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Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa

Anukampa Questions and Answers

Anukampa Summary

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 3 Muhyadheen Mala Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 4 Chapter 3 Muhyadheen Mala

Muhyadheen Mala Questions and Answers

Muhyadheen Mala Summary

Students can Download Chapter 3 Motion in a Straight Line Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 3 Motion in a Straight Line

Plus One Physics Motion in a Straight Line One Mark Questions and Answers

Engineering Physics MCQ PDF arranged chapterwise! Start practicing now for exams, online test.

Question 1.
Which of the followig curves does not represent motion in one dimension?

Answer:
(b) In one dimensional motion, the body can have at a time one value of velocity but not two values of velocities.

Question 2.
Free fall of an object (in vacuum) is a case of motion with
(a) Uniform velocity
(b) Uniform acceleration
(c) Variable acceleration
(d) Uniform speed
Answer:
(b) Uniform acceleration:
Free fall of an object (in vacuum) is a case of motion with uniform acceleration.

Question 3.
The area under velocity-time graph fora particle in a given interval of time represents
(a) velocity
(b) acceleration
(c) work done
(d) displacement
Answer:
(d) displacement:
Area under velocity-time graph represents displacement of a particle in a given interval of time.

Question 4.
The velocity-time graph of a body moving in straight line is shown in the figure. The displacement and distance travelled by the body in 6s are respectively

(a) 8, 16m
(b) 16m, 8m
(c) 16m, 16m,
(d) 8m, 8m
Answer:
(a) 8, 16m

Displacement is equal to area under the velocity time graph with proper sign.
∴ Displacement = 4 × 2 – 2 × 2 + 2 × 2 = 8m
Distance is equal to total area under the speed time graph.
∴ Distance = 4 × 2 + 2 × 2 + 2 × 2 = 16m.

Question 5.
A cartravels half the distance with constant velocity of 40 kmph and the remaining half with a constant velocity 60 kmph. The average velocity of the car in kmph is
(a) 40
(b) 45
(c) 48
(d) 50
Answer:
(c) 48
Average velocity = ν_av = \(\frac{s}{\frac{s}{80}+\frac{s}{120}}\) = 48kmph.

Question 6.
Two objects A and B travel from P to Q through two different paths as shown in figure. If both A and B takes the same time interval to travel from P to Q, then which of the following statements are true?

(a) A and B have same speed.
(b) A and B have same velocity
(c) A and B have same average velocity.
(d) The speed of A is greater than that of B
(e) The speed of B isgreaterthanthatofA
Answer:
(d) The speed of A is greater than that of B.

Question 7.
The acceleration of a moving object is equal to the
(a) gradient of a displacement-time graph
(b) gradient of a velocity-time graph
(c) area below a speed-time graph
(d) area below a displacement – time graph
(e) area below a velocity-time graph
Answer:
(b) Gradient of a velocity-time graph.

Question 8.
A ball is thrown vertically upwards and comes back. Which of the following graph represents the velocity-time graph of the ball during its flight?

Answer:

Question 9.
The magnitude of average velocity is equal to average speed. In which case this condition is satisfied?
Answer:
When a particle is moving with constant velocity, the magnitude of its average velocity is equal to average speed.

Question 10.
Can a body be said to be at rest as well as in motion at the.same time?
Answer:
Yes, rest and motion are relative terms. A body at rest with respect to one body may be in motion with respect to another body.

Question 11.
What conclusion can you draw if the average velocity is equal to instantaneous velocity?
Answer:
The particle is moving with constant velocity.

Question 12.
Two cars are moving in such a way that their relative velocity is zero. Which of the following graph represent this situation?

Answer:
(a) b

Question 13.
The speed-time graph is shown in figure. Is it possible.

Answer:
No speed cannot be negative.

Question 14.
Why the speed of the object can never be negative?
Answer:
Speed is distance covered per unit time. Since distance cannot be negative, speed cannot be negative.

Question 15.
Is it possible that the velocity of an object be in a direction other than the direction of acceleration? If yes, give an example.
Answer:
Yes. A body is moving with decreasing velocity.

Question 16.
Is it possible to have the rate of change of velocity constant while the velocity itself changes both in magnitude and direction? If yes, give an example.
Answer:
Yes. Projectile motion.

Question 17.
If the acceleration of the particle is constant in magnitude but not in direction, what type of path does the body flow?
Answer:
Circular path.

Question 18.
Two stones of different sizes are dropped simultaneously from the top of a building. Which stone would reach earlier? Why?
Answer:
Both reach at ground simultaneously. Acceleration is same for both stones.

Question 19.
A piece of paper and iron piece are dropped simultaneously from the same point in vacum. Which one will reach at ground earlier?
Answer:
Both reach at ground simultaneously.

Question 20.
Is it possible that your cycle has a southward velocity but northward acceleration? If yes, give an example.
Answer:
Yes, when brakes are applied to a moving cycle, the directions of velocity and acceleration becomes opposite.

Plus One Physics Motion in a Straight Line Two Mark Questions and Answers

Question 1.
An ant is moving through a graph paper along x-axis. A boy observes that the ant covers 1mm in every second.

1. What type of motion is this?
2. When the boy is in school bus he observe the speedometer of the bus. Which speed is observed by the speedometer?

Answer:

1. Uniform motion or uniform velocity
2. Instantaneous speed (Ratio of the displacement to small interval of time).

Question 2.
Match the following

Answer:
a. (c)
b. (d)
c. (b)
d. (a)

Question 3.
Some examples of motion are given below. State in each case if the motion is one, two or three dimension

1. A Kite flying on a windy day.
2. A speeding car on a long straight highway.
3. A carrom coin rebounding from the side of the board.
4. A planet revolving around its star.

Answer:

1. 3 Dimensional motion
2. 1 Dimensional motion
3. 2 Dimensional motion
4. 2 Dimensional motion

Plus One Physics Motion in a Straight Line Three Mark Questions and Answers

Question 1.
Two bodies start moving in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of 40 m/s and the second starts from rest with a constant acceleration of 4 m/s².

1. What is uniform speed?
2. Find the time that elapses before the second catches the first body.

Answer:
1. A body is said to be uniform if it travels equal displacements in equal intervals of time.

2. Distance travelled by first body in a time t
S₁ = Vt
S₁ = 40 × t
Distance travelled by second body in a time t
S₂ = ut + \(\frac{1}{2}\) at²
S₂ = \(\frac{1}{2}\) × 4 × t²
When these two bodies meet,
S₁ = S₂
40 × t = \(\frac{1}{2}\) × ²
t = 20 s.

Question 2.
Velocity time graph of a moving object is shown below.

1. What is the acceleration of the object?
2. Draw displacement – time graph for the above motion shown in the graph.

Answer:
1. Acceleration = 0

2.

Question 3.
Two straight lines drawn on the same displacement time graph make angles 30° and 60° with time axis respectively in the figure.

1. Which line represents greater velocity?
2. What is the ratio of the velocity of line A to the velocity of line B?

Answer:
1. B

2.

Question 4.
A particle starts from rest and its acceleration a plotted against time t is shown below.

1. This body is at
   • constant acceleration
   • variable acceleration
   • constant velocity
   • rest
2. Plot the corresponding velocity (V) against time (t)
3. Plot the corresponding displacement (S) against time (t)

Answer:
1. Constant acceleration

2.

