CBSE Class 9 – Page 119

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D

January 5, 2022June 22, 2018 by Prasanna

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D.

Other Exercises

Question 1.
Solution:
(i) Radius of sphere = 3.5cm
(a) Volume = \(\frac { 4 }{ 3 } \) πr³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q1.1

Question 2.
Solution:
Let r be the radius of the sphere and volume = 38808 cm³
∴\(\frac { 4 }{ 3 } \) πr³ = 38803
=> \(\frac { 4 }{ 3 } \) x \(\frac { 22 }{ 7 } \) r³ = 38803
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q2.1

Question 3.
Solution:
Let r be the radius of the sphere
∴ Volume = \(\frac { 4 }{ 3 } \) πr³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q3.1

Question 4.
Solution:
Surface area of a sphere = 394.24 m²
Let r be the radius, then 4πr² = 394.24
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q4.1

Question 5.
Solution:
Surface area of sphere = 576π cm²
Let r be the radius, then 4r² = 576π
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q5.1

Question 6.
Solution:
Outer diameter of shell = 12cm,
Outer radius (R) = \(\frac { 12 }{ 2 } \) = 6cm
and inner diameter = 8cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q6.1

Question 7.
Solution:
Length of cuboid of (l) = 12cm
Breadth (b) = 11cm
and height (h) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q7.1

Question 8.
Solution:
Radius of sphere (r) = 8cm
Volume = \(\frac { 4 }{ 3 } \)πr³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q8.1

Question 9.
Solution:
Radius of solid sphere (R) = 3cm.
Volume = \(\frac { 4 }{ 3 } \)π(R)³ = \(\frac { 4 }{ 3 } \)π(3)³ cm³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q9.1

Question 10.
Solution:
Radius of metallic sphere (R) = 10.5cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q10.1

Question 11.
Solution:
Diameter of a cylinder = 8cm
Radius (r) = \(\frac { 8 }{ 2 } \) = 4cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q11.1

Question 12.
Solution:
Diameter of sphere = 6cm
Radius (R) = \(\frac { 6 }{ 2 } \) = 3cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q12.1

Question 13.
Solution:
Diameter of sphere = 18cm
Radius (R) = \(\frac { 18 }{ 2 } \) = 9cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q13.1

Question 14.
Solution:
Diameter of the sphere = 15.6 cm
Radius (R) = \(\frac { 15.6 }{ 2 } \) = 7.8 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q14.1

Question 15.
Solution:
Diameter of the canonball = 28cm
Radius (R) = \(\frac { 28 }{ 2 } \) = 14 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q15.1

Question 16.
Solution:
Given,
Radius of spherical big ball (R) = 3cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q16.1

Question 17.
Solution:
Ratio in the radii of two spheres = 1:2
Let radius of smaller sphere = r then,
radius of bigger sphere = 2r
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q17.1

Question 18.
Solution:
Let r1 and r2 be the radii of two spheres
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q18.1

Question 19.
Solution:
Radius of the cylindrical tub = 12cm.
First level of water = 20cm
Raised water level = 6.75cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q19.1

Question 20.
Solution:
Radius of the ball (r) = 9cm.
Volume of ball = \(\frac { 4 }{ 3 } \)πr³
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q20.1

Question 21.
Solution:
Given,
Radius of hemisphere of lead (r) = 9cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q21.1

Question 22.
Solution:
Given,
Radius of hemispherical bowl (r) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q22.1

Question 23.
Solution:
External radius of spherical shell (R) = 9cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q23.1

Question 24.
Solution:
Inner radius (r) = 4 cm
Thickness of steel used = 0.5
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D Q24.1

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

January 5, 2022June 22, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

Other Exercises

Question 1.
Write the complement of each of the following angles:
(i) 20°
(ii) 35°
(iii) 90°
(iv) 77°
(v) 30°
Solution:
We know that two angles are complement to each other if their sum is 90°. Therefore,
(i) Complement of 20° is (90° – 20°) = 70°
(ii) Complement of 35° is (90° – 35°) = 55°
(iii) Complement of 90° is (90° – 90°) = 0°
(iv) Complement of 77° is (90° – 77°) = 13°
(v) Complement of 30° is (90° – 30°) = 60°

