Please refer to Atoms And Molecules Class 9 Science Exam Questions provided below. These questions and answers for Class 9 Science have been designed based on the past trend of questions and important topics in your class 9 Science books. You should go through all Class 9 Science Important Questions provided by our teachers which will help you to get more marks in upcoming exams.
Class 9 Science Exam Questions Atoms And Molecules
Class 9 Science students should read and understand the important questions and answers provided below for Atoms And Molecules which will help them to understand all important and difficult topics.
1 Mark Questions
Question. Atomic radius is measured in nanometers and
(a) 1nm = 10-10m
(b) 1m = 10-10nm
(c) 1m = 10-9nm
(d) 1nm = 10-9m
Answer.
C
Question. Symbol of Iron is :-
(a) Ir
(b) I
(c) Fe
(d) None of these
Answer.
C
Question. Atomicity of chlorine and Argon is
(a) Diatomic and Monoatomic
(b) Monoatomic and Diatomic
(c) Monoatomic and Monoatomic
(d) Diatomic and Diatomic
Answer.
A
Question. Molecular mass of water (H2O)is
(a) 18g
(b) 8g
(c) 33g
(d) 34g
Answer.
A
Question. 1 Mole of a compound contains –
(a) 6.023 × 1023 atoms
(b) 6.023× 1024 atoms
(c) 60.23 × 1023 atoms
(d) 6.023 × 1025 atoms
Answer.
A
Question. Oxygen is –
(a) Monovalent
(b) Bivalent
(c) Trivalent
(d) Tetravalent
Answer.
A
1 Mark Questions
Question. 10gm of silver nitrate solution is added to 10 gm of sodium chloride solution. What change in mass do you expect after the reaction and why?
Answer. No change in mass will take place because mass can neither be created nor destroyed in a chemical reaction as law of conservation of mass states.
Question. What is the order of size of atoms?
Answer. The atomic radius is of the order of 10–10 m.
Question. Write the atomicity of the following molecules:
(i) H2SO4 (ii) CCl4
Answer. Atomicity is the number of atoms present in a molecule.
(i) H2SO4 = 7 atoms are present
(ii) CCl4 = 5 atoms are present.
Question. Define the term mole.
Answer. The mole is the amount of the substance which contains same number of particles (atoms/molecules/ formula units, etc) as there are C-12 isotope atoms in 12 gram.
Question. What is molar mass? What are its units?
Answer. The mass of one mole of substance ( i.e. Avogadro’s number of particles = 6.023 x 1023) is called its molar mass. Its unit is gram per mole (g / mol) or kilogram per mole (kg / mol).
Question. Define atomicity.
Answer. Atomicity is defined as number of atoms present in a molecule.
e.g. Atomicity of NH3 is 4 because there is one atom of N and three atoms of H.
Question. Give the definition of a cation and an anion.
Answer. Cation- It is the positively charged ion, e.g., Na+ which is attracted towards cathode (negative electrode) in an electric field.
Anion – It is the negatively charged ion, e.g., Cl–which is attracted towards anode (positive electrode) in an electric field.
Question. An element X has a valency 3. Write the formula of its oxide.
Answer. Using criss – cross method of finding chemical formula
Valency of X = +3
Valency of O = -2
After cross over valencies, formula will be X2O3.
2 Marks Questions
Question. Calculate the number of moles is 52 g of He (Helium)
[At mass : O = 16 u, He = 4 u]
Answer.
Question. If 12 gm of carbon is burnt in the presence of 32 gm of oxygen, how much carbon dioxide will be formed ?
Answer. The law of conservation of mass states that mass can neither be created nor destroyed or mass of reactants is always equal to that of products.
C + O2 → CO2
Carbon + Oxygen → Carbon dioxide
Mass of reactants = 12 + 32 = 44 g
Mass of Product (CO2) = 44 g
(One mole of C reacts with 1 mole of oxygen to form 1 mole of CO2)
Question. How do atoms exist?
Answer. The atoms of noble gases (eg He, Ne etc.) are unreactive and exist in the free state (as single atoms).But, atoms of most of the elements are very reactive and usually exist either in the form of molecules or ions.