3.

Question 5.
Displacement is a vector quantity which distance is a scalar quantity.

1. Distinguish between scalar and vector quantities.
2. An athlete runs along a circular track of radius 50m. Find the distance travelled and the displacement of the athlete when he coveres % of the circle.
3. What is the distance travelled by a body in a time t having an initial velocity u and moving with uniform acceleration ‘a’?

Answer:
1. A physical quantity having both magnitudes and direction is called vecter. A physical quantity having only magnitude is called scalar.

2.

Displacement AD

Distance, ABCD = AB + BC + CD

Distance = \(\frac{3}{2}\) πr

3. c = ut + \(\frac{1}{2}\) at²

Question 6.
“The aerial distance between two towers is 4km. But speedometer of car shows 5.6km when travel from one tower to another”

1. By reading this statement explain the concept of distance and displacement.
2. What is the numerical ratio of displacement of object to distance? Explain.
3. A particle is moving along a circular trace of radius ‘R’. What is the distance travelled and displacement of the particle in half revolution?

Answer:
1. Distance os the length of the path covered by the object. It is a scalar quantity. Displacement is the length between initial point and final point.

2. \(\frac{\text { displacement }}{\text { distance }} \leq 1\)
For the straight-line path, displacement is equal to the distance travelled. But for the curved path displacement is less than the distance travelled.

3. distance = πR
displacement = R + R = 2R.

Plus One Physics Motion in a Straight Line Four Mark Questions and Answers

Question 1.
A stone is thrown upwards from the ground with a velocity ‘u’.

1. What is the maximum height attained by the stone?
2. Check the correctness of the equation obtained in (a) using the method of dimensional analysis.
3. Draw the position-time graph of the stone during its return journey. (g = 10m/s²)

Answer:
1. V² = u² + 2as
0 = u² + 2gH
H = u²/2g

2. H = \(\frac{u^{2}}{2 g}\)
When we write the above equation in terms of dimension, we get

This means that, H= \(\frac{u^{2}}{2 g}\) is dimensionally correct.

3.

Question 2.
Gopal dropped an apple from the top of his flat at a height of 10m. He told his sister Seetha on the ground below that it will reach the ground in 2 seconds after he drops it.

1. Can she catch it after 2 seconds?
2. Derive suitable relation for time of fall.
3. Draw the velocity-time graph of the above body (assume the body rebounds from the floor)

Answer:
1. No.

2. S = ut + 1/2at²
h = 0 + 1/2gt²
If ball is dropped from a height ‘h’, we can write.
\(\sqrt{\frac{2 h}{g}}\) = t.

3.

Question 3.
A car of mass 1000 kg starts from rest at t = 0 and under goes acceleration as shown in figure.

1. Draw the corresponding velocity-time graph.
2. What is the retarding force acting on the car?
3. What is the total distance travelled by the car during t = 0 to t = 4 sec.

Answer:
1.

2. Retarding acceleration, a = -3m/s²
∴ Retarding force F = ma
= 1000 × 3 = 3000N.

3. Area of velocity time gives distance.
∴ Area = \(\frac{1}{2}\)bh = \(\frac{1}{2}\) × 4 × 6
Distance = 12m.

Question 4.
A tow rope used to pull the car of mass 700kg will break if the tension exceeds 1500N.

1. Calculate the maximum acceleration with which the car can be pulled through a level road
2. Calculate the minimum time required to bring the car to work station 500m away from the break point

Answer:
1. T = ma
1500= 700 × a
a = \(\frac{1500}{700}\) = 2.14 m/s².

2. S = ut + 1/2 at²
500= 0 + 1/2 × (2.14) × t²
t = \(\sqrt{\frac{2 \times 500}{2.14}}\) = 21.61sec.

Question 5.
1. Figure shows the position-time graph of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t ≤ 0 and on a parabolic path fort> 0? Justify your answer.

2. Can a body have an acceleration without velocity? Justify your answer with a physical situation.
3. The table given below shows the velocity of a car at different times.

a. Draw acceleration time graph
b. Find the distance travelled by the car in 6 sec.
Answer:
1. No. initially the body remains at rest and then the body moves with constant acceleration in a straight line.

2. Yes, considerthe oscillation of simple pendulum. At extreme position, velocity becomes zero and acceleration is non – zero value.

3. The velocity of a car at different times:
a. acceleration a = \(\frac{16-4}{1-0}\) = 5 m/s²
This value (acceleration is constant through-out the motion. Hence the acceleration time graph will be a straight line parallel to time axis.

b. S = ut + \(\frac{1}{2}\) + 2at²
= 11 × 6 + \(\frac{1}{2}\) × 5 × 6²
S = 156m.

Question 6.
The relative velocity of a body A with respect to a body B is the time rate at which body A changes its position with respect to body B.

1. If V_A and V_B are the velocities of A and B moving in opposite directions, what is the relative velocity of A with respect to B?
2. Two trains along the same straight rails are moving with constant velocity of 60 km/h and 30 km/b towards each other. If at time t = 0, the distance between them is 90km, find the time when they collide.
3. The velocity-time graph of two bodies A and B make angles of 30° and 60° with the time axis, what is the ratio of their acceleration?

Answer:
1. V_BA = V_A + V_B

2. Relative velocity = 60 + 30
= 90km/h
∴ Hence t = \(\frac{\text { displacement }}{\text { relative velocity }}\)
= \(\frac{90}{90}\) = 1h.

3. Slope of velocity time graph gives acceleration. Hence

Plus One Physics Motion in a Straight Line Five Mark Questions and Answers

Question 1.
A balloon is ascending at the rate of 14 ms^-1 and at a height of 98 m above the ground. A stone is dropped from it.

1. State whether the motion of the balloon is accelerated or retarded.
2. After how much time does the stone reach the ground?
3. Determine the velocity with which the stone strikes the ground.

Answer:
1. The motion of balloon is uniform motion. It has neither acceleration nor retardation.

2. u = 14m/s, a = -9.8, S = -98 m.
S = ut + \(\frac{1}{2}\) at²
98 = 14 t – \(\frac{1}{2}\) × 9.8 t²
4.9 t² – 14 t – 98 = 0
Solvingthisweget t = 6.123 sec

3. ν² = u² + 2as
^ν2 = (14)² + 2 x ^–9.8 x ^–98
= 196 + 1920.8
^ν2 = 2116.8
^ν2 = \(\sqrt{2116.8}\) = 45.99 m/s.

Question 2.
A particle is moving along x-axis with a uniform positive acceleration.

1. Draw the position-time graph for its motion.
2. Obtain the expression for the displacement by drawing velocity-time graph.
3. A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of the tower of height 25m from the ground. How long does it remain in air?(g = 10 ms^-2)

Answer:
1.

2.

Consider a body moving with an acceleration ‘a’. Let ‘u’ be the initial velocity at t = 0 and final velocity ‘v’ at t = t. The area of velocity-time graph gives displacement. This is a quadratic equation. Hence t can be found using this formula

Question 3.

1. State the difference between speed and velocity. Can a body move with uniform speed but with variable velocity? Explain with the help of an example.
2. Show that a body thrown vertically upwards returns with the same magnitude of velocity.

