Explore topic-wise InterviewSolutions in .

(a) A solution in which at a given temperature, no more solute can be dissolved by the solvent is called a saturated solution. The concentration of such solutions is the maximum at a given temperature.
To illustrate this, take a glass or a beaker and add some common salt to it. Fill the beaker with water and stir it well. One now gets a salt solution. Continue adding salt slowly to the beaker along with stirring. At some point of time, the solution reaches a state where even though salt is added and stirred, some of the salt stays undissolved. Then one can say that the solution has reached its saturation level at that temperature.
(b) A solution in which at a given temperature, more solute can be dissolved by the solvent is called an unsaturated solution. Unsaturated solutions can be made saturated by adding more solute to it. In the activity mentioned in (a), the solution obtained in the beginning is an unsaturated solution.
(c) The suspension is a heterogeneous mixture of 2 or more components. The particles forming the suspension are visible to the naked eye. In this mixture, the solute particles do not dissolve in the solvent, instead, they remain suspended in the solvent medium.When left undisturbed, the solute particles tend to settle down at the bottom of the container.

Eg: Mixture of sand and water forms a suspension.

a) Strong acid : The acid which undergoes 100% ionisation is called strong acid. 

e.g.: HCl, H₂SO₄ 

b) Strong base : The base which undergoes 100% ionisation is called strong base. 

e.g.: NaOH, KOH 

c) Weak acid: The acid which undergoes less than 100% ionisation is called weak acid. 

e.g.: CH₃COOH, H₂CO₃ 

d) Weak base: The base which undergoes less than 100% ionisation is called weak base. 

e.g.: NH₄OH, Mg(OH)₂ 

At the pH value of 1 the pH paper strip is Red, at the pH value of 7 the pH paper strip is green and at the pH value of 14 the pH paper strip is blue in colour.

Sulphuric acid is a strong acid amongst all of the above.

According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.

This mixture contains three components: sand, water and mustard oil. Now, sand is a solid which is insoluble in water as well as mustard oil. Water and mustard oil are immiscible liquids. 

(a) The mixture of sand, water and mustard oil is filtered. Sand is left on the filter paper as residue. Water and mustard oil collect as filtrate. 

(b) The filtrate containing water and mustard oil is put in a separating funnel. Water forms the lower layer and mustard oil forms the upper layer in separating funnel. The lower layer of water is fun out first by opening the stop-cock of the separating funnel. Mustard oil remains behind in the separating funnel and can be removed separately.

Separation of mixture of two liquids the miscible liquid (which mix together) in all proportion and form single layer) and immiscible liquid (which do not mix with each other and form a separate layers) are separated by the two methods).

• By the fractional distillation (for miscible liquid) 

To separate a mixture of two or more miscible liquids for which the difference in boiling points is less than 25 K, fractional distillation process is used, for example, for the separation of different gases from air, different factions from petroleum product etc. The apparatus is similar to that for simple distillation, except that a fractionating column is fitted in between the distillation flask and the condenser.

 A simple fractionating column is a tube packed with glass beads. The beads provide surface for the vapor to cool and condense repeatedly.

Thermal energy is the energy of a body arising from motion of its atoms or molecules.

At 273.15K and 1atm pressure every gas has molar volume = 22.4L.

√0.1764 = √(1764/10000)

The given system of equations is: 

2x – 3y – 7 = 0 

(k+2)x – (2k+1)y – 3(2k-1) = 0 

The above equations are of the form 

a₁ x + b₁ y − c₁ = 0 

a₂ x + b₂ y − c₂ = 0 

Here, a₁ = 2, b₁ = -3, c₁ = -7 

a₂ = (k+2), b₂ = -(2k+1), c₂ = -3(2k-1) 

So according to the question, 

For unique solution, the condition is 

\(\frac{a_1 }{ a_2} = \frac{b_1 }{ b_2} = \frac{c_1}{ c_2}\) 

