gaseous atom”.
eg. Na – 498. 5 KJ/mol.

Question 19.
What is redox reaction? Give one example.
Answer:
A chemical reaction in which both oxidation and reduction processes occur simultaneously are known as redox reaction.
eg. 2Na(s) + Cl₂ (g) → 2NaCl(s)

Question 20.
Give any two uses of heavy water.
Answer:

1. It is used as a mild bleaching agent.
2. Used an an antiseptic in surgery.

Question 21.
At equilibrium the concentration of N₂ = 3.0 × 10^-3 M, O₂ = 4.2 × 10^-3 M and NO = 2.8 × 10^-3 M in a sealed tube. What will be K_c for the reaction N₂(g)+ O₂(g) ⇌ 2NO(g)?
Answer:
K_c = \(\frac{\left[\mathrm{NO}_{2}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{N}_{2}\right]}\)
= \(\frac{2.8 \times 10^{-3}}{10^{-3} \times(3.0)(4.2) \times 10^{-3}}\)
= 0.622

Question 22.
What is resonance effect?
Answer:
The permanent polarity is produced by the interaction of lone pair and n electrons in conjugate system.

Question 23.
In sulphur estimation 0.157 g of organic compound gave 0.4813 g of barium sulphate calculate the percentage of sulphur in the compound.
Answer:
Mol. mass of BaSO₄ = 233
233 g of BaSO₄ contain 32 g of sulphur
0.483 g of BaSO₄ contain =
% Sulphur = \(\frac{32 \times 0.483 \times 100}{233 \times 0.157}\) = 42.10%

Part – C

III. Answer any FOUR of the following questions : ( 4 × 4 = 16 )

Question 24.
(a) What is chromatography?
Answer:
Chromatography: It is important technique which separates mixtures into their compounds and also tests the purity of compounds.

(b) Explain how is an organic compound separated by distillation method?
Answer:
Distillation is method used to separate constituents of a liquid mixture which differ in their boiling points. Distillation is a process which involves two steps.

1. Vapourisation: Liquid is converted into vapours.
2. Condensation: Vapours are condensed again into liquid.

Question 25.
(a) Explain in brief how is nitrogen detected using sodium fusion extract?
Answer:

Experiment	Observation	Inference
Organic compound + sodium fusion extract + boiled with FeSO4 + conc. H2SO4	Blue colour	Compound is Nitrogen

(b) Write the structure of 1,2 – Dibromobenzene.
Answer:

Question 26.
(a) Explain mechanism of chlorination of methane.
Answer:
The mechanism involves three steps:
(1) Initiation : A chlorine molecule absorbs energy from sunlight to form free radicals.

(2) Propagation : A chlorine free radical reacts with methane forming a HCl and methyl free radical.
CH₄ + Cl^• → CH₃^•+ HCl
(b) The methyl free radical attack chlorine molecule forming a methyl chloride and chlorine free radical
CH₃^•+ Cl₂ → CH₃Cl + Cl
Termination : The two free radical combines to form product.
Cl^• + Cl^• → Cl (Chlorine)
CH₃^•+CH₃^•→ C₂H₆ (Ethane)
CH₃^•+ Cl^• → CH₃Cl (Methyl Chloride)

(b) Give one use of ethene.
Answer:
Ethene is used for artificial ripening of fruits.

Question 27.
(a) How is ethyne prepared from calcium carbide?
Answer:
When calcium carbide react with water to form ethyne.
CaC₂ + 2H₂O → CH ≡ CH +Ca(OH)₂

(b) Mention the formula of Huckle rule.
Answer:
Huckel formula = (4n + 2) π electron

Question 28.
(a) Explain mechanism of nitration of benzene.
Answer:

Mechanism: Step 1: Generation of electrophile
OH – NO₂ + H – HSO₄ → NO₂⁺+ HSO₄^–+ H₂O
Step 2 : Attack of electrophile on benzene ring to form carbocation.