Answer:
1. Speed is scalar quantity but velocity is a vector quantity. If a body moving along the circumstance of a circle with uniform speed, its velocity changes continuously with time.

2.

Consider a body projected upward from a point A with a velocity ‘u’. If the body reaches at B, the displacement becomes zero. Hence time taken to reach at B can be found.
S = ut + 1/2 at²
0 = ut + 1/2 × ^–10 t²
ut = 5t²
t = \(\frac{u}{5}\) ____(1)
The velocity at B can be found using the formula
V_B = u + at _____(2)
substitute eq(1) in eq(2)
V_B = u + ^–10\(\frac{u}{5}\)
= u – 2u
V_B = -u
∴ V_A = -V_B

Plus One Physics Motion in a Straight Line NCERT Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a) (b)

Question 2.
The position time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in the following figure Choose the correct entries in the brackets as follows:

1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/different) time
5. (A/B) overtakes (B/A) on the road (once/twice).

Answer:
1. It is clear from the graph that OQ > OP. So, A lives closer to the school than B.

2. The position-time graph of A starts from the origin (t = 0) while the position-time graph of B starts from C which indicates that B started later than A after a time interval OC. So, A started earlier than B.

3. The speed is represented by the steepness (or slope) of the position-time graph. Since the position-time graph of B is steeper than the position-time of graph A, therefore, we conclude that B is faster than A.

4. Corresponding to both P and Q, the time interval is the same, i.e., OD. This indicates that both A and B reach their homes at the same time.

5. The position-time graphs intersect at the point K. This indicates that B crosses A. Since there is only one point of intersection, therefore, the two cyclists cross each other only once.

Question 3.
A car moving along a straight highway with speed of 126 kmh^-1 is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer:
Initial velocity,
u = 126kmh^-1 = 126 × \(\frac{5}{18}\) ms^-1 = 35m^-1
Final velocity, ν = 0;
Distance, S = 200m
Using ν² – u² = 2aS,
0² – 35 × 35 = 2a × 200
or
a = \(-\frac{35 \times 35}{400}\)ms^-2
= 3.06ms^-2
So, the retardation of the car is 3.06 ms^-2
Using ν = u + at,
0 = 35-3.06 × t
or
3.06t = 35
t = \(\frac{35}{3.06}\)s = 11.4s.

Question 4.
On a two-lane road, car A is travelling with a speed of 36kmh^-1 Two cars B and C approach car A in opposite directions with a speed of 54 kmh^-1 each. At a certain instant, when the distance AB is equal to AC, both being I km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:
ν_A = 36kmh^-1
= 36 × \(\frac{5}{18}\)ms^-1 = 10ms^-1
ν_B = ν_C = 54ms^-1
= 54 × \(\frac{5}{18}\)ms^-1 = 15ms^-1

Relative velocity of B w.r.t. A, ν_BA = 5 ms^-1
Relative velocity of C w.r.t.A, ν_CA= 25ms^-1
Time taken by C to cover distance AC
= \(\frac{1000 m}{25 m s-1}\) = 40s
Now, forB, 1000 = 5 × 40 + \(\frac{1}{2}\)a × 40 × 40
On simplification, a = 1 ms^-2.

Question 5.
Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion.

1. with zero speed at an instant may have non-zero acceleration at that instant,
2. with zero speed may have non-zero velocity,
3. with constant speed must have zero accelera¬tion,
4. with positive value of acceleration must be speed-ing up.

Answer:

1. True
2. False
3. True
4. False

For (1), consider a ball thrown up. At the highest point, speed is zero but the acceleration is non-zero, For (2), if a particle has non-zero velocity, it must have speed, For (3), if the particle rebounds instantly with the same speed, it implies infinite acceleration which is physically impossible. For (4), true only when the chosen positive direction is along the direction of motion.

Question 6.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh^-1. What is the magnitude of average velocity, and average speed of the man over the following intervals of time:

1. 0 to 30 min,
2. 0 to 50 min,
3. 0 to 40 min?

Answer:
Average speed overthe interval of time from 0 to 30min
= \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}=\frac{2.5 \mathrm{km}}{\frac{1}{2} \mathrm{h}}\)
= 5kmh^-1
Magnitude of average velocity overthe interval of time from 0 to 30 min is 5kmh^-1. This is because the “distance travelled” and the ‘magnitude of displacement’ over the interval of time from 0 to 30 min are equal. Distance covered from 30 to 50 minutes
= 7.5kmh^-1 × \(\frac{20}{60}\) h = 2.5km
Total distance covered from 0 to 50 minute
= 2.5 km+ 2.5 km = 5km
Total time = 50min = \(\frac{50}{60}\) h = \(\frac{5}{6}\) h
Average speed overthe interval of time from 0 to 50min
= \(\frac{5 \mathrm{km}}{5 / 6 \mathrm{h}}\) = 6kmh^-1
The displacement overthe interval of time from 0 to 50 min is zero. So the magnitude of average velocity is zero. Distance covered from 30 to 40 min
= 7.5kmh^-1 × \(\frac{1}{6}\) h = 1.25km
Total distance covered from 0 to 40 minute
= 2.5 km + 1.25 km = 3.75km
Average speed overthe interval of time from 0 to 40min
= \(\frac{3.75 \mathrm{km}}{\frac{40}{60} \mathrm{h}}\) = 5.625kmh^-1
The “magnitude of displacement” is (2.5 -1.25) km, i.e., 1.25 km.
Time interval = \(\frac{2}{3}\)h
The ‘magnitude of average velocity’ = \(\frac{1.25 \mathrm{km}}{2 / 3 \mathrm{h}}\),
= 1.875 kmh^-1.

Question 7.
Look at the graphs (a) to (d) in the following figure carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of the particle.

Answer:
None of the four graphs can represent one-dimensional motion of the particle. In fact, all the four graphs are impossible.

1. A particle cannot have two different positions at the same time.
2. A particle cannot have velocity in opposite directions at the same time.
3. Speed is always positive (non-negative).
4. Total path length of a particle can never decrease with time.

Note: The arrows on the graphs are meaningless.

Question 8.
The x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
No, wrong, x-t plot does not show the trajectory of a particle. Context: A body is dropped from a tower (x = 0) at t = 0.

Question 9.
The velocity-time graph of a particle in one-dimensional motion is shown below. Which of the following formulae are correct for describing the motion of the particle over the time interval from t₁ to t₂?

(a) x(t₂) = x(t₁) + ν(t₁)(t₂ – t₁) + \(\frac{1}{2}\) a(t₂ – t₁)²
(b) ν(t₂) = ν(t₁) + a(t₂ – t₁)
(c) ν_average = [x(t₂) – x(t₁)]/(t₂ – t₁)
(d) a_average = [ν(t₂) – ν(t₁)]/(t₂ – t₁)
(e) x(t₂) = x(t₁)+ ν_av (t₂ – ₁) +\(\frac{1}{2}\) a_av(t₂ – t₁)²
(f) x(t₂) – x(t₁) = Area underthe ν – 1 curve bounded by t-axis and the dotted lines.
Answer:
(c), (d), (f)
Explanation: It is clear from the shape of ν – t graph that acceleration of the particle is not uniform between time intervals t₁ and t₂. [Note that the given ν – t graph is not straight.] The equations (a), (b) and (e) represent uniform acceleration.