Question. What is an atom?
Answer. The word ‘atom’ means ‘indivisible’ – something which cannot be divided further. So, atoms are the smallest particles of matter which may or may not be capable of free existence and takes part in a chemical reaction.
Question. What is atomicity?
Answer. Atomicity is the number of atoms present in any compound/molecule of any substance. For eg, in Ammonia (NH3) there is 1 atom of N and 3 atoms of hydrogen so its atomicity is 4.
Question. Calculate the molar mass of the following substances.
(a) Ammonia
(b) Phosphorus molecule
(a) Molar mass of ammonia (NH3) = 1 ×14 + 3 × 1 = 17 u
(b) Molar mass of phosphorus molecule (P4) = 4 ×31 =1 24 u
Question. What is the difference between cation and anion?
Answer. Ionic compounds are made up of positive and negative ions. When an atom loses an electron, it becomes positively charged or cation where as when atom gains electron, it becomes negatively charged or anions.
Question. Find out the number of moles in 100 g of water.(Atomic masses of O =16, H = 1)
Answer. 1 mole of water (H2O) = Gram molecular mass of H2O = 18 g
Thus, 18 g of H2O = 1 mole of H2O
∴ 100 g of H2O =
1 / 18 x 100 moles = 5.56 moles.
Question. Calculate the mass of 0.5 mole of silver.( atomic mass of Ag = 108 )
Answer. 1 mole of silver (Ag) = Gram atomic mass of Ag = 108 g
∴ 0.5 mole of silver
108 / 1 x 0.5 = 54 = g.
Question. What is a “parmanu”?
Answer. Maharishi Konad first postulated that matter is divisible that is, if we keep breaking any matter, we will get smaller particles and finally, the particles may be so small that it cannot be divide further. These indivisible particles were named as “parmanu
Question. Calculate the number of moles in 17 gm of H2O2. (Atomic weight of H = 1 u, O = 16 u).
Answer. Atomic mass of H = 1
Atomic mass of O = 16
Molecular mass of H2O2 = (2 × 1) + (2 × 16) = 34 gm
Since 34 gm of hydrogen peroxide ( H2O2) = 1 mole
Question. Calculate percentage composition of glucose (C6H12O6).
Answer. (C = 12, H = 1, O = 16)
Molecular mass of glucose (C6H12O6) = 6 × 12 + 1 × 12 + 6 × 16
= 72 + 12 + 96 = 180 g
3 Marks Questions
Question. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer.
Question. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer.
Question. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S2
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
Answer. (a) Ethyene = C2H2 = 12×2 + 1×2 = 24 + 2 = 26 u = 26 g
(b) Sulphur molecular = S8 = 32 x 8 = 256 u = 256 g
(c) Phosphorus molecule = P4 = 31 x4 = 124 u = 124 g
(d) Hydrochloric acid = HCl = 1+ 35.5 = 36.5 u = 36.5 g
(e) Nitric acid = HNO3 = 1 + 14 + (16×3) = 15 + 48 = 63 u = 63 g
Question. 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 ( )?
Answer. (a) Atomic mass of nitrogen is 14 u.
therefore 1 mol of N = 14g
(b) Atomic mass of aluminium = 27u
therefore 1 mol of Al = 27g and so 4 mol of Al = 27×4 = 108g
(c) molecular mass of = 23×2 + 32 + 16×3 = 46 + 32 + 48 = 126 u
therefore 1 mol of has weight/mass 126g.
hence, 10 mol of = 10×126 = 1260g
Question. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Answer. (a) molecular mass of O2 = 32 u= 32g(1 mole)
since 32 g of =1mole then 12g of O2 = 1×12/32 =0.375mole.
(b) molecular mass of H2O = 1×2 + 16 = 18 u= 18g(1mole)
20g H2O = 1×20/18 = 1.11mole.
(c) molecular mass of CO2 = 12 + 16×2 = 12 + 32 = 44 u= 44g (1mole)
22g of CO2 = 1×22/44 = 0.5mole.
Question. State the Postulates of Dalton Theory?