Step 3 : The carbocation loses proton to form nitrobenzene

(b) Which element is detected by kjeldhal method?
Answer:
Nitrogen

Question 29.
(a) What is ozone hole? Explain.
Answer:
One of the reason of ozone hole by a class of compounds called freons. Due to several kinds of human activities involving freons they will diffuse into ozone layer of the stratosphere.
They undergo photodissociation as follows:

The chlorine free radical react with ozone to form chlorine monoxide radical and O₂
O₃ + Cl^• → ClO^• + O₂
Reaction of chlorine monoxide radicals with oxygen generates chlorine radicals
ClO^• + O → O₂ + Cl^•
The chlorine radicals are continuously regenerated causes damage of ozone layer. This is a ozone hole.

(b) Explain green house effect.
Answer:
The heating of atmosphere due to absorption of IR radiation emitted by earth by the gases of atmosphere is called greenhouse effect.

Green house gases absorb large amount of energy from sunlight and transfer it as heat by which earth gets heated. But they do not readily radiate back the earth’s heat energy to the space. Instead they reflect back a part of the heat energy to the lower atmosphere.

Part – D

Answer any FIVE of the following questions : ( 5 × 5 = 25 )

Question 30.
(a) Why size of cation is smaller than its parent atom?
Answer:
A cation is formed by the loss of electron which decrease in size due to more attraction by nucleus, because of increase in nuclear charge.

(b) Which of these isoelectronic species -Al³⁺ or Mg²⁺ has lower size?
Answer:
Higher charges lower the size.
Al³⁺ has lower size than Mg²⁺.

(c) How is a chemical bond formed?
Answer:
A bond is formed by the attraction force which holds various constituent together in different chemical species.

Question 31.
(a) Define bond length.
Answer:
The average distance between the centres of the nucleus of the two bonded atoms in a molecule.

(b) Explain sp³ hybridisation in methane.
Answer:
The electronic configuation of carbon in the excited state is 1s¹ 2s¹ 2P_x¹  2P_y¹ 2P_z¹. One 2s orbital and 3p orbitals hybridise to give four hybrid orbitals. The hybrid orbitals are directed towards the comer of a regular tetrahedron. The angle between any two orbitals is 109°28′. The four sp³ hybrid orbitals of carbon atom overlap axially with s-orbitals of four hydrogen atom to give C – H σ bonds.
The C – H σ bonds is sp³ – s bond. H – C – H bond angle is 109°28′. The shape of the molecule is tetrahedral.

Question 32.
(a) Write electronic configuration of lithium molecule.
Answer:
(σ₁1s)² (σ^x1s)² (σ2s)²

(b) Why Helium (He₂) molecule does not exist.
Answer:
Helium molecule has bond order equal to zero, hence molecule does not exist.

(c) Define bond order.
Answer:
It is the number of covalent bonds holding the atoms in the molecule.

Question 33.
(a) Give any three postulates of molecular orbital theory.
Answer:

1. The molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper symmetry.
2. The number of molecular orbitals formed is equal to the number of combining atomic orbitals.
3. The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital.

(b) Differentiate between σ bond and π bond.
Answer:

Sigma bond	Pi bond
1. Sigma bond formed by the axial combination of orbitals.	1. Pi bond formed by the head to head combination.
2. More efficient	2. Less efficient

Question 34.
(a) Mention rules used to balance a chemical reaction by oxidation number method.
Answer:

1. Oxidation number of a mono atomic ion is same as the charge on it.
2. Oxidation number of metals are positive and those of non metals are negative.
3. In neutral compounds, the sum of oxidation number of all the atoms is equal to zero.
4. In case of polyatomic ion, the sum of the oxidation numbers of all the atoms of the ion is equal to charge on the ion.

(b) Give an example of covalent hydride.
Answer:
CH₄, NH₃, HF, H₂O

Question 35.
(a) How is temporary hardness of water removed?
Answer:

1. By boiling of water.
2. By Clark’s method : In this method lime is added to the hard water. It precipitates out CaCO₃ and Mg(OH)₂ which can be filtered off.
   Ca(HCO₃) + Ca(OH)₂ → 2CaCO₃ ↓ +2H₂O

(b) Give one biological importance of sodium and potassium ion.
Answer:
Biological importance of sodium.
1. Transmit nerve signals.