Question 2.
Write the supplement of each of the following angles:
(i) 54°
(ii) 132°
(iii) 138°
Solution:
We know that two angles are supplement to each other if their sum if 180°. Therefore,
(i) Supplement of 54° is (180° – 54°) = 126°
(ii) Supplement of 132° is (180° – 132°) = 48°
(iii) Supplement of 138° is (180° – 138°) = 42°

Question 3.
If an angle is 28° less than its complement, find its measure.
Solution:
Let required angle = x, then
Its complement = x + 28°
∴ x + x + 28° = 90° ⇒ 2x = 90° – 28° = 62°
∴ x = \(\frac { { 62 }^{ \circ } }{ 2 }\) = 31°
∴ Required angle = 31°

Question 4.
If an angle is 30° more than one half of its complement, find the measure of the angle.
Solution:
Let the measure of the required angle = x
∴ Its complement = 90° – x
∴ x = \(\frac { 1 }{ 2 }\) (90° – x) + 30°
2x = 90° – x + 60°
⇒ 2x + x = 90° + 60°
⇒ 3x = 150°
⇒ x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°
∴ Required angle = 50°

Question 5.
Two supplementary angles are in the ratio 4 : 5. Find the angles.
Solution:
Ratio in two supplementary angles = 4 : 5
Let first angle = 4x
Then second angle = 5x
∴ 4x + 5x = 180
⇒ 9x = 180°
∴ x = \(\frac { { 180 }^{ \circ } }{ 9 }\) = 20°
∴ First angle = 4x = 4 x 20° = 80°
and second angle = 5x
= 5 x 20° = 100°

Question 6.
Two supplementary angles differ by 48°. Find the angles.
Solution:
Let first angle = x ”
Then second angle = x + 48°
∴ x + x + 48° = 180°⇒ 2x + 48° = 180°
⇒ 2x = 180° – 48° = 132°
x= \(\frac { { 132 }^{ \circ } }{ 2 }\) =66°
∴ First angle = 66°
and second angle = x + 48° = 66° + 48° = 114°
∴ Angles are 66°, 114°

Question 7.
An angle is equal to 8 times its complement. Determine its measure.
Solution:
Let the required angle = x
Then its complement angle = 90° – x
∴ x = 8(90° – x)
⇒ x = 720° – 8x ⇒ x + 8x = 720°
⇒ 9x = 720° ⇒ x = \(\frac { { 720 }^{ \circ } }{ 9 }\) = 80°
∴ Required angle = 80°

Question 8.
If the angles (2x – 10)° and (x – 5)° are complementary angles, find x.
Solution:
First complementary angle = (2x – 10°) and second = (x – 5)°
∴ 2x – 10° + x – 5° = 90°
⇒ 3x – 15° = 90° ⇒ 3x = 90° + 15° = 105°
∴ x = \(\frac { { 105 }^{ \circ } }{ 3 }\) = 35°
∴ First angle = 2x – 10° = 2 x 35° – 10°
= 70° – 10° = 60°
and second angle = x – 5 = 35° – 5 = 30°

Question 9.
If an angle differ from its complement by 10°, find the angle.
Solution:
Let required angle = x°
Then its complement angle = 90° – x°
∴ x – (90° – x) = 10
⇒ x – 90° + x = 10°⇒ 2x = 10° + 90° = 100° 100°
⇒ x = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50°
∴ Required angle = 50°

Question 10.
If the supplement of an angle is two-third of itself Determine the angle and its supplement.
Solution:
Let required angle = x
Then its supplement angle = 180° – x
∴ (180°-x)= \(\frac { 2 }{ 3 }\)x
540° – 3x = 2x ⇒ 2x + 3x = 540°
⇒ 5x = 540°⇒ x = \(\frac { { 540 }^{ \circ } }{ 5 }\) = 108°
-. Supplement angle = 180° – 108° = 72°

Question 11.
An angle is 14° more than its complementary angle. What is its measure?
Solution:
Let required angle = x
Then its complementary angle = 90° – x
∴ x + 14° = 90° – x
x + x = 90° – 14° ⇒ 2x = 76°
⇒ x = \(\frac { { 76 }^{ \circ } }{ 2 }\) = 38°
∴ Required angle = 38°

Question 12.
The measure of an angle is twice the measure of its supplementary angle. Find its measure.
Solution:
Let the required angle = x
∴ Its supplementary angle = 180° – x
∴ x = 2(180°-x) = 360°-2x
⇒ x + 2x = 360°
⇒ 3x = 360°
⇒ x = \(\frac { { 360 }^{ \circ } }{ 3 }\) = 120°
∴ Required angle = 120°

Question 13.
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Solution:
Let required angle = x
Then its complement angle = 90° – x
and supplement angle = 180° – x
∴ 3(90° – x) = 180° – x
⇒ 270° – 3x = 180° – x
⇒270° – 180° = -x + 3x => 2x = 90°
⇒ x = 45°
∴ Required angle = 45°

Question 14.
If the supplement of an angle is three times its complement, find the angle.
Solution:
Let required angle = x
Then its complement = 90°- x
and supplement = 180° – x
∴ 180°-x = 3(90°-x)
⇒ 180° – x = 270° – 3x
⇒ -x + 3x = 270° – 180°
⇒ 2x = 90° ⇒ x = \(\frac { { 90 }^{ \circ } }{ 2 }\) =45°
∴ Required angle = 45°

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

January 5, 2022June 22, 2018 by Prasanna

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C.

Other Exercises

Question 1.
Solution:
Radius of base (r) = 35cm
and height (h) = 84cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q1.1

Question 2.
Solution:
Height of cone (h) = 6cm
Slant height (l) = 10cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q2.1

Question 3.
Solution:
Volume of right circular cone = (100 π) cm³
Height (h) = 12cm.
Let r be the radius of the cone
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q3.1

Question 4.
Solution:
Circumference of the base = 44cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q4.1

Question 5.
Solution:
Slant height of the cone (l) = 25cm
Curved surface area = 550 cm²
Let r be the radius
πrl = curved surface area
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q5.1

Question 6.
Solution:
Radius.of base (r) = 35cm.
Slant height (l) = 37cm.
We know that
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q6.1

Question 7.
Solution:
Curved surface area = 4070 cm²
Diameter of the base = 70cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q7.1

Question 8.
Solution:
Radius of the conical tent = 7m
and height = 24 m.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q8.1

Question 9.
Solution:
Radius of the first cone (r) = 1.6 cm.
and height (h) = 3.6 cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q9.1

Question 10.
Solution:
Ratio in their heights =1:3
and ratio in their radii = 3:1
Let h1,h2 he their height and r1,r2 be their radii, then
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q10.1
The ratio between their volumes is 3:1
hence proved

Question 11.
Solution:
Diameter of the tent = 105m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q11.1

Question 12.
Solution:
No. of persons to be s accommodated =11
Area to be required for each person = 4m²
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q12.1

Question 13.
Solution:
Height of the cylindrical bucket (h) = 32cm
Radius (r) = 18cm
Volume of sand filled in it = πr²h
= π x 18 x 18 x 32 cm³
= 10368π cm³
Volume of conical sand = 10368 π cm³
Height of cone = 24 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q13.1

Question 14.
Solution:
Let h be the height and r be the radius of the cylinder and cone.
Curved surface area of cylinder = 2πrh
and curved surface area of cone = πrl
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q14.1

Question 15.
Solution:
Diameter of the pillar = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q15.1

Question 16.
Solution:
Height of the bigger cone (H) = 30cm
By cutting a small cone from it, then volume of smaller cone = \(\frac { 1 }{ 27 } \) of volume of big cone
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.1
Let radius and height of the smaller cone be r and h
and radius and height of the bigger cone be R and H.

Hence at the height of 20cm from the base it was cut off. Ans.

Question 17.
Solution:
Height of the cylinder (h) = 10cm.
Radius (r) = 6cm.
Height of the cone = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q17.1

Question 18.
Solution:
Diameter of conical vessel = 40cm
Radius (r) = \(\frac { 40 }{ 2 } \) = 20cm
and depth (h) = 24cm.
.’. Volume = \(\frac { 1 }{ 3 } \) πr²h
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q18.1

Hope given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

January 5, 2022June 22, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

Other Exercises

Question 1.
How many least number of distinct points determine a unique line?
Solution:
At least two distinct points determine a unique line.

Question 2.
How many lines can be drawn through both of the given points?
Solution:
Through two given points, one line can be drawn.

Question 3.
How many lines can be drawn through a given point?
Solution:
Through a given point, infinitely many lines can be drawn.

Question 4.
in how many points two distinct lines can intersect?
Solution:
Two distinct lines can intersect at the most one point.

Question 5.
In how many points a line, not in a plane, can intersect the plane?
Solution:
A line not in a plane, can intersect the plane at one point.

Question 6.
In how many points two distinct planes can intersect?
Solution:
Two distinct planes can intersect each other at infinite number of points.

Question 7.
In how many lines two distinct planes can intersect?
Solution:
Two distinct planes intersect each other in one line.

Question 8.
How many least number of distinct points determine a unique plane?
Solution:
Three non-collinear points can determine a unique plane.

Question 9.
Given three distinct points in a plane, how many lines can be drawn by joining them?
Solution:
Through three given points, one line can be drawn of they are collinear and three if they are non-collinear.

Question 10.
How many planes can be made to pass through a line and a point not on the line?
Solution:
Only one plane can be made to pass through a line and a point not on the line.

Question 11.
How many planes can be made to pass through two points?
Solution:
Infinite number of planes can be made to pass through two points.

Question 12.
How many planes can be made to pass through three distinct points?
Solution:
Infinite number of planes can be made of they are collinear and only one plane, if they are non-collinear.

Hope given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

January 5, 2022June 22, 2018 by Prasanna

NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Question 1.
Which of the following correctly represents 360 g of water ?
(i) 2 moles of H₂O
(ii) 20 moles of water
(iii) 6.022 x 10²³ molecules of water
(iv) 1.2044 x 10²⁵ molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv).
Mass of water
Correct Answer:
(d)
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 1

More Resources

Question 2.
Which of the following statements is not true about atoms ?
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.
Correct Answer:
(a) Atoms of inert gas elements can exist independently. However, atoms of all other elements cannot exist independently.

Question 3.
The chemical symbol for nitrogen gas is :
(a) Ni
(b) N₂
(c) N⁺
(d) N
Correct Answer:
(b).

Question 4.
The chemical symbol for sodium is :
(a) So
(b) Sd
(c) NA
(d) Na.
Correct Answer:
(d ).

Question 5.
Which of the following would weigh the maximum ?
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 2
Correct Answer:
(c).
(a) 0.2 x 342 g = 68.4 g
(b) 2 x 44 g = 88 g
(c) 2 x 100 g = 200 g
(d) 10 x 18 g = 180 g.

Question 6.
Which of the following has maximum number of atoms ?
(a) 18 g of H₂O
(b) 18 g of O₂
(c) 18 g of CO₂
(d) 18 g of CH₄.
Correct Answer:
(d).
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 3

Question 7.
Which of the following contains maximum number of molecules ?
(a) 1g CO₂
(b) 1g N₂
(c) 1g H₂
(d) 1g CH₄.
Correct Answer:
(c).
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 4

Question 8.
Mass of one atom of oxygen is :
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 5
Correct Answer:
(a).

Question 9.
3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 x 10²³
(b) 6.09 x 102²²(c) 6.022 x 10²³
(d) 6.022 x 10²¹.
Correct Answer:
(a).
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 6

Question 10.
A change in the physical state can be brought about :
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change.
Correct Answer:
(c).

Short Answer Questions

Question 11.
Which of the following represents a correct chemical formula ? Name it.
(a) CaCl
(b) BiPO₄
(c) NaSO₄
(d) NaS.
Answer:
(b). Both Bi and PO₄ are trivalent ion (Bi³⁺ and PO₄^3- ).
It is known as bismuth phosphate.

Question 12.
Write the chemical formulae for the following compounds :
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate.
Answer:
(a) CuBr₂
(b) Al(NO₃)₃
(c) Ca₃(PO₄)₂
(d) Fe₂S₃
(e) HgCl₂
(f) Mg(CH₂COO)₂.

Question 13.
Write the chemical formulae of all the compounds that can be formed by the combination of following ions :
Cu²⁺, Na⁺, Fe³⁺, Cl^–, SO₄^2-, PO₄^3- .
Answer:
For Cu²⁺ ion : CuCl₂, CuSO₄, Cu₃(PO₄)₂
For Na⁺ ion : NaCl, Na₂SO₄, Na3PO₄
For Fe³⁺ um :FeCl₃, Fe₂(SO₄)₃, FePO₄.

Question 14.
Write the cations and anions present (if any) in the following compounds :
(a) CH₃COONa
(b) NaCl
(c) H₂
(d) NH₄NO₃.
Answer:
(a) CH₃O^– (anion), Na⁺ (cation)
(b) Na⁺ (cation), Cl^– (anion)
(c) Molecular compound (no ions)
(d) NH₄⁺(cation), NO₃^– (anion).

Question 15.
Give the formulae of the compounds formed from the following sets of elements :
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen.
Answer:
(a) CaF₂
(b) H₂S
(c) NH₃
(d) CCl₄
(e) Na₂O
(f) CO and CO₂.

Question 16.
Which of the following symbols of elements are incorrect ? Give their correct symbols :
(a) Cobalt Co
(b) Carbon C
(c) Aluminium Al
(d) Helium He
(e) Sodium So
Answer:
(a) Incorrect; correct symbol is Co
(b) Incorrect; correct symbol is C
(c) Incorrect; correct symbol is Al
(e) Incorrect; correct symbol is Na.

Question 17.
Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
(a) Ammonia
(b) Carbon monoxide
(c) Hydogen chloride
(d) Aluminium fluoride
(e) Magnesium sulphide.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 8

Question 18.
State the number of atoms present in each of the following chemical species :
(a) CO₃^2-
(b) PO₄^3-
(c) P₂O₅
(d) CO.
Answer:
(a) Four
(b) Five
(c) Seven
(d) Two.

Question 19.
What is the fraction of mass of water (H₂O) due to neutrons ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 9

Question 20.
Does the solubility of a substance change with temperature ? Explain with the help of an example.
Answer:
In most of the cases, the solubility of a substance in a particular solvent such as water increases with the rise in temperature. For example, more of sugar can dissolve in a particular volume of water by increasing the temperature. .

Question 21.
Classify each of the following on the basis of their atomicity.
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 10
Answer:
(a) 2
(b) 3
(c) 3
(d) 8
(e) 4
(f) 4
(g) 14
(h) 3
(i) 2
(j) 5
(k) 1
(l) 1.

Question 22.
You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting ?
Answer:
Dissolve both sugar and salt separately in water taken in two glass beakers. Pass electric current through both. In case, solution is conducting, it represents a salt dissolved in water. If it fails to conduct electricity, it represents a sugar solution.

Question 23.
Calculate the number of moles of magnesium present in a magnesium ribon weighing 12 g. Molar atomic mass of magnesium is 24 g mol^-1.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 11

Long Answer Questions

Question 24.
Verify by calculating that :
(a) 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have mole ratio of 3 : 5.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 12

Question 25.
Find the ratio by mass of the combining elements in the following compounds :
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 13
Answer:

Question 26.
Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 16

Question 27.
The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.
Answer:
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Question 28.
Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 18

Question 29.
The mass of one steel screw is 4.11 g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 x 10²⁴ kg). Which one of the two is heavier and by how many times ?
Answer:
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Question 30.
A sample of vitamin C is known to contain 2.58 x 10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample ?
Answer:
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Question 31.
Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight.
(a) Whose container is heavier ?
(b) Whose container has more number of atoms ?
Answer:
(a) Molar atomic mass of carbon = 12 g
No. of moles of carbon carried by Raunak = 5
Mass of 5 moles of carbon = (12 x 5) = 60 g
Molar atomic mass of sodium = 23 g
No. of moles of sodium carried by Krish = 5
Mass of 5 moles of sodium = (23 x 5) = 115 g
This shows that the container carried by Krish is heavier.
(b) Since both the containers have same number of moles, the number of atoms present in these containers are also same i.e. 5 x N_A atoms.

Question 32.
Fill in the missing data in the given table
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Answer:

Question 33.
The visible universe is estimated to contain 10²² stars. How many moles of stars are present in the visible universe ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 23

Question 34.
What is the SI prefix for each of the following multiples and submultiples of a unit ?
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Answer:
(a) kilo
(b) deci
(c) centri
(d) micro
(e) nano
(f) pico

Question 35.
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Answer:

Question 36.
Compute the difference in masses of 10³ moles each of magnesium atoms and magnesium ions.
Answer:
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Question 37.
Which has more number of atoms ?
100 g of N₂ or 100 g of NH₃
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 28

Question 38.
Compare the number of ions present in 5.85 g of sodium chloride.
Answer:
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Question 39:
A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 31

Question 40.
What are ionic and molecular compounds ? Give examples.
Answer:
Simple ionic compounds are of binary nature. It means that both the positive and negative ions have one atom only. The symbols of these ions are written side by side with their valencies at their bottom. A common factor if any, is removed to get a simple ratio of the valencies of the combining atoms. The criss-cross method is then applied to arrive at the final chemical formula of the compound. Let us write the formulae of a few simple ionic compounds. For example,
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The molecular compounds also called covalent compounds. The compounds listed above are heteroatomic in nature. It means that different elements participate in these compounds. They may be homoatomic also which means that these are formed from the atoms of the same element. For example, hydrogen molecule (H₂), chlorine molecule (Cl₂), oxygen molecule (0₂), nitrogen molecule (N₂), etc.
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Question 41.
Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1 x 10^-28 g). Which one is heavier ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 34

Question 42.
A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Answer:
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Question 43.
A sample of ethane (C₂H₆) gas has the same mass as 1.5 x 10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of gas contain ?
Answer:
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Question 44.
Fill in the blanks
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ………….. .
(b) A group of atoms carrying a fixed charge on them is called ………….. .
(c) The formula unit mass of Ca₃(PO₄)₂ is ………….. .
(d) Formula of sodium carbonate is ………… and that of ammonium sulphate is …………… .
Answer:
(a) Law of conservation of mass
(b) Polyatomic ions
(c) 3 x Atomic mass of Ca + 2 x Atomic mass of P + 8 x Atomic mass of O.
3 x 40 + 2 x 31 + 8 x 16 = 120 + 62 + 128 = 310 u
(d) Na₂CO₃, (NH₄)₂SO₄.

Question 45.
Complete the following crossword puzzle by using the name of the chemical elements. Use the data given in table.
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Answer:

Question 46.
Write the formulae for the following and calculate the molecular mass for each one of them.
(a) Custic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt.
Answer:
(a) KOH = 39 + 16 + 1 = 56 u
(b) NaHCO₃ = 23 + 1 + 12 + 3 x 16 = 84 u
(c) CaCO₃ = 12 + 12 x 3 x 16 = 100 u
(d) NaOH = 23 + 16 + 1 = 40 u
(e) C₂H₅OH = 2 x 12 + 6 x 1 + 16 = 46 u
(f) NaCl = 23 + 35.5 = 58.5 u

Question 47.
In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex sereies of reactions to give a molecule of glucose having a molecular formula C₆H₁₂O₆ and a molecule of oxygen with molecular formula O₂. How many grams of water would be required to produce 18 g of glucose ? Compute the volume of water so consumed assuming the density of water to be 1 g cm^-3.
Answer:
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