Students can Download Chapter 7 Equilibrium Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 7 Equilibrium

Plus One Chemistry Equilibrium One Mark Questions and Answers

Question 1.
Equilibrium in a system having more than one phase is called _________
Answer:
heterogeneous

Question 2.
Addition of a catalyst to a chemical system at equilibrium would result in
a) Increase in the rate of forward reaction
b) Increase in the rate of reverse reaction
c) A new reaction path
d) Increase in the amount of heat evolved in the reaction
Answer:
c) A new reaction path

Question 3.
With increase in temperature, equilibrium constant of a reaction
a) Always increases
b) Always decreases
c) May increase or decrease depending upon the number of moles of reactants and products
d) May increase or decrease depending upon whether reaction is exothermic or endothermic
Answer:
d) May increase or decrease depending upon whether reaction is exothermic or endothermic

Question 4.
Water is a conjugate base of ____________ .
Answer:
H₃O⁺

Question 5.
Which of the following substances on dissolving in water will give a basic solution?
a) Na₂CO₃
b) Al₂(SO₄)₃
c) NH₄Cl
d) KNO₃
Answer:
a) Na₂CO₃

Question 6.
Choose the correct answer for the reaction,
N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g); ∆_rH = -91.8 kJ mol^-1 The concentration of H₂(g) at equilibrium can be increased by
1) Lowering the temperature
2) Increase the volume of the system.
3) Adding N₂ at constant volume.
4) Adding H₂ at constant volume.
Answer:
2) and 4) are correct.

Question 7.
Conjugate base of a strong acid is a
Answer:
Weak base

Question 8.
The expression forostwald dilution law is
Answer:
Ka = Cα²

Question 9.
The hydroxyl ion concentration in a solution having p^H = 4 will be
Answer:
10^-14

Question 10.
A mono protic acid in 1M solution is 0.01 % ionized the dissociation constant of this acid is
Answer:
10^-8

Question 11.
A certain buffer solution contains equal concentration of x^– and Hx. The Kafor Hx is 10^-6 p^H of buffer is
Answer:
6

Question 12.
In the equilibrium reaction
CaCO_3(s) → CaO_(s) + CO_2(g) the equilibrium constant is given by —–
Answer:
PCO₂

Question 13.
Congugate base of a strong acid is a _________ .
a) Strong base
b) Strong acid
c) Weak acid
d) Weak base
e) Salt
Answer:
d) Weak base

Question 14.
The species acting both as bronsted acid and base is _________ .

Answer:
b) HSO₄^–

Question 15.
P^H of .01 M KOH solution will be _________ .
Answer:
12

Plus One Chemistry Equilibrium Two Mark Questions and Answers

Question 1.
“High pressure and low temperature favours the formation of ammonia in Haber’s process.” Analyse the statement and illustrate the conditions using Le-Chatliers principle?
Answer:
The given statement is correct.
N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g); ∆_rH =-91.8 kJ mol^-1 Since the number of moles decreases in the forward reaction a high pressure of ~ 200 atm is applied. Since the forward reaction is exothermic the optimum temperature of ~ 700 K is employed for maximum yield of ammonia.

Question 2.
“Chemical equilibrium is dynamic in nature”. Analyse the statement and justify your answer.
Answer:
At equilibrium the reaction does not stop. Both forward and backward reactions are taking place at equal rates. Thus, at equilibrium two exactly opposite changes occur at the same rate. Hence, chemical equilibrium is dynamic in nature.

Question 3.
Pressure has no influence in the following equilibrium: N₂(g) + O₂(g) \(\rightleftharpoons \) 2NO(g)

1. Do you agree with this?
2. What is the reason for this?

Answer:

1. Yes.
2. Here the total number of moles of the reactants is equal to that of the products. Hence pressure is having no influence in this equilibrium.

Question 4.
During a class room discussion a student is of the view that the value of equilibrium constant can be influenced by catalyst.

1. Do you agree with the statement?
2. Justify the role of catalyst in an equilibrium reaction?

Answer:

1. No.
2. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression. It only helps to attain the equilibrium state in a faster rate.

Question 5.
What is the equilibrium constant (K) in the following cases?

1. Reaction is reversed.
2. Reaction is divide by 2.
3. Reaction is multiplied by 2.
4. Reaction is splitted into two.

Answer:

1. 1/K
2. √K
3. K²
4. K₁K₂

Question 6.
1. What is homogeneous equilibrium?
2. Suggest an example for this.
Answer:
1. The equilibrium in which the reactants and products are in the same phase,

2. N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g)
In this equilibrium, the reactants and products are in the gaseous phase.

Question 7.
The equilibrium constants for two reactions are given. In which case the yield of product will be the maximum?
For first reaction: K₁ = 3.2 × 10^-6
For second reaction: K₂ = 7.4 × 10^-6
Answer:
Higher the value of K, greater will be the yield of product. So maximum yield will be in the second case.

Question 8.
Write an expression for equilibrium constant, Kc forthe ‘ reaction, 4NH₃(g) + 5O₂(g) \(\rightleftharpoons \) 4NO(g) + 6H₂O(g)
Answer:

Question 9.
1. State Henry’s law.
2. Suggest an example fora gas in liquid equilibrium.
Answer:
1. The mass of a gas dissolved in a given mass of a solvent at any temperature is directly proportional to the pressure of the gas above the solvent.

2. Equilibrium between the CO₂ molecules in the. gaseous state and the CO₂ molecules dissolved in water under pressure,
CO₂(g) \(\rightleftharpoons \) CO₂(in solution)

Question 10.
1. What is heterogeneous equilibrium?
2. Suggest an example forthis.
Answer:
1. Equilibrium in a system having more than one phase is called heterogeneous equilibrium

2. Equilibrium between solid Ca(OH)₂ and its saturated solution:
Ca(OH)₂(s) + (aq) \(\rightleftharpoons \) Ca₂+(aq) + 2OH^–(aq)

Question 11.
For the equilibrium 2SO₃(g) → 2SO₂(g) + O₂(g), K_c at 47 °C 3.25 × 10^-9 mol per litre. What will be the value of K_p at this temperature (R = 8.314 J K^-1mol^-1).
Answer:
R = 8.314 J K^-1 mol^-1 ∆n = 3 – 2 = 1
T = 47 °C = 273 + 47 = 320 K
K_c = 3.25 × 10^-9
K_p = K_c (RT)^∆n
= 3.25 × 10^-9 (8.314 × 320)¹
= 3.25 × 10^-9 × 8.314 × 320 = 8.65 × 10^-12

Learn how to find Kp from pressures, and see examples that walk through sample problems step-by-step for you to improve your chemistry knowledge and skills.

Question 12.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ions in it.
Answer:
pH= -log[H⁺] = 3.76
log[H⁺] = – 3.76
[H⁺] = antilog of (- 3.76) = 1.738 × 10^-4 mol L^-1

Question 13.
The equilibrium constant can be expressed in terms of partial pressure as well as concentration.
1. Give the relation between K_p and K_c.
2. What is the relation between K_p and K_c for the reaction, N₂(g) + O₂(g) \(\rightleftharpoons \) 2NO(g)?
Answer:
1. K_p = K_c(RT)^∆n, where ∆n = (number of moles of gaseous products) – (number of moles of gaseous reactants).

2. Here, ∆n = 2 – 2 = 0
K_p = K_c(RT)^∆n , K_p = K_c (RT)°
∴ K_p = K_c

Question 14.
1. Explain Arrhenius concept of acids and bases with suitable examples.
2. How proton exists in aqueous solution? Give reason.
Answer:
1. According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions, H⁺(aq) and bases are substances that produce hydroxyl ions, OH^–(aq). Forexample, HCl is an Arrhenius acid and NaOH is an Arrhenius base.

2. In aqueous solution the proton bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H₃O⁺(aq). This is because a bare proton, H+ is very reactive and cannot exist freely in aqueous solutions.

Question 15.
1. What is an acidic buffer?
2. Suggest an example for an acidic buffer.
Answer:
1. An acidic buffer is a buffer solution having pH less than 7. It is prepared by mixing a weak acid and its salt formed with a strong base.

2. Mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) is an example for an acidic buffer. Its pH is around 4.75.

Question 16.
1. What is a basic buffer?
2. Suggest an example for basic buffer.
Answer:
1. A basic buffer is a buffer solution having pH greater than 7. It is prepared by mixing a weak base and its salt formed with a strong acid.

2. Mixture of ammonium hydroxide (NH₄OH) and ammonium chloride (NH4CI) is an example for a basic buffer. Its pH is around 9.25.

Plus One Chemistry Equilibrium Three Mark Questions and Answers

Question 1.
The concentration of reactant and products for the reaction, H₂(g) + l₂(g) \(\rightleftharpoons \) 2Hl(g) are recorded as follows:

Reactant or Product	Molar Concentration
H2	0.080
l2	0.060
HI	0.490

a) Write down the expressions for equilibrium constant of the above reaction.
b) Calculate the equilibrium constant at the temperature 298 K if [Hl] = 0.49 M, [H₂]=0.08 M and [l₂]=0.06 M.
Answer:

Question 2.
1. When equilibrium is reached in a chemical reaction?
2. What is the influence of molar concentration in a reaction at equilibrium?
3. Write the expression for equilibrium constant for the decomposition of NH4CI by the reaction,
NH₄Cl \(\rightleftharpoons \) NH₃+HCl
Answer:
1. When the rate of forward reaction is equal to rate of backward reaction, the chemical reaction is said to be in equilibrium.

2. Rate of chemical reaction is directly proportional to the product of molar concentration of the reactants.

3. \(\mathrm{K}=\frac{\left[\mathrm{NH}_{3}\right] \mathrm{HCl}}{\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}\)

Question 3.
1. What is meant by K_p?
2. How K_p is related to K_c?
Answer:
1. K_p is the equilibrium constant in terms of the partial pressures of the reactants and products (Pressure should be expressed in bar as standard state is 1 bar). It is used for reactions involving gases.

2. K_p = K_c (RT)^∆n
where R = universal gas constant, T = absolute temperature and ∆n = number of moles of gaseous product(s) – number of moles of gasesous reactant(s).

Question 4.
a) What do you mean by equilibrium constant?
b) Write any two characteristics of equilibrium constant.
c) Write an expression for equilibrium constant of the reaction, 2SO₂(g) + O₂(g) \(\rightleftharpoons \) SO₃(g).
Answer:
a) Equilibrium constant at a given temperature is the ratio of product of molar concentrations of the products to that of the reactants, each concentration term being raised to the respective stoichiometric coefficients in the balanced chemical equation.

b) 1. The value of equilibrium constant is independent of the initial concentrations of the reactants and products.
2. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

Question 5.
2NO₂(g) \(\rightleftharpoons \) N₂O₄(g); ∆H = -52.7 kJ mol^-1
1. What change will happen if we increase the temperature?
2. What is the effect of increase in pressure in the above equilibrium?
3. What happens when N₂O₄ is removed from the reaction medium?
Answer:
1. Since the forward reaction is exothermic, on increasing temperature the rate of backward reaction (endothermic reaction) increases.

2. Since the number of moles decreases in the forward reaction, on increasing pressure, the rate of forward reaction increases.

3. Rate of forward reaction increases.

Question 6.
Consider this reaction:
CO(g) + 2H₂(g) \(\rightleftharpoons \) CH₃OH(g); ∆_rH = -92 kJ mol^-1
Explain the influence of the following on the basis of
Le Chatelier’s principle.
1. Decrease in pressure.
2. Increase in temperature.
3. Increase in the partial pressure of hydrogen.
Answer:
1. On decreasing pressure the reaction shifts in the direction in which there is increase in the number of moles. Thus, the rate of backward reaction increases on decreasing pressure.

2. On increasing temperature, the rate of endothermic reaction increases. Here, backward reaction is endothermic. Hence, on increasing temperature the rate of backward reaction increases.

3. Hydrogen, being a reactant increase in its partial pressure increases the rate of forward reaction.

Question 7.
The equilibrium showing dissociation of phosgene gas is given below:
COCl₂(g) \(\rightleftharpoons \) CO(g) + Cl₂(g)
When a mixture of these three gases at equilibrium is compressed at constant temperature, what happens to
1. The amount of CO in mixture?
2. The partial pressure of COCl₂?
3. The equilibrium constant for the reaction?
Answer:
1. Amount of CO decreases, because the system favours the reaction in which number of moles decreases with increase of pressure i.e., backward reaction.

2. Increases.
3. Equilibrium constant remains the same since temperature is constant.

Question 8.
The equilibrium constant of the reaction H₂(g) + l₂(g) \(\rightleftharpoons \) 2Hl(g) is 57 at 700 K. Now, give the equilibrium constants for the following reactions at the same temperature:

Answer:

Question 9.
1. What are buffer solutions?
2. Which of the following are buffer solutions?
NaCl + HCl
NH₄Cl + NH₄OH
HCOOH + HCOOK
3. What is the effect of pressure on the following equilibria?
i) Ice \(\rightleftharpoons \) Water
ii) N₂(g) + O₂(g) \(\rightleftharpoons \) 2N0(g)
Answer:
1. These are solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali.

2. NH₄Cl + NH₄OH

3. i) When pressure is increased the melting point of ice decreases and hence the rate of forward reaction will increase.
ii) Pressure has no effect in this equilibrium because there is no change in the number of moles of the gaseous reactants and products.

Question 10.
The aqueous solution of the compounds NaCl, NH₄Cl and CH₃COONa show different pH.

1. Identify the acidic, basic and neutral solution among them.
2. The concentration of hydrogen ion in a soft drink is 4 × 10^-4. What is its pH?

Answer:

1. Acidic-aqueous solution of NH₄Cl Neutral – aqueous solution of NaCl Basic – aqueous solution of CH₃COONa
2. pH = – log[H⁺] = – log[4 × 10^-4] = 3.398

Question 11.
1. What is pH? What is its significance?
2. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3. What is its pH?
Answer:
1. pH is a logarithmic scale used to express the hydronium ion concenration in molarity more conveniently. The pH of a solution is defined as negative logarithm to the base 10 of the activity of hydrogen ion. pH = — log \({ { a }_{ { H }^{ + } } }\) = —log[H⁺]

2. [H⁺] = 3.8 × 10^-3
pH = -log[H⁺]
= -log [3.8 × 10^-3] = -(-2.42) = 2.42

Question 12.
1. State the Le-Chatelier’s principle.
2. Apply the above principle in the following equilibrium and predict the effect of pressure.
CO(g) + 3H₂(g) \(\rightleftharpoons \) CH₄(g) + H₂O(g)
Answer:
1. The Le Chateliers principle states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change.

2. On increasing pressure the rate of forward reaction increases. This is because number of moles decreases in the forward reaction. In other words, the value of Q_c decreases on increasing pressure. As Q_c < K_c, the reaction proceeds in the forward direction.

Question 13.
1. Explain Lewis concept of acids and bases.
2. Why does BF₃ act as a Lewis acid?
Answer:
1. A Lewis acid can be defined as a species which accepts electron pair and a Lewis base is a species which donates an electron pair.

2. In BF₃, the boron atom is electron deficient and it accepts a lone pair of electron. So it acts as a Lewis acid.

Question 14.
1. How the value of AG influence the direction of an equilibrium process?
2. The equilibrium constant for a reaction is 8. What will be the value of ∆G at 27 °C?
Answer:
1. If ∆G is negative, then the reaction is spontaneous and proceeds in the forward direction.
If ∆G is positive, then reaction is considered non- spontaneous. instead, as reverse reaction would have a negative ∆G, the products of the forward reaction shall be converted to the reactants.
If ∆G is 0, reaction has achieved equilibrium. At this point, there is no longer any free energy to drive the reaction.
2.

Question 15.
1. What is common ion effect?
2. Suggest an example for this effect.
Answer:
1. Common ion effect may be defined as the suppression of the dissociation of a weak electrolyte by the addition of some strong electrolyte containing a common ion.

2. The dissociation equilibrium of NH₄OH is shifted towards left in presence of NH₄Cl having the common ion, NH₄⁺.

Question 16.
1. Predict whether an aqueous solution of (NH₄)₂SO₄ is acidic, basic or neutral?
2. Justify your answer.
Answer:
1. An aqueous solution of (NH₄)₂SC₄ is acidic in nature.

2. (NH₄)₂SO₄ is formed from weak base, NH₄OH, and strong acid, H₂SO₄. In water, it dissociates completely
(NH₄)₂SO₄(aq) → 2NH₄⁺ (aq) + SO₄^2- (aq)
NH₄⁺ ions undergoes hydrolysis to form NH₄OH and H⁺ ions.
NH₄+(aq) + H₂O(l) \(\rightleftharpoons \) NH₄OH(aq) + H⁺(aq)
NH₄OH is a weak base and therefore remains almost unionised in solution. This results in increased H⁺ ion concentration in solution making the solution acidic.

Question 17.
1. What are sparingly soluble salts? Suggest an example.
2. Define solubility product constant, K_sp.
3. Obtain the relation between solubility product constant (K_sp) and solubility (S), of a solid salt of general formula M_x^p+ X_y^q-.
Answer:
1. Sparingly soluble salts are those salts with solubility less than 0.01 M.
e.g. BaSO₄

2. The solubility product of a sparingly soluble salt at a given temperature is defined as the product of the concentrations of its ions in the saturated solution, with each concentration term raised to the power equal to the number of times the ion occurs in the equation representing the dissociation of the electrolyte.

3. The equilibrium in the saturated solution of the salt can be represented as,

Question 18.
The Le Chatelier’s principle is applicable to physical and chemical equilibria.
1. What are the factors which can influence the equilibrium state of a system?
2. Explain the factors affecting the chemical equilibrium on the basis of Le Chatelier’s principle taking Haber’s process for the manufacture of ammonia as an example.
Answer:
1. The following factors can influence the equilibrium state of a system:

• Change in concentration of the reactants or products.
• Change in temperature.
• Change in pressure.
• Addition of inert gas.
• Presence of catalyst.

2. N₂(g) + 3H₂(g) \(\rightleftharpoons \) 2NH₃(g); ∆_rH = -91.8 kJ mol^-1
When concentration of N₂ or H₂ is increased, a good yield of NH₃ can be achieved. The rate of forward reaction can also be increased by removing NH₃ from the reaction mixture.

When pressure is increased, the system will try to decrease pressure and for this system will proceed in that direction where there is minimum number of moles i.e., forward reaction. Thus, a good yield of NH₃ can be achieved by increasing pressure.

Since the formation of NH₃ is an exothermic reaction, a good yield of NH₃ can be achieved by decreasing the temperature. But if the temperature is decreased to very low value the reactant molecules do not have sufficient energy to interact. Hence, an optimum temperature of 500°C is used.

Question 19.
1. Soda water is prepared by dissolving CO₂ in water under high pressure. What is the principle involved in this process?
2. At 1000 K, equilibrium constant Kc for the reaction 2SO₃(g) \(\rightleftharpoons \) 2SO₂(g) + O₂(g) is 0.027. What is the value of K_p at this temperature?
Answer:
1. Henry’s law
2. K_p =K_c(RT)^∆n
∆n = 3.2 = 1
K_p = 0.027 × (0.0831 × 1000)¹ = 2.2437

Question 20.
1. For the reaction PCL \(\rightleftharpoons \) PCl₃ +Cl_c
i) Write the expression of K_c.
ii) What happens if pressure is increased?
2. Write the conjugate acid and base of the following species:
i) H20 ii) HCO;
3. Name the phenomenon involved in the preparation of soap by adding NaCI.
Answer:
1. i) \(\kappa_{c}=\frac{\left[\mathrm{PC}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
ii) If we increase the pressure the system will try to decrease the pressure. For this system will proceed in the direction where there is minimum number of moles, i.e., rate of backward reaction increases by decreasing the pressure.

2. i) Conjugate acid of H₂O is H₃O⁺
Conjugate base of H₂O is OH”
ii) Conjugate acid of HCO,” is H₂CO₃ Conjugate base of HCO₃^– is CO₃^2-

3. Common ion effect.

Question 21.
a) The pH of black coffee is 5.0. Calculate the hydrogen ion concentration.
b) The K_sp of barium sulphate is 1.5 × 10^-9. Calculate the solubility of barium sulphate in pure water.
Answer:

Question 22.
1. What is conjugate acid-base pair?
2. Illustrate with an example.
Answer:
1. The acid-base pairthat differs only by one proton is called conjugate acid-base pair. Such acid-base pairs are formed by loss or gain of a proton.

2. Consider the ionization of hydrochloric acid in water.

HCl(aq) acts as an acid by donating a proton to H₂O molecule which acts as a base because it accepts the proton. The species H₂O⁺ is produced when water accepts a proton from HCl. Therefore, Cl^– is the conjugate base of HCl and HCl is the conjugate acid of Cl^–. Similarly, H₂O is the conjugate base of H₂O⁺ and H₃O⁺⁺ is the conjugate acid of the base H₂O.

Question 23.
1. What are the applications of equilibrium constant?
2. What is meant by reaction quotient, Q_c?
3. Predict the direction of net reaction in the following cases:
i) Q_c < K_c
ii) Q_c > K_c
iii) Q_c = K_c
Answer:
1. The applications of equilibrium constant are:
• To predict the extent of a reaction on the basis of its magnitude.
• To predict the direction of the reaction.
• To calculate equilibrium concentrations.

2. Reaction quotient, Q_c at a given temperature is defined as the ratio of the product of concentrations of the reaction products to that of the reactants, each concentration term being raised to their individual stoichiometric coefficients in the balanced chemical equation, where the concentrations are not necessarily equilibrium values.

3. i) When Q_c > K_c, the reaction will proceed in the
direction of reactants (reverse reaction), i.e., net reaction goes from right to left.
ii) When Q_c < K_c, the reaction will proceed in the direction of products (forward reaction), i.e., net reaction goes from left to right.
iii) When Q_c = K_c, the reaction mixture will be at equilibrium, i.e., no net reaction occurs.

Question 24.
Solubility product helps to predict the precipitation of salts from solution.
1. Find the relation between solubility (S) and solubility product (K_sp) of calcium fluoride and zirconium phosphate.
2. The solubility product of two sparingly soluble salts XY₂ and AB are 4 × 10^-15 and 1.2 × 10^-16 respectively. Which salt is more soluble? Explain.
Answer:
1. The equilibrium in the saturated solution of calcium fluoride can be represented as,
CaF₂(s) \(\rightleftharpoons \) Ca²⁺(aq) + 2F^–(aq)
K_sp = [Ca²⁺][F^–]² = S.(2S)₂ = 4S³
The equilibrium in the saturated solution of zirconium phosphate can be represented as,
Zr₃(PO₄)₄(s) \(\rightleftharpoons \) 3Zr⁴⁺(aq) + 4PO₄^3-(aq)
K_sp = [Zr⁴⁺]³[PO₄^3-]⁴ = (3S)³.(4S)⁴ = 6912S⁷

2. XY₂ is more soluble than AB.

Question 25.
a) How common ion effect can influence the solubility of ionic salts?
b) What is the application of common ion effect in gravimetric estimation?
Answer:
1. In a salt solution, if we increase the concentration of any one of the ions, according to Le Chatelier’s principle, it should combine with the ion of its opposite charge and some of the salt will be precipitated till K_sp = Q_sp. Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till K_sp = Q_sp.

2. The common ion effect is used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation.

Plus One Chemistry Equilibrium Four Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2.
In Contact process, SO₃ is prepared by the oxidation of SO₂ as per the following reaction:
2SO₂(g) + O₂(g) \(\rightleftharpoons \) 2SO₃(g); ∆H = -189.4
a) What happens to the rate of forward reaction when i) temperature is increased?
ii) pressure is decreased?
iii) a catalyst V₂O₅ is added?
b) Calculate the pH of 0.01 M H₂SO₄ solution. Also, calculate the hydroxyl ion concentration in the above solution.
Answer:
1. D When temperature is increased, the rate of forward reaction decreases since it is exothermic.
ii) When pressure is decreased the rate of forward reaction decreases since it is associated with decrease in number of moles.
iii) When a catalyst V₂O₅ is added the rate of both forward and backward reactions are increased by the same extent and equilibrium is reached earlier.
2.

Question 3.
Calculate the [H⁺] in the following biological fluids whose pH are given in brackets.
i) Human muscle fluid (6.83)
ii) Human stomach fluid (1.22)
iii) Human blood (7.38)
iv) Human saliva (6.4)
Answer:
i) pH =-log[H+] = 6.83
log[H⁺] = – 6.83
[H⁺] = antilog (- 6.83) = 1.48 × 10^-7 mol L^-1
ii) [H⁺] = antilog (- 1.22) = 6.03 × 10^-2 mol L^-1
iii) [H⁺] = antilog (- 7.38) = 4.17 × 10^-8 mol L^-1
iv) [H⁺] = antilog (- 6.4) = 3.98 × 10^-7 mol L^-1

Question 4.
The pH value of a solution determines whether it is acidic, basic or neutral in nature.
1. The concentration of hydrogen ion in the sample of a soft drink is 3.8 × 10^-3 mol/L. Calculate its pH. Also predict whether the above solution is acidic, basic or neutral.
2. The dissociation constants of formic acid (HCOOH) and acetic acid (CH₃COOH) are 1.8 × 10^-4and 1.8 × 10^-4 respectively. Which is relatively more acidic? Justify your answer.
Answer:
1. pH = – log[H⁺] = – log[3.8 × 10^-3] = 2.42
Since pH is less than 7, it is an acidic solution,

2. HCOOH is more acidic.
K_a value is directly proportional to the acid strength, i.e., greater the K_a value, stronger is the acid.

Question 5.
a) Write the expression for Henderson – Hasselbalch equation for i) An acidic buffer & ii) A basic buffer.
b) Calculate the pH of a solution which is 0.1 M in
CH₃COOH and 0.5 M in CH₃COONa. Ka for CH₃COOH is 1.8 × 10^-6.
Answer:

Plus One Chemistry Equilibrium NCERT Questions and Answers

Question 1
A liquid is in equilibrium with its vapour in a sealed containerat a fixed temperature. The volume of the container is suddenly increased. (3)
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
a) Vapour pressure decreases due to increase in volume.
b) Rate of evaporation remains same and rate of condensation decreases.
c) Finally the same vapour pressure is restored and the rate of evaporation becomes equal to the rate

Question 2.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH? (2)
Answer:
pH = -log[H⁺]
= -log (3.8 × 10^-3) = 2.42

Question 3.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. (2)
Answer:
H= -log [H⁺]
or log [H⁺] = -3.76 = -4.24
[H⁺] = antilog (-3.76) = 1.74 × 10^-4M

Question 4.
The ionization constants of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base. (3)
Answer:
The relation between ionization constant of an acid and that of its conjungate base is Ka x Kb = Kw

Question 5
The ionization constant of nitrous acid is 4.5 x 10-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. (2)
Answer:
Sodium nitrite is a salt of strong base and weak acid. Its degree of hydrolysis, h is given by the relation.

Question 6
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine. (2)
Answer:
Pyridinium hydrochloride is a salt of a weak base (pyridine) and a strong acid (HCl). The pH of an aqueous solution of this salt is given by the relation:

Kerala Plus One Maths Model Question Paper 3

Time Allowed: 2 1/2 hours
Cool off time: 15 Minutes
Maximum Marks: 80

General Instructions to Candidates :

• There is a ‘cool off time’ of 15 minutes in addition to the writing time .
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read instructions carefully .
• Read questions carefully before you answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• .Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Questions 1 to 7 carry 3 score each. Answer any 6.

Question 1.
a. Write set is the subset of all the given sets?
(a) {1,2,3,….}
(b) {1}
(c) {0}
(d) {}
b. Write down the power set of A = {1,2,3}

Question 2.
In any triangle ABC, prove that
a(SinB- sinC) + b(sinC- sinA) +c(sinA- sinB) = 0

Question 3.
a. Solve x² + 2 = 0
b. Find the multiplicative inverse of 2-3i
a. x² + 2 = 0

Question 4.
a. Solve

b. Find the graphical solution of the above inequality.