Answer. The postulates of Dalton theory are
a) All matter is made of vary tiny particles called atom
b) Atoms are indivisible particle; they cannot be created or destroyed during a chemical reaction
c) Atoms of a \given element are identical in mass and chemical properties
d) Atoms of different elements have different mass and chemical properties.
e) Atom combines in the ratio of their whole number to form compounds
f) The relative number and kinds of atoms are constant in a compound.
Question. Find the percentage of water of crystallization in feSO4. 7H2O.
Answer. The RMM of FeSO2 .7 O= 55.9 + 32.0 + 4 × 16 + 7(18)
= 55.9 + 32 + 64 + 126
= 277.9 g/mol.
277.9 g/mol of FeSO4 contain 126g of water
∴ 100g of crystal will contain 100 x 126 / 277.9 of water of crystallization
This is 45.34 of water of crystallization
The amount of water of crystallization in FeSo4 .7HO2 = 45.34% by mass.
Question. 2.42g of copper gave 3.025g of a black oxide of copper, 6. 49g of a black oxide, on reduction with hydrogen, gave 5.192g of copper. Show that these figures are in accordance with law of constant proportion?
Answer. The percentage of copper is first oxide
= 2.43 x 100 / 3.025 = 80.0
The percentage of copper is second oxide = 5.192 x 100 / 6.49
= 80.02
As the percentage of copper in both the oxides is same, thence law of constant composition is verified.
Question. A compound was found to have the following percentage composition by mass Zn = 22.65%, S = 11.15%, H = 4.88%, O = 61.32%. The relative molecular mass is 287g/mol. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallizations.
Answer.
Question. Which element will be more reactive and why → the element whose atomic number is 10 or the one whose atomic number is 11?
Answer. Element with atomic number 11 is more reactive than the one with atomic number 10 because electronic configuration of atomic number 11 will be 2, 8, 1 so, it has to loose only 1e- from its outermost shall to be stable which is more easy than the element with atomic number 10 because its electronic configuration is 2, 8 and has 8e- in the outermost shell and
hence is already stable.
Question. What are the failures of Dalton Atomic theory?
Answer. Failures of Dalton Atomic Theory are :-
1) Atom is not the smallest particle as it is made up of protons, neutrons and electrons.
2) Atom’s mass can be cornered to energy (E = mc2) and hence can be created and destroyed.
3) Atoms of one element have been charged into atoms of another element through artificial transmutation of elements.
4) Atoms of same element need not resemble each other in all respects as isotopes (Different of same element) exist.
5) Atoms of different elements need not differ in all respects as isobars (same forms o ta different elements) exist.
Question. John placed 10 moles of sulphur molecules (S ) and 5 moles of glucose (C6H12O6) in the two different pans of a physical balance. Find which pan of the balance would be heavier, support your answer with calculations.
Answer. (atomic mass : O = 16 u, C = 12 u, H = 1 u, S = 32 u)
Molecular mass of sulphur molecules ( S8)
= 32 × 8 = 256 gm
Molecular mass of glucose (C6H12O6)
= (6 × 12) + ( 1 × 12) + ( 6 × 16)
= 72 + 12 + 96 = 180
Weight of 10 moles of S8 =10 × 256 = 2560
Weight of 5 moles of glucose = 5 × 180 = 900
Therefore, 10 moles of sulphur would weigh more.
Question. Find out the mass of an atom of copper (atomic mass of Cu = 63.5).
Answer. 1 mole of Cu atoms = Gram atomic mass of Cu = 63.5 g
Since 1 mole of Cu atoms = 6.022 × 1023 atoms of Cu
6.022 × 1023 atoms of Cu weigh = 63.5 g
∴ 1 atom of Cu will weigh =
63.5 / 6.022 x 1023g
= 10.54 x 10-23g
Question. 1022 atoms of an element ‘X’ are found to have a mass of 930 mg. Calculate the molar mass of the element ‘X’.
Answer. Molar mass of an element is the mass of Avogadro’s number of atoms.