Biological importance of potassium.
1. Activates many enzymes.

(c) Name the radioactive element of alkali metals.
Answer:
Francium.

Question 36.
(a) Potassium is lighter than sodium? Why?
Answer:
Potassium has unusual increase of size (or) density of K = 0.86 g/cm³, density of Na = 0.97 g/cm³,
Hence potassium is lighter than sodium.

(b) Give any two uses of sodium carbonate.
Answer:

1. Water softening
2. Laundering and cleaning.

(c) What happens when a piece of aluminium is added to dilute hydrochloric acid.
Answer:
Aluminium dissolves in HCl to liberate hydrogen gas.
2Al + 6HCl → 2AlCl₃ + 3H₂ ↑

Question 37.
(a) How is borax prepared?
Answer:
Borax can also be obtained by neutralisation of boric acid with Na₂CO₃
4H₃BO₃ + Na₂CO₃ → Na₂B₄O₇ + 6H₂O + CO₂. ↑

(b) Write a short note on fullerene.
Answer:
Fullerens are made by heating the graphite in an electric arc in the presence of inert gases. Fullerens are cage like molecules. It contains twenty six membered rings and twelve five membered rings. All the C-atoms are equal, they will undergo sp² hybridisation. Each carbon atom form three sigma bonds with other three C-atoms. This ball shaped molecule has 60 vertices. It also contain both single and double bonds with C-C distances of 143.5 pm and 138.3 pm.

(c) Concentrated nitric acid can be transported in aluminium container. Give a reason.
Answer:
Conc. HNO₃ reacts with aluminium forms a protective layer called passive. It does not react further.

Part – E

Answer any FIVE of the following questions : ( 5 × 5 = 25 )

Question 38.
(a) Define Gay-Lussac’s law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to its temperature.

(b) What is SI unit of luminous intensity?
Answer:
Unit – Candela or Cd.

(c) Define ‘mole’.
Answer:
Mole : It is the amount of substance that contain as many particles (atoms, ions, molecules) exactly 12 g of the C¹² isotopes.

Question 39.
(a) What is meant by average atomic mass?
Answer:
The susbtance which has isotopes, and found relative abundance.
eg 12C = 98.892
13C = 1.108
14C = 2 × 10^-10

(b) What is density? How is it calculated?
Answer:
Density of a substance is the amount of mass per unit volume.

(c) What are the fundamental particles of atom.
Answer:
Electron, proton and neutron.

Question 40.
(a) Mention the postulates of Bohr model of an atom?
Answer:

1. The electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits.
2. The energy of an electron in the orbit does not change with time.
3. The frequency of radiation absorbed (or) emitted when transition ocurs between two stationary states that differ in energy by ΔE is given by
   
4. The angular momentum of an electron in a given stationary state can be expressed as mvr = \(\frac{\mathrm{nh}}{2 \pi}\)

(b). What is the vaiue of Rydberg’s constant.
Answer:
R = 2.18 x 1O^-18 J.

Question 41.
(a) Mention postulates of kinetic molecular theory of gases.
Answer:

1. All gases are made up of very large numbers of minute particles called molecules.
2. Intermolecular forces of attraction (or) repulsion are negligible.
3. The pressure exerted by a gas is due to the collisions made by the gas molecules on the walls of the container.
4. The average kinetic energy of the molecules is directly proportional to the absolute temperature.
5. The molecules are involved in rapid, random movement. During their motion, they collide with each other and also against the walls of the container.

OR (Internal choice)

(a) What is an ideal gas? Derive ideal gas equation?
Answer:
Ideal gas which obey’s Boyle’s law and Charle’s law at all temperatures or PV = RT
Derivation
\(\left(\mathrm{v} \propto \frac{1}{\mathrm{P}}\right) \mathrm{V} \propto \mathrm{T}\)
= V ∝ \(\frac{1}{\mathrm{P}}\) × P
= V = \(\frac{\mathrm{RT}}{\mathrm{P}}\) (R = gas constant)
PV = RT or
PV = nRT

(b) Explain surface tension.
Answer:
Surface tension = \(\frac{\text { Force }}{\text { Area }}\)
The molecules in liquid state on surface experience a net downward force and have more energy than the molecules in bulk, which don’t experience any net force.
If the surface of the liquid is increased by pulling a molecule from the bulk.

