NCERT Solutions For Class 9 Science Chapter 3 Atoms And Molecules
|1. Exercise Questions|
|2. Intext Questions|
NCERT Solutions For Class 9 Science Chapter 3 Atoms And Molecules on this step-by-step answer guide . In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those NCERT Solutions. It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science Atoms And Molecules so you can be searching for assist from them. Students have to solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Atoms And Molecules below and prepare for your tests easily.
NCERT Solutions for class 9 Science
Chapter 3 Atoms and Molecules
Q.1 . A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen .
Calculate the percentage composition of the compound by weight .
Ans . Mass of the given sample compound = 0.24g
Mass of boron in the given sample compound = 0.096g
Mass of oxygen in the given sample compound = 0.144g % composition of compound = % of boron and % of oxygen
Therefore % of boron = ( mass of boron ) / ( mass of the sample compound ) x 100 = 40 %
Therefore % of oxygen = ( mass of oxygen ) / ( mass of the sample compound ) x 100 = 60 %
Q. 2. When 3.0 g of carbon is burnt in 8.00 g oxygen , 11.00 g of carbon dioxide is produced . What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen ? Which law of chemical combinations will govern your answer ?
Ans . Carbon + Oxygen → Carbon dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide .
If 3 g of carbon is burnt in 50 g of oxygen , then 3 g of carbon will react with 8 g of oxygen .
The remaining 42 g of oxygen will be left unreactive .
In this case also , only 11 g of carbon dioxide will be formed .
The answer is governed by the law of constant proportions .
Q. 3. What are polyatomic ions ? Give examples .
Ans . A polyatomic ion is a group of atoms carrying a charge ( positive or negative ) .
For example , ammonium ion ( NH4 + ) , hydroxide ion ( OH – ) , carbonate ion ( CO3 – ) , sulphate ion ( SO4 – ) .
Q. 4. Write the chemical formula of the following :
( a ) Magnesium chloride .
( b ) Calcium oxide .
( c ) Copper nitrate .
( d ) Aluminium chloride .
( e ) Calcium carbonate .
( f ) Hydrogen chloride .
Ans . ( a ) Magnesium chloride → MgCl₂
( b ) Calcium oxide → CaO
( c ) Copper nitrate → Cu ( NO3 ) 2
( d ) Aluminium chloride → AlCl3
( e ) Calcium carbonate → CaCO3
( f ) Hydrogen chloride → HCI
Q. 5. Give the names of the elements present in the following compounds :
( a ) Quick lime ( b ) Hydrogen bromide
( c ) Baking powder ( d ) Potassium sulphate
Q. 6. Calculate the molar mass of the following substances :
( a ) Ethyne , C₂H₂ .
( b ) Sulphur molecule , S8 .
( c ) Phosphorus molecule , P4 ( atomic mass of phosphorus = 31 ) .
( d ) Hydrochloric acid , HCl .
( e ) Nitric acid , HNO3 .
Ans . ( a ) Molar mass of ethyne ,
C₂H₂ = 2 × 12 + 2 × 1 = 26g
( b ) Molar mass of sulphur molecule ,
S8 = 8 x 32 = 256 g
( c ) Molar mass of phosphorus molecule ,
P4 = 4 x 31 124 g
( d ) Molar mass of hydrochloric acid ,
HCl = 1 + 35.5 = 36.5 g
( e ) Molar mass of nitric acid ,
HNO3 = 1 + 14 + 3 x 16 = 63 g
Q. 7 . What is the mass of
( a ) 1 mole of nitrogen atoms ?
( b ) 4 moles of aluminium atoms ( Atomic mass of aluminium = 27 ) ?
( c ) 10 moles of sodium sulphite ( Na₂SO3 ) ?
Ans . ( a ) 1 mole of nitrogen atoms = 1 x mass of N = 1 x 14 = 14g
( b ) 4 moles of aluminium atoms = 4 x 27 = 108 g
( c ) 10 moles of sodium sulphite ( Na₂SO3 ) = 10 × Mass of Na₂SO3 .
Mass of Na₂SO3 = 2 × Mass of N a atom + 1 × Mass of S atom + 3 x Mass of O atom = 2 x 23 + 1 x 32 + 3 x 16 = 46 + 32 + 48 = 126g / mol
∴ 10 moles of Na₂SO3 = 10 × 126 = 1260 g
Q. 8. Convert into mole :
( a ) 12 g of oxygen gas .
( b ) 20 g of water .
( c ) 22 g of carbon dioxide .
Q. 9. What is the mass of :
( a ) 0.2 mole of oxygen atoms ?
( b ) 0.5 mole of water molecules ?
Ans . ( a ) Mass of one mole of oxygen atoms = 16 g
Then , mass of 0.2 mole of oxygen atoms
= 0.2 x 16 g = 3.2 g
( b ) Mass of one mole of water molecule = 18g
Then , mass of 0.5 mole of water molecules = 0.5 x 18 g = 9 g
Q. 10 . Calculate the number of molecules of sulphur ( S8 ) present in 16 g of solid sulphur .
Ans . 1 mole of solid sulphur ( S8 ) = 8 x 32 g = 256 g .
i.e. , 256g of solid sulphur contains = 6.022 × 1023 molecules
Q. 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide .
( Hint : The mass of an ion is the same as that of an atom of the same element . Atomic mass of Al = 27 u )
Ans . 1 mole of aluminium oxide ( Al₂O3 ) = 2 × 27 + 3 x 16 = 102 g
i.e. , 102 g of Al₂O3 = 6.022 x 1023 molecules of Al₂O3
1g of Al₂O3 contains = 3.011 × 1020 molecules of Al₂O3
The number of aluminium ions ( A1³+ ) present in one molecule of aluminium oxide is 2 .
Therefore , The number of aluminium ions ( A1³+ ) present in = 3.11 x 1020 molecules ( 0.051 g ) of aluminium oxide ( Al₂O3 ) = 2 × 3.11 × 1020
= 6.022 × 1020
Q. 1. In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid . The products were 2.2 g of carbon dioxide , 0.9 g water and 8.2 g of sodium ethanoate . Show that these observations are in agreement with the law of conservation of mass .
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water . [ NCERT Q. 1 , Page 32 ]
Ans . In a reaction , sodium carbonate reacts with ethanoic acid to produce sodium ethanoicate , carbon dioxide , and water .
Sodium carbonate + ethanoic acid → sodium ethanoate + Carbon dioxide + Water
Mass of sodium carbonate = 5.3 g ( Given )
Mass of ethanoic acid = 6 g ( Given )
Mass of sodium ethanoate = 8.2 g ( Given )
Mass of carbon dioxide = 2.2 g ( Given )
Mass of water = 0.9 g ( Given )
Now , total mass before the reaction = ( 5.3 + 6 ) g = 11.3 g
and total mass after the reaction = ( 8.2 + 2.2 + 0.9 ) g = 11.3 g
Therefore , total mass before the reaction = Total mass after the reaction
Hence , the given observations are in agreement with the law of conservation of mass .
Q. 2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water . What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas ? [ NCERT Q. 2 , Page 33 ]
Ans . The ratio of hydrogen and oxygen by mass to form water is 1 : 8 . Then , the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g . Thus , the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 x 3 = 24 g .
Q. 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass ? [ NCERT Q. 3 , Page 33 ]
Ans . The postulate of Dalton’s atomic theory which is a result of the law of conservation of mass is ” Atoms are indivisible particles , which can neither be created nor destroyed in a chemical reaction ” .
Q. 4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions ? [ NCERT Q. 4 , Page 33 ]
Ans . The postulate of Dalton’s atomic theory which can explain the law of definite proportion is ” The relative number and kind of atoms in a given compound remain constant ” .
Q. 5. Define atomic mass unit . [ NCERT Q. 1 , Page 35 ]
Ans . Mass unit equal to exactly one – twelfth the mass of one atom of carbon – 12 is called one atomic mass unit . It is written as ‘ u ‘ .
Q. 6. Why is it not possible to see an atom with naked eyes ? [ NCERT Q. 2 , Page 35 ]
Ans . The size of an atom is so small that it is not possible to see it with naked eyes . Also , atom of an element does not exist independently .
Q. 7. Write down the formula of
( i ) Sodium oxide
( ii ) Aluminium chloride
( iii ) Sodium sulphide
( iv ) Magnesium hydroxide [ NCERT Q. 1 , Page 39 ]
Ans . ( i ) Sodium oxide → Na₂O
( ii ) Aluminium chloride → AlCl3
( iii ) Sodium sulphide → Na₂S
( iv ) Magnesium hydroxide → Mg ( OH ) 2
Q. 8. Write down the name of compounds represented by the following formulae :
( i ) Al₂ ( SO4 ) 3 .