1022 atoms of the element have mass = 930 mg = 930 / 1000 g = 0.930 g
∴ 6.022 × 1023 atoms will have mass =
0.930 / 1022 x 6.022 x 1023g = 56.0g
∴ Molar mass of the element = 56 g mol.
Question. If we burn a piece of paper its mass becomes less after burning. Is there any exception of the law of conservation of mass ?
Answer. In case of burning of a paper, it seems that the mass reduces. But the paper changes into ash, CO2 and water vapour. The CO2 and water vapour being gaseous escape into the environment leaving the ash behind. If we calculate the mass of carbon dioxide and water vapour, the total mass of the reactant and the product will remain the same, thus verifying the law. Thus, there is no exception present which is against the law of conservation of mass.
Question. Calculate the mass of SO2 gas which will contain the same number of molecules as present in 4.4 g of CO2.
Answer. Molar mass of CO = 12 + 2 × 16 g mol1 = 44 g mol .
No. of moles in 4.4 g CO2 =0.1 mole of CO2 will contain the same number of molecules as present in 0.1 mole of SO2 (because equal number of moles contains equal number of molecules). Thus, we have to calculate mass of 0.1
mole of SO2.
Molar mass of SO2 = 32 + 2 × 16 g mol = 64 g mol-1
Mass (m) = n × M = 0.1 × 64 g = 6.4 g.
Question. Which will contain larger number of atoms, 1 g of gold or 1 g of silver ? Explain with reason. (Atomic masses of Gold = 197 u, Silver = 108 u)
Answer. No. of moles (n) in 1 g of gold = m/M = 1/197
No. of moles (n) in 1 g of silver = m/M = 1/108
As
1/108 > 1/197
Greater the number of moles, greater is the number of atoms present.
Hence silver will contain more number of atoms.
Question. What does symbol means?
Answer. A symbol represents an atom of an element which may consist of a single letter or a combination of letters derived either from the English name or Latin name of the element. It was J.J. Barzelius of Sweden who proposed that the first letter (or the first letter and another letter) of the name of an element be used as its symbol.
For e.g. :- Symbol of Carbon is C (first letter of name)
Symbol of Calcium is Ca (first and second letter of name)
Symbol of Chlorine is Cl (first and third letter of name)
Symbol of Iron is Fe (first and second letter of its latin name Ferrum)
Question. What is molecular mass?
Answer. The molecular mass of substance (element or compound) is the number of times the molecule of the substance is heavier than 1/12th the mass of an atom of Carbon -12.
OR
Molecular mass is equal to sum of the atomic masses of all the atoms present in one molecule of the substance.
For eg , molecular mass of H2SO4 = Mass of 2 ‘H’ atoms + Mass of ‘S’ atom + Mass of 4 ‘O’ atom = 2 × (1) + 32 + 4(16) = 2 + 32 + 64 = 98 u.
Question. What are ions?
Answer. An atom is made up of subatomic particles namely electrons, protons and neutrons. Electrons carries 1 unit negative charge and proton carries 1 unit positive charge. Neutron is a neutral particle. As every atom contains equal number of electron and proton which balance the charges and results in an electrically neutral atom. When this neutral atom loses or gains electron, it get overall electric charge. This electrically charged atom is known as ‘ion’. For e.g. Na+, Cl—
Question. What are the differences between sodium atom and sodium ion?
Answer. When Sodium atom loses 1 electron it changes into sodium ion. Differences between atom and ion are as follows:
Question. (a) What is Avogadro constant ?
(b) Calculate the number of particles present in 56 gm of N2 molecule.
Answer. (a) The number of atoms or molecules in 12 gm of carbon-12 ( in a mole of a substance) is called as Avogadro constant (6.022 × 1023 )
(b) Number of molecules (N) = Giteh mass x Avogadro’s number / molar mass Molar mass of 1 mole of N2 = 14 × 2 =28
N = 56/28 x 6.022 x 1023
N = 12.044 x 1023
Hence there are 12.044 x 1023 present in 56 gm of N2
Question. Nitu presented a silver lamp to her mother on her birthday. The lamp contained 3.011 × 1023 atoms of silver in it. What is the mass of silver lamp and the cost of it if 1 gm silver costs Rs 60.