Question 42.
(a) State Charle’s law.
Answer:
Charles law : states that volume of a given mass of a gas is directly proportional to temperature at constant pressure.
V ∝ T at constant pressure.

(b) Write Van der Waal’s equation.
Answer:
Van der Waal’s equation
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T\)

(c) State first law of thermodynamics and write its mathematical statement.
Answer:
First law of thermodynamics “The energy of an isolated system is constant”.
∆U = q + w

Question 43.
(a) Calculate AG° for conversion of oxygen to ozone; \(\frac{3}{2}\) O₂(g) → O₂(g) at 298 K,
if K_p = 2.47 × 10^-29.
Answer:
ΔG° = -2.303RT log K_p
= -2.303 × 8.314 × 298 log 2.47 × 10^-29.
= 1.63 kJ/mole

(b) How is ΔU measured calorimetrically.
Answer:
The heat of combustion of a substance at constant volume is measured by calorimetric method.
The bomb is a closed container made of a heavy steel. The bomb is coated inside with gold to prevent oxidation of steel during combustion reaction. A known mass of the substance whose heat of combustion is to be determined is taken in the platinum cup. The bomb is filled with oxygen at a pressure of 20-25 atm. The bomb is closed with a tight screw cap.

The bomb is surrounded by water bath taken in a insulated outer vessel. The initial temperature of water bath is noted. Let it be t₁°C. The combustion is initiated by passing electric current through platinum filament. The heat evolved during combustion raises the temperature of water outside the bomb calorimeter. The highest temperature recorded by thermometer is noted. Let it be t₂°C.

Calculation

1. Let the mass of the substance taken be mg.
2. Let the molecular mass of the susbtance taken be ‘M’
3. Let the heat capacity of calorimeter be Q.
4. Let Δt be the rise in temperature.
5. Then Δu, the heat of combustion at constant volume.

Calculated by using the equation
\(\Delta \mathrm{U}=\mathrm{Q} \times \Delta \mathrm{t} \frac{\mathrm{M}}{\mathrm{m}} \mathrm{J}\)

Question 44.
(a) What is buffer solution?
Answer:
The property of certain solution which resist the change in pH on addition of small amount of acid (or) alkali to it. .

(b) State Lechatelier principle.
Answer:
If a system in equilibrium is subjected to a change of concentration, temperature (or) pressure, the equilibrium shifts in a direction so as to undo the effect of the change imposed.

(c) Derive P_H + P_OH = 14
Answer:
K_w = [H₃O⁺][OH^–] = 10^-14
Taking log of minus sign
-log K_w = – log(H₃O⁺) (OH^–)
-log K_w = – log[H₃O⁺] – log [OH^–]
PK_w = P_H + P_OH = 14
pH + pOH = 14

Question 45.
(a) Define ionic product of water.
Answer:
Ionic product of water may be defined as product of molar concentration of hydrogen ion and hydroxyl ion concentration at a given temperature
K_w = [H₃O⁺][OH^–] = 10^-14

(b) Derive an equation for dissociation constant of weak acid.
Answer:
Let us consider a weak acid HX dissociate in aqueous solution and attains equilibrium

where a is the extent of ionisation.
\(\mathrm{K}_{\mathrm{a}}=\frac{\left(\mathrm{H}_{3} \mathrm{O}^{+}\right)\left(\mathrm{X}^{-}\right)}{(\mathrm{HX})}=\frac{\mathrm{C} \alpha \cdot \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}\)
\(\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}\)
where Ka is called dissociation constant of acid.

Students can Download 1st PUC Chemistry Model Question Paper 3 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
2. Write balanced chemical equations and draw labeled diagram wherever allowed.
3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

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

Question 1.
‘Cis platin’ a medicine used in the treatment of which disease?
Answer:
Cancer/Cancer Tumour.