( ii ) CaCl₂
( iii ) K₂SO4
( iv ) CaCO3 . [ NCERT Q. 2 , Page 39 ]
Ans . ( i ) Al₂ ( SO4 ) 3 → Aluminium sulphate
( ii ) CaCl₂ → Calcium chloride
( iii ) K₂SO4 → Potassium sulphate
( iv ) CaCO3 → Calcium carbonate
Q. 9 . What is meant by the term chemical formula ? [ NCERT Q. 3 , Page 39 ]
Ans . The chemical formula of a compound means the symbolic representation of the composition of a compound .
From the chemical formula of a compound , we can know the number and kinds of atoms of different elements that constitute the compound .
For example , from the chemical formula CO₂ carbon dioxide , we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound , carbon dioxide .
Q. 10. How many atoms are present in a
( i ) H₂S molecule , and
( ii ) PO4 ion ? [ NCERT Q. 4 , Page 39 ]
Ans . ( i ) In a H₂S molecule , three atoms are present ; two of hydrogen and one of sulphur .
( ii ) In a PO4 ion , five atoms are present ; one of phosphorus and four of oxygen .
Q. 11. Calculate the formula unit masses of ZnO , Na₂O , K₂CO3 , given atomic masses of Zn = 65 u , Na = 23 u , K = 39 u , C = 12 u , and O = 16 u . [ NCERT Q. 2 , Page 40 ]
Ans . Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65+ 16 = 81 u
Formula unit mass of Na₂O = 2 x Atomic mass of Na + Atomic mass of O
= 2 x 23 + 16 = 62 u
Formula unit mass of K₂CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O
= 2 x 39 + 12 + 3 x 16 = 138 u
Q. 12. If one mole of carbon atoms weighs 12 g , what is the mass ( in g ) of 1 atom of carbon ? [ NCERT Q. 1 , Page 42 ]
Ans . One mole of carbon atoms weighs 12 g ( Given )
i.e. , mass of 1 mole of carbon atoms = 12 g
Then , mass of 6.022 x 1023 number of carbon atoms = 12 g
Therefore , mass of 1 atom of carbon
Q. 13. Which has more number of atoms , 100 grams of sodium or 100 grams of iron ( given , atomic mass of Na = 23 u , Fe = 56 u ) ? [ NCERT Q. 2 , Page 42 ]
Ans . Atomic mass of Na = 23 u ( Given )
Then , gram atomic mass of Na = 23 g
Now , 23 g of Na contains = 6.022 x 1023 number of atoms
Again , atomic mass of Fe = 56 u ( Given )
Then , gram atomic mass of Fe = 56 g
Now , 56 g of Fe contains = 6.022 x 1023 number of atoms
Benefits of NCERT Solutions For Class 9 Science Chapter 3 Atoms And Molecules
NCERT Solutions For Class 9 Science Chapter 3 Atoms And Molecules contains extremely important points, and for each chapter, each concept has been simplified to make it easier to remember and increase your chances of achieving excellent exam results. Exam Preparation References Here are some tips on how these solutions can help you prepare for the exam.
- This helps students solve many of the problems in each chapter and encourages them to make their concepts more meaningful.
- NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules encourage you to update your knowledge and refine your concepts so that you can get good results in the exam.
- These NCERT Solutions For Class 9 Science Chapter 3 Atoms And Molecules are the best exam materials, allowing you to learn more about your week and your strengths. To get good results in the exam, it is important to overcome your weaknesses.
- Most of the questions in the exam are formulated in a similar way to NCERT textbooks. Therefore, students should review the solutions in each chapter in order to better understand the topic.
- It is free of cost.
Tips & Strategies for Class 9 Exam Preparation
- Plan your course and syllabus and make time for revision
- Please refer to the NCERT solution available on the cbsestudyguru website to clarify your concepts every time you prepare for the exam.
- Use the cbsestudyguru learning app to start learning to successfully pass the exam. Provide complete teaching materials, including resolved and unresolved tasks.
- It is important to clear all your doubts before the exam with your teachers or Alex (an Al study Bot).
- When you read or study a chapter, write down algorithm formulas, theorems, etc., and review them quickly before the exam.
- Practice an ample number of question papers to make your concepts stronger.
- Take rest and a proper meal. Don’t stress too much.
Why opt for cbsestudyguru NCERT Solutions for Class 9 ?
- cbsestudyguru provide NCERT Solutions for all subjects at your fingertips.
- These solutions are designed by subject matter experts and provide solutions to every NCERT textbook questions.
- cbsestudyguru especially focuses on making learning interactive, effective and for all classes.
- We provide free NCERT Solutions for class 9 and all other classes.