Atomic mass of Ag = 108 u, N = 6.022 × 1023 per mole
Answer. Atomic mass of silver(Ag) = 108 u
since mass of 6.022 × 1023 atoms of silver(Ag) = 108 u
Hence mass of 3.011 × 1023 atoms of silver (Ag) = 108 x 3.011 x 1023 / 6/.022 x 1023 = 54 gm
Cost of 1 gm silver = 60
Therefore, cost of silver lamp = 60 × 54 = Rs. 4240
Question. Give an account of the ‘mole concept’.
Answer. Mole is a unit to express amount of any chemical compound that contains atoms/molecules/ions equal to atoms present in 12 gram of carbon-12 isotope. This number of atoms is fixed and is equal to 6.023 × 1023 (Avagadro’s number). Thus a mole expresses a collection of 6.023 × 1023 atoms, molecules, ions, etc. For e.g., 1 mole of oxygen atom refers to 6.023 × 1023 atoms of oxygen.
Question. Calculate the ratio by number of atoms for Magnesium sulphide.
Answer. [Atomic mass of Mg => 24, S => 32]
Atomic mass of Mg = 24
Atomic mass of S = 32
Question. Calculate the mass of 1 molecule of oxygen gas.
Answer. [ Atomic mas of O = 16 u, N = 6.022 × 1023 mol-1]
Weight of 1 mole of oxygen molecules = 32g of oxygen
Number of atoms in 1 mole of oxygen = 6.022 × 1023 molecules
Because 6.022 × 1023 molecules of O2 weigh 32 g
Therefore, 1 molecule of O2 will weigh = 32 / 6.022 x 1023
= 5.313 × 10-23 g
Question. Calculate the molar mass of the following substances.
Answer. (a) Ammonia
(b) Phosphorus molecule
(a) Molar mass of ammonia (NH3) = 1 ×14 + 3 × 1 = 17 u
(b) Molar mass of phosphorus molecule (P4) = 4 × 31 = 124 u
Question. How many molecules are present in 34 gm of ammonia ?
Molecular mass of ammonia (NH ) = 1 (N) + 3(H3) = 1(14) + 3(1) = 14 + 3 = 17
Since 17 gm of ammonia contains 6.022 × 1023 molecules
Therefore 34 gm of ammonia (NH3)contains = 2× 6.022 × 1023 = 12.044 × 1023 molecules.
Question. Which of the following has larger number of hydrogen atoms ?
Answer. (i) One mole of ammonia ( NH3) (ii) Two moles of hydrochloric acid ( HCl)
1 mole of NH3 contains 3 moles of H-atoms = 3 × N0 atoms
2 moles of HCl contains 2 moles of H-atoms = 2 × N0 atoms.
Thus, 1 mole of NH3 contains larger number of H-atoms than 2 moles of HCl.
Question. Define and explain atomic mass of an element.
Answer. Atomic mass of an element is the relative mass of its atoms compared with the mass of an single atom of C-12 isotope taken as 12 units. For example, the atomic mass of oxygen is 16 which indicates that an atom of oxygen is 16 times heavier than 1/12th of a 12C atom.
Atomic Mass Unit (amu) = 1th/12 the mass of 12C atom
5 Marks Questions
Question. Verify by calculating that 5 mole of CO2 and 5 mole of H2S do not have the same mass.
Answer. (Atomic mass of C = 12 u, O = 16 u, H = 1 u, S = 32 u)
Molecular mass of CO = 1(C) + 2(O)
= 1(12) + 2(16)
= (1 × 12) +(2 × 16)
= 44 u
5 mole of CO2 = 5 × 44
= 220 gm
Molecular mass of H2S = 2(H) + 1(S)
= (2 × 1) + (1 × 32)
= 34 u
Hence, both (CO2 and H2S) do not have same mass.
Question. What are the major drawbacks of Dalton’s atomic theory?
Answer. The major drawbacks of Dalton’s atomic theory are as follows: –
• This theory was based on the concept of indivisibility of atom but recent studies show that atom can be further divided into electrons, protons and neutrons.