Question 2.
Write the mathematical expression for Boyle’s law.
Answer:
V ∝ \(\frac{1}{P}\) at constant temperature.

Question 3.
Give the example for gaseous reversible reaction for which Kp = Kc
Answer:
H_2(g) + I_2(g) ⇌ 2HI_(g)

Question 4.
Which group elements in the periodic table are called Noble gases?
Answer:
18th group.

Question 5.
What is the oxidation number of the element in its free state?
Answer:
0 or Zero.

Question 6.
Write the general electronic configuration of First group elements.
Answer:
ns¹

Question 7.
Why is boric acid considered as weak acid?
Answer:
Due to small size of B and presence of only six electrons in the valence shell i.e. protonic acid.

Question 8.
Mention the structure of SiO₄^4-.
Ans.
Tetrahedral structure.

Question 9.
Give an example for non-benzenoid compound.
Answer:
Pyridine

Question 10.
Write the expanded form of ‘CNG
Answer:

Part – B

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

Question 11.
Convert 37°C to °F.
Answer:
°F = \(\frac{9}{5}\) + °C +32 = \(\frac{9}{5}\) X 37 + 32 = 98.6°

Question 12.
Under what conditions of temperature and pressure real gases tend to behave ideally?
Answer:
Low pressure and high temperature.

Question 13.
What is dipole moment? What is its SI unit.
Answer:
It is the product of +ve or negative charge and the distance between the centre of them, μ = e x d
SI unit coulomb meter or cm.

Question 14.
Write any two diagonal relationship between Beryllium and Aluminium.
Answer:
(a) Both forms covalent compounds and soluble in organic compounds.
(b) Both BeCl₂ and AlCl₃ act as Lewis acids.

Question 15.
Give the reaction for the synthesis of water gas and producer gas.
Answer:

Question 16.
How is chloromethane converted to methane?
Answer:
Chloromethane heated with zinc and NaOH to methane.

Question 17.
Illustrate Markovnikov’s rule with an example.
Answer:

The Br^– of HBr is added to double bonded carbon as containing less of OH atoms to give 2–bromopropanol.

Question 18.
What is BOD? What is its significance?
Answer:
The amount of oxygen consumed by the microorganisms in decomposing the organic matter present in water is called BOD. It is a water quality parameter to know the amount of organic matter available for bacteria.

Part – C

III. Answer any FIVE of the following questions; carrying THREE marks: ( 5 x 3 = 15 )

Question 19.
(a) Give reason : Ionic radius of F^– is more than atomic radius of F.
Answer:
Due to more number of electrons in F^– than F.

(b) How does ionization enthalpy varies down the group?
Answer:
Ionisation enthalpy decreases down the group due to increase in atomic size of atoms.

(c) Ionization enthalpy of nitrogen is more than that of oxygen. Give reason.
Answer:
Due to completely half filled p-orbitals (1s² 2s² 2p³) in nitrogen but in oxygen atom p-orbital is more than half-filled.

Question 20.
With the help of MOT write the energy level diagram of hydrogen molecule. What is its bond order and predict magnetic property.
Answer:
BO = \(\frac{1}{2}\) (N_BMO N_ABMO)
= \(\frac{1}{2}\)(2 – 0 ) = 1
It is diamagnetic due to absence of unpaired electrons in the molecule.

Question 21.
Calculate the formal charge of each oxygen atom of ozone molecule
Answer:

Question 22.
(a) Give any two differences between sigma and pi bonds.
Answer:
Sigma Bond

1. Head to Head overlapping of AO.
2. It is strong bond due to large overlapping.
3. Free rotation is possible.

Pi Bond

1. Lateral (sideways) overlapping of AO.
2. It is weak bond due to small overlapping.
3. Free rotation about n bond is not possible.

(b) What is the magnetic nature of oxygen molecule?
Answer:
It is paramagnetic due to presence of unpaired electrons in π_px π_py antibonding molecular orbitals.