\(\frac{2}{(k+2)}\) = \(\frac{−3}{ −(2k+1)}\) = \(\frac{−7}{−3(2k−1)}\) 

\(\frac{2}{(k+2)}\) = \(\frac{−3}{ −(2k+1)}\) and \(\frac{−3}{ −(2k+1)}\) = \(\frac{−7}{−3(2k−1)}\) 

⇒ 2(2k+1) = 3(k+2) and 3 x 3(2k−1) = 7(2k+1) 

⇒ 4k + 2 = 3k + 6 and 18k – 9 = 14k + 7 

⇒ k = 4 and 4k = 16 

⇒ k = 4

Hence, the given system of equations will have infinitely many solutions, if k = 4.

The given system of equations is: 

2x + 3y – 2= 0 

(k+2)x + (2k+1)y – 2(k-1) = 0 

The above equations are of the form 

a₁ x + b₁ y − c₁ = 0 

a₂ x + b₂ y − c₂ = 0 

Here, a₁ = 2, b₁ = 3, c₁ = -5 

a₂ = (k+2), b₂ = (2k+1), c₂ = -2(k-1) 

So according to the question, 

For unique solution, the condition is 

\(\frac{a_1 }{ a_2} = \frac{b_1 }{ b_2} = \frac{c_1 }{ c_2}\) 

\(\frac{2}{ (k+2)} = \frac{3}{ (2k+1)} = \frac{−2}{ −2(k−1)}\)

2/ (k+2) = 3/ (2k+1) and 3(2k+1) = 22(k−1) 

⇒2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1) 

⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1 

⇒ k = 4 and k = 4 

Hence, the given system of equations will have infinitely many solutions, if k = 4.

(i) ∠ABC = ∠y = 100^o (opposite angles are equal in a parallelogram)

∠x + ∠y = 180^o (sum of adjacent angles is = 180^o in a parallelogram)

∠x + 100^o = 180^o

∠x = 180^o – 100^o

= 80^o

∴ ∠x = 80^o ∠y = 100^o∠z = 80^o (opposite angles are equal in a parallelogram)

(ii) ∠RSP + ∠y = 180^o (sum of adjacent angles is = 180^o in a parallelogram)

∠y + 50^o = 180^o

∠y = 180^o – 50^o

= 130^o

∴ ∠x = ∠y = 130^o (opposite angles are equal in a parallelogram)

∠RSP = ∠RQP = 50^o (opposite angles are equal in a parallelogram)

∠RQP + ∠z = 180^o (linear pair)

50^o + ∠z = 180^o

∠z = 180^o – 50^o

= 130^o

∴ ∠x = 130^o ∠y = 130^o ∠z = 130^o

(iii) In ΔPMN

∠NPM + ∠NMP + ∠MNP = 180° [Sum of all the angles of a triangle is 180°]

30° + 90° + ∠z = 180°

∠z = 180°-120°

= 60°

∠y = ∠z = 60° [opposite angles are equal in a parallelogram]

∠z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]

∠z = 60°

∠z + ∠LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

60° + 90°+ ∠x = 180°

∠x = 180°-150°

∠x = 30°

∴ ∠x = 30^o ∠y = 60^o ∠z = 60^o

(iv) ∠x = 90° [vertically opposite angles are equal]

In ΔDOC

∠x + ∠y + 30° = 180° [Sum of all the angles of a triangle is 180°]

90° + 30° + ∠y = 180°

∠y = 180°-120°

∠y = 60°

∠y = ∠z = 60° [alternate interior angles are equal]

∴ ∠x = 90^o ∠y = 60^o ∠z = 60^o

(v) ∠x + ∠POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

∠x + 80° = 180°

∠x = 180°-80°

∠x = 100°

∠y = 80° [opposite angles are equal in a parallelogram]

∠SRQ =∠x = 100°

∠SRQ + ∠z = 180° [Linear pair]

100° + ∠z = 180°

∠z = 180°-100°

∠z = 80°

∴ ∠x = 100^o ∠y = 80^o ∠z = 80^o

(vi) ∠y = 112° [In a parallelogram opposite angles are equal]

∠y + ∠VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠z + 40° + 112° = 180°

∠z = 180°-152°

∠z = 28°

∠z =∠x = 28° [alternate interior angles are equal]

∴ ∠x = 28^o ∠y = 112^o ∠z = 28^o

No, Diagonals of a rhombus bisect each other at 90°. Where, diagonals must be perpendicular.