• This theory states that all the atoms of an element have exactly same mass. But, now it i s known that atoms of the same element with different atomic mass (i.e. isotopes) also exist.
• This theory states that atoms of different elements have different masses but even atoms of different element can also have the same mass (i.e. isobars).
• This theory states that substances made up of same kind of atoms have similar properties. But, charcoal, graphite and diamond are all made up of carbon atoms but have different physical properties.
Question. What is formula unit mass?
Answer. Ionic compounds like NaCl consist of a very large but equal number of Na ions and Cl ions arranged in a definite order in the crystal lattice. Thus, the actual formula of NaCl should be written as (Na+ Cl–)n where n is a very large number. The formula of NaCl, written as Na+ Cl– represents only the simplest formula and not the actual formula. The formula Na+ Cl– is, therefore, called one formula unit and its mass is called as formula unit mass.
(NaCl consists of equal number of Na+ and Cl– ions) ( Formula unit of NaCl)
Question. Calculate the mass of 0.5 mole of sugar (C12H22O11 ) (Atomic masses of C = 12, H = 1, O = 16)
Answer. 1 mole of sugar (C12H12O11 ) = Gram molecular mass of C12H12O11
= 12 × 12 + 22 × 1 + 11 × 16 g = 342 g
∴ 0.5 mole of sugar =
342 / 1 x 0.5 = 171g.
Step 2. Calculation of number of atoms
Number of atoms = No. of moles × Avogadro’s number
i.e., N = n × N0 =
1 / 197 x 6.022 x 1023
= 3.06 × 1021 atoms.
Question. Calculate the mass in milligrams of 1021 atoms of U-238.
Answer.
m (mass ) = n × M ..(2)
Placing the value of ‘n’ from equation (1) into equation (2)
Since 1 g = 1000 mg
Hence 0.395 g = 395 mg
Question. Calculate the number of molecules present in 1 litre of water assuming that density of water is 1 g mL-1.
Answer. 1 Litre of water = 1000 mL-1
Mass of 1000 mL of water = Volume × Density = 1000 mL × 1 g mL-1 = 1000 g
No. of moles in 1000 g of H2O =
m/M = 1000/18 = 55.55 moles
1 Litre of water = 1000 mL
Mass of 1000 mL of water = Volume × Density = 1000 mL × 1 g mL-1 = 1000 g
No. of moles in 1000 g of H2O =
m/M = 1000/18 = 55.55 moles
No. of molecules (N) = n × No = 55.55 × 6.022 × 1023.
= 3.345 × 1025 molecules
Question. In formation of CO2 Carbon and oxygen react. They do so in the same ratio 3: 8.What mass of oxygen would be required to react completely with 6.0 g of carbon ? Calculate the mass of carbon dioxide gas so formed ?
Carbon and oxygen combine as per law of constant proportions, i.e. 3 g of carbon will always combine with 8 g of oxygen or 3 g carbon(C) combines with oxygen (O) = 8 g
∴ 6 g carbon(C) will combine with oxygen (O) = 8 / 3 x 6 = 16g
Total amount of carbon dioxide( CO2) = 6 + 16 = 22 g
Question. Give the names of the elements present in the following compounds.
(a) Quicklime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
(e) Marble
(a) Quicklime – CaO (Calcium oxide)
Elements – Calcium, oxygen
(b) Hydrogen bromide —— HBr
Elements – Hydrogen, bromine
(c) Baking powder — NaHCO3 (Sodium hydrogen carbonate)
Elements – Sodium , hydrogen , carbon , oxygen
(d) Potassium sulphate – K2SO4
Elements- Potassium , sulphur , oxygen
(e) Marble – CaCO (Calcium carbonate)
Elements- Calcium, carbon, oxygen
Question. Find the number of moles of
(a) 48 g of oxygen gas
(b) 22 g of CO2 gas
(a) Molecular weight of Oxygen( O ) = 2 × 16 = 32 u
i.e. mass of 1 mole of O = 32 g
32 g of O = 1 mol
∴ 48 g of O2 =
1 x 48 / 32 = 3 / 2 = 1.5 mol
number of moles =
given mass / molar