Question 23.
Balance the following redox reaction using oxidation number method.
MnO₂ + Br^– → Mn²⁺ +Br₂ + H₂O (In Acidic medium)
Answer:

Step II – Oxidation Reaction x 2
Reduction Reaction x 1

Balance the O atoms (add to RHS) and balance the H+ ions (To LHS)
MnO₂ + 2B^– + 4H⁺ → Mn⁺² +2Br₂ +2H₂O

Question 24.
(i) Explain the laboratory preparation of dihydrogen.
Answer:
In the laboratory, dihydrogen is prepared by adding dil. H2S04 to granulated zinc.
i.e. Zn + H₂SO₄ (dil) → ZnSO₄ + H₂
The H₂ gas liberated is collected by downward displacement of water.

(ii) Give an example of ionic hydride.
Answer:
LiH/NaH/CaH₂/BaH₂ etc.

Question 25.
Give any three points of differences between lithium and other alkali metals.
Answer:

1. Small size of lithium and its ion.
2. High charge to size ratio (polarising power).
3. High ionization enthalpy and low electropositive character of lithium.
4. Absence of d-orbitals.

Question 26.
(i) Write the molecular formula of silica.
Answer:
SiO₂

(ii) Write the partial structure of silicone.
Answer:

(iii) Name the catalyst used in gasoline production.
Answer:
ZSM-5.

Part – D

IV. Answer any FIV E of the following questions’carrying FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 gram of this substance is subjected to complete combustion.
Answer:
Given % carbon = 69%, H = 4.8%, w = 0.20g

(b) Give an example each for element and compound.
Answer:
Element = sodium/Na.
Compound = NaOH/sodium hydroxide.

Question 28.
(a) Write the significance of quantum numbers n, I and m.
Answer:
(i) Principal quantum number (n): Determines energy and size of the orbits (electrons).
(ii) Azimuthal Q.N. (1): Determines the shape of the suborbitals (s,p,d,f).
(iii) Magnetic Q.N. (m): Determines the orientation of the orbitals.

(b) State Pauli’s exclusion principle.
Answer:
For two electrons of an atom the values of n, 1 and m but s is different.
OR
No two electrons of an atom can have the same values of n, 1 and m are same 3 is different.

(c) Write the electronic configuration of copper (Z = 29).
Ans. Cu29= 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Question 29.
(a) The FM station of All India Radio, Hassan, broadcast on a frequency of 1020 kilohertz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter.
Answer:
Given γ = 1020 kilohertz = 1020 x 1000 hertz
C = 3 x10⁸ ms^-11 λ = ?

(b) What is the maximum number of electrons present in third main energy level?
Answer:
Third Main energy level = M r /.
∴ No. of electrons in M shell = 18e.

Question 30.
(a) Give any three postulates of kinetic molecular theory of gases.
Answer:

1. ll gases are made up of with many number of tiny particles called molecules.
2. Gas molecules are separated from each other by large distance. Hence the volume of the gas molecules is negligible compared to total volume of the gas.
3. There is no force of attraction between the gas molecules.
4. The average kinetic energy of gas molecules is directly proportional to Kelvin temperature.

(b) On a ship sailing in Pacific Ocean where temperature is 23.4°C, a balloon is filled with 2L air. What will be the volume of the balloon when the ship reaches 1 the Indian Ocean where temperature is 26.1°C?
Answer:
V₁= 2L V₂ = ?
T₁= 23.4°C + 273 = 296.4 K
T₂ = 26.1°C + 273K = 299.1K

Question 31.
(a) The combustion of one mole benzene takes place at 298K and 1 atm. After combustion, CO₂(g) and H₂O (l) are produced and 3267.0 kJ of beat is liberated. Calculate the standard enthaipy of formation of benzene.
Given : Standard enthalpy of formation of CO₂(g) and H₂O (1) are – 393.5 kJ mol^-1 and -285.0 kJ mol^-1 respectively.
Answer:
Required equation : 6C (s) + 3H₂(g) → C₆H₆ (l); ∆H_f = ?
(i) C₆H₆(l) + \(\frac{15}{2}\)O₂(g) → 6CO₂(g) + 3H₂O(l); ∆H_c = -3267KJ
(ii) C(s) + O₂(g) → CO₂ (g); ∆H_f = -393.5 KJ
(iii) H₂(s) + \(\frac{1}{2}\)O₂(g) → H₂O(l); ∆H_f= -285 KJ
reverse the equation (i) + (ii) 6 + (iii) * 3

(b) Write the mathematical expression for First law of thermodynamics.
Answer:
∆U = q +w or ∆U = q – P∆V

(c) Give an example of isolated system.
Answer:
Tea placed in a thermos flask.