(i) A rhombus is a parallelogram in which adjacent sides are equal.

(ii) A square is a rhombus in which one angle is right angle.

(iii) A rhombus has all its sides of equal length.

(iv) The diagonals of a rhombus bisect each other at right angles.

(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

In parallelogram BDEF

BD = EF [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF [In a parallelogram opposite sides are equal]

Since, BD = EF = DC

So, BD = DC

πx + y = 9 For x = 0,

π(0) + y = 9

⇒ y = 9

π(1) + y = 9

⇒y = 9 − π

π(2) + y = 9

⇒ y = 9 − 2π

π(−1) + y = 9

⇒ y = 9 + π

⇒ (−1, 9 + π) is a solution of this equation.

R : A → A, A = {1, 2, 3,4}

R = {(a, b) / |a – b| ≥ 0}

= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

∴ R = A × A

R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}

Here, y > x + 1

When x = 1 and y = 2, 2 ≯  1 + 1

When x = 1 and y = 4, 4 > 1 + 1

When x = 1 and y = 6, 6 > 1 + 1

When x = 2 and y = 2, 2 ≯  2 + 1

When x = 2 and y = 4, 4 > 2 + 1

When x = 2 and y = 6, 6 > 2 + 1

∴ R₄ = {(1, 4), (1, 6), (2, 4), (2, 6)}

Domain (R₄) = {1, 2}

Range (R₄) = {4, 6}

R : A → A, A = {1, 2, 3,4}

R = {(a, b)/a – b = 10} = { }

R = {(a, b) / a ∈ N, a < 5, b = 4}

∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}

Range (R) = {b / b = 4} = {4}

R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}

Since a ∈ Z and |a| < 3, a < 3 and a > -3

∴ -3 < a < 3

∴ a = -2, -1, 0, 1, 2

b = |a – 1|

When a = -2, b = 3

When a = -1, b = 2

When a = 0, b = 1

When a = 1, b = 0

When a = 2, b = 1

Domain (R) = {-2, -1, 0, 1, 2}

Range (R) = {0, 1, 2, 3}

A = {1, 2, 3}, B = {4, 5, 6}

∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}

Since R₃ ⊆ A × B, R is a relation from A to B.

Domain (R₃) = Set of first components of R = {1, 2, 3}

Range (R₃) = Set of second components of R = {4, 5, 6}

A = {1, 2, 3}, B = {4, 5, 6}

∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}

Since (4, 2) ∈ R₄ , but (4, 2) ∉ A × B, R₄ ⊄ A × B

∴ R₄ is not a relation from A to B.

R = {(a, a² ) / a is a prime number less than 15}

∴ a = 2, 3, 5, 7, 11, 13

∴ a = 4, 9, 25, 49, 121, 169

∴ R = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}

∴ Domain (R₁) = {a/a is a prime number less than 15}

= {2, 3, 5, 7, 11, 13}

Range (R₁ ) = {a /a is a prime number less than 15}

= {4, 9, 25, 49, 121, 169}

A = {1, 2, 3}, B = {4, 5, 6}

∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

R₂ = {(1, 5),(2, 4),(3, 6)}

Since R₂ ⊆ A × B,

R₂ is a relation from A to B.

Domain (R₂) = Set of first components of R₂ = {1, 2, 3}

Range (R₂ ) = Set of second components of R = {4, 5, 6}

A = {1, 2, 3}, B = {4, 5, 6}

∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

R₁ = {(1, 4), (1, 5), (1, 6)}

Since R ⊆ A × B,

R₁ is a relation from A to B.

Domain (R₁) = Set of first components of R = {1}

Range (R₁) = Set of second components of R = {4, 5, 6}

A = {-1, 1}

∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

A = {1, 2, 3} and B = {2, 4}

A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}

B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}

B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}

∴ (A × B) ∩ (B × A) = {(2, 2)}

{(x, y) / x²+ y² = 100, where x, y ∈ W}

We have, x² + y² = 100

When x = 0 and y = 10,

x² + y² = 0² + 10² = 100

When x = 6 and y = 8,

x² + y² = 6² + 8² = 100

When x = 8 and y = 6,

x² + y² = 8² + 6² = 100

When x = 10 and y = 0,

x² + y² = 10² + 0² = 100

∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}

(i) B ∩ C = {5, 6}

A × (B ∩ C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

∴ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) B ∪ C = {4, 5, 6}

A × (B ∪ C)= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

∴ A × (B ∪ C) = (A × B) ∪ (A × C)

Correct option is (D) 24N

6N = {6x : x ∈ N} = {6, 12, 18, 24, 30, ……}

8N = {8x : x ∈ N} = {8, 16, 24, 32, ……}

∴ 6N ∩ 8N = {24, 48, 72, …..}

= {24x : x ∈ N}

= 24N

(C) Transitive

For any a ∈ N, a ≯ a

∴ (a, a) ∉ R

∴ > is not reflexive. 

For any a, b ∈ N, if a > b, then b ≯ a.

∴ > is not symmetric.

For any a, b, c ∈ N,

if a > b and b > c, then a > c

∴ > is transitive.

n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5

(i) n(A ∪ B) = n(X) – [n(A ∪ B)’] = n(X) – n(A’ ∩ B’)

= 50 – 5

= 45

(ii) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)

= 35 + 20 – 45

= 10

(iii) n(A’ ∩ B) = n(B) – n(A ∩ B)

= 20 – 10

= 10

(iv) n(A ∩ B’) = n(A) – n(A ∩ B)

= 35 – 10

= 25

(D) A subset of A × B

A = {1, 2, 3}

The power set of A is given by

P(A) = {{Φ}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

H ≡ C is not a Bronsted-Lowry aci

OH^- is not both a Bronsted-Lowry acid and a Bronsted-Lowry bas.

(E)  Liquid NH₃

H₂ gas is liberated. It burns with pop sound when burning candle is brought near the gas.

2NaOH + Zn → Na₂ZnO₂ + H₂ 

Explanation- When metal react to base then metallic salt and hydrogen (H₂) gas are formed.

Balancing of chemical equation-

(i) Here one Na atom is in left side and 2 Na atom are in right side. So to equate the Na atoms, we will multiply in NaOH by 2.

2NaOH + Zn → Na₂ZnO₂ + H₂ 

(ii) And we shall check that other atoms are equal or not. If the other atoms are equal to both side then the equation is balanced. If the other atoms are not equal then we shall make them equal. 

Now Na atoms are equal, Zn atoms are equal, H atoms are equal and O atoms are also equal on both sides. So the chemical equation is balanced.

(A)    NaNH₂

Higher the concentration, lower will be pH of the solution.

1 M HCl has a higher concentration of H⁺ ions than 1 M CH₃COOH because HCl is a strong acid then CH₃COOH hence 1M HCL produces more amount of H⁺ ions than that produced by 1 M CH₃COOH.

1 M HCl has higher concentration of H⁺ ions.

HCl → H⁺ +Cl^-

1 M HCI has a higher concentration of H⁺ ions because when HCI dissolves in water it dissociates completely into ions while CH₃COOH is a weak acid and does not dissociate into ions completely.

Answer is one H⁺