Question 5.
a. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of atmost 3 girls,
b. Find n if ²ⁿC₃: ⁿC₃ = 12 : 1

Question 6.
a. The new coordinates of the points (2, 5) if the origin is shifted to the point (1, 1) by translation of axes.
A) (3, 1) B(1, 3) C (-1, 3) D (3, -1)
b. Find what the equation x² + xy – 3y²– y + 2 = 0 becomes when the origin is shifted to the point (1,1)?

Question 7.
Evaluate :

Questions from 8 to 17 carry 4 score each. Answer any 8.

Question 8.

Question 9.
Consider the statement 1.3 + 2.3² + 3.3³

a. Verify whether the statement is true for n =1.
b. Prove the result by using principle of mathematical induction.

Question 10.
a. Express (1 + i)³ + (1 – i)³ in a + ib form.
b. Find the polar form of the complex number – 1 – i.

Question 11.
solve the following linear inequalities graphically:

Question 12.
a. “P_r =

b. The letters of the word FATHER be permuted and arranged in a dictionary, find the rank of the word FATHER?

Question 13.
a. The distance of the point P(1,-3) from the line 2y – 3x = 4 is
A) 13
B)  \(\frac { 7 }{ 13 } \sqrt { 3 } \)
C) √13
D) None of these
b. Reduce the equation√3x + y + 8 = 0 in to normal form. Find the values of P and ω
imagee

Question 14.
a. A conic with e = 0 is known as
A) a parabola
B) an ellipse
C) a hyperbola
D) a circle
b. Consider the circle x² + y² + 8x + 10y – 8 = 0
i) Find the centre C and radius ‘r’.
ii) Find the equation of the circle with centre at C and passing through the point (1, 2)

Question 15.
a. What is the perpendicular distance from the point P(6,7,8) from XY plane.
A) 8
B) 7
C) 6
D) 9
b. Find the equation of the set of points P, the sum of whose distances from A(4,0,0) and B(-4,0,0) is equal to 10.

Question 16.
a. Convert 20°40‘ into radian measure.
b. If sin x = \(\frac { 12 }{ 13 } \) and x is an acute angle, find the value of cos 2x.
c. Prove that
\(\frac { sinx-sin\quad 3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } \)
= 2 sin x.

Question 17.
a. Write the component statements of the following statement: All prime numbers are either even or odd
b. Verify by the method of contradiction, p = √7 is irrational

Questions from 18 to 24 carry 6 score each. Answer any 5.

Question 18.
a. Write the interval (6,12) in the set-builder form.
b. Draw the Venn diagram of the following sets :
i) A’ ∩ B’
ii)A – B
c. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis? How many like tennis only and not cricket?

Question 19.
a. Write the number of terms in the expansion of (a – b)²ⁿ
b. Find the general term in the expansion of (x²– yx)¹², x ≠ 0
c. Find the coefficient of x⁶ y³ in the expansion of (x + 2y)⁹

Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits.

Question 20.
a. The common ratio of the G.P is \(\frac { -4 }{ 5 } \) and the sum to infinity is \(\frac { -80 }{ 9 } \). Find the first term.
b. Evaluate:

c. Find the sum of first n terms of the series : 0.6+0.66+0.666+…….to n terms.

Question 21.
a. Find the derivative of xⁿ from first principles.

Question 22.
a. Find the point of intersection of the lines 2x + y = 5 and x + 3y + 8 = 0.
b. Find the equation of a line passing through the point of intersection of the above lines and parallel to the line 3x + 4y =7.
c. Find the distance between these two paralle lines.

Question 23.
a. Find the Mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from mean is ………

b. Calculate mean, variance and standard deviation of the following distribution:

Question 24.
a. In aleap year the probability of having 53 Sundays is ……..

b. Events E and F are such that P (not E or not F) = 0.25, state whether E and F are mutually exclusive.
c. How many points are there in a sample space, if a card is selected from a pack of 52 cards.

Answer

Answer 1.
a. d) { }
b. P(A)={{1,2,3}, {1,2}, {2,3}, {1,3}, {1}, {2}, {3}, ø}

Answer 2.
LHS = a (sinB-sinC) + 6(sinC-sinA) +c (sinA-sinB)
= 2R sin A(sinB-sinC) + 2R sin B (sinC-sinA)+2R sinC (sinA – sinB)
= 2R [sin A sin B – sinAsinC + sinB sinC – sinBsinA + sinCsinA – sin C sinB]
= 2R x 0 = 0 = RHS

Answer 3.

Answer 4.

Answer 5.

Answer 6.
a. B) (1, 3) since [2 – 1, 5 – 2]
b. Let the coordinates of a point P changes from (x,y) to (X,Y) when origin is shifted to (1,1)
∴ x = X + 1, y = Y + 1 Substituting in the given equation, we get
(X+1)² + (X + 1) (7 + 1) -3 (7 + 1)² -(7 + 1) + 2 = 0
⇒ X²+2X + 1 + XY + X+ Y+ 1-3 (P + 27 + 1) – (7+ 1) + 2 – 0
⇒ X² – 3Y² + XY + 3X- 6Y = 0
∴ Equation in new system is
X² – 3Y² + XY + 3X -6Y = 0

Answer 7.

Answer 8.

Answer 9.

Answer 10.

Answer 11.

Answer 12.
a. D
b. The alphabetical order of the word ‘ FATHER are A, E, F, H, R, T No. of words beginning with

Answer 13.

Answer 14.
a. D
b. Comparing x² + y¹ – 8x +10 y – 12 = 0 with

Answer 15.
a. A. 8 [ || perpendicular distance of a point from XY plane = z coordinate
b. Let P(x, y, z) be the point such that PA + PB = 10

Answer 16.

Answer 17.
a. p: All prime numbers are even. q: All prime numbers are odd
b. Let us assume that √7 is a rational number
∴ √7 = , where a and b are co-prime, i.e. a and b have no common factors, which implies that 7b² – a² ⇒ 7 divides a.
∴ there exists an integer ‘k’ such that a = 7k
a² = 49k² ⇒ 7b² = 49k²⇒b²= 7k² 7 divides b.
i.e., 7 divides both a and b, which is contradiction to our assumption that a and b have no common factor.
∴ our supposition √7 is wrong, is an irrational number.

Answer 18.
a. {x : x ∈ R, 6 < x ≤ 12}

Answer 19.
a. Number of terms in expansion can be given by 2n + 1
b. General term can be given by

Answer 20.

Answer 21.

Answer 22.
a.  2x + y =5 ……………. (1)
x + 3y = -8………………….. (2)
(1) x 3 – (2) ⇒
6x + 3y = 15 x + 3y = -8
(-) (-) (+)
————-
5x  = 7 7
∴ x = 7/5

b. Slope of the required line, m = \(-\frac { 3 }{ 4 } \)
|| slope of the parallel line Equation of the line is y – y = m (x – x₁)

Answer 23.
a. b


c. The parallel lines 15x + 20y – 57 = 0 and
3x + 4y – 7 = 0 ⇒15x + 20y – 35=0 x ing by 5
Parallel distance

Answer 24.
a. (b) 7/3
b. (not E or not F)
= P(E ‘∪ F’) = P(E ∩ F)’ = 1 = 1P(E ∩ F)
⇒ 0.25 = 1 -P(E n F) ⇒ p(E ∩ F) = 1 -0.25 = 0.75 ≠ 0
⇒ E and F are not mutually exclusive
c. 52

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