Question 32.
(a) What are the values for ArH° and ArS° for reaction to be spontaneous at all
temperatures?
Answer:
When ∆_rH° = -ve and ∆_rS° = +ve the process is spontaneous at all temperture.

(b) Write the relationship between
(i) Enthalpy (H) and infernal energy (U) (ii) C_p and C_v.
(iii) Free energy (G) Enthalpy (H) and Entropy (S).
Answer:
(i) ∆H = ∆U + P∆V
(ii) C_p – C_v= R for 1 mole of an ideal gas
C_p – C_v =nR for n moles of an ideal gas
(iii) ∆G = ∆H – T∆S

Question 33.
(a) The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium 500K. [Ni] = 1.5 x 10^-2 M, [H₂] = 3.0 x 10^-2 M and [NHJ = 1.2 x 10^-2 m. Calculate the equilibrium constant.
Answer:
Given [N₂] = 1.5 x 10^-2 M, [H₂] = 3.0 x 10^-2 M and [NH₃] = 1.2 x 10^-2m
For N₂(g) + 3H₂ (g) → 2NH₃ (g)

b) Define common ion effect
Answer:
It is the process of decreasing (or suppression) the dissociation of a weak electrolyte by adding a strong electrolyte having a common ion is called common ion effect.

(c) What is the relationship between [HsO+] and [OH-] for neutral solution?
Answer:
[H₃O₊] = [OH^–].

Question 34.
(a) What is acid and bases according to Arrhenius concept?
Answer:
Arrhenius Acid : Substances which give hydrogen ion in water are called Arrhenius Acid. Example: HCl. HNO₃
Arrhenius base : Substances which gives hydroxyl ion (OH-) in water are called Arrhenius bases.
Example: NaOH, KQH, NH₄OH, etc.

(b) Give an example for (i) so.M-vapoisr equilibrium (ii) liquid-vapour equilibrium,
(i) I₂(s) → I₂ (vapour) / NH₄Cl(s)⇌ NH₄Cl(vapour) (ii) H₂O(l) → H₂O(v).

(c) Write the equilibrium K. expression for H2 + I2
Answer:
K_c = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}\)

Part – E

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

Question 35.
(a) For the molecule CH₃CH₂CH₂OH
(i) Identify functional group
Answer:
Functional group = – OH

(ii) Write the bond line forarulbu
Answer:
Bond line formula =

(iii) Write the succeeding homologue.
Succeeding homologue = CH₃CH₂CH₂CH₂OH

(b) Explain inductive effect with a suitable example.
Answer:
The permanent displacement of a (sigma) electrons along the saturated carbon chain away/ towards the group/atom attached at the end of the chain.

Question 36.
(a) An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Answer:
Given % carbon = 69%, H = 4.8%, w = 0.20g

(b) What are nucleophiles? Give an example.
Answer:
Chemical species having -ve charge reacts with nucleus of an atom/molecule is called nucleophile, e.g: H^–, CN^–, H₂O, NH₃, CP^–, Br^– etc.

Question 37.
(a) Explain the mechanism of chlorination of methane.
Answer:
Mechanism of chlorination of methane :
Step I : Chain initiation: Methane and Cl₂ heated in UV light, Cl₂ undergoes homolytic fission giving chlorine free radicals,

Step II: Chain propagation: Cl reacts with methane gives methyl free radical,

Methyl free radical reacts with Cl2 molecule gives C1‘ radical, i.e. CH₃ + Cl₂ → CH₃Cl + Cl’ These are repeated several times.

Step III: Chain termination:
The reaction goes to the end when the free radicals reacts with each other.

(b) Write the Newman’s projections of ethane.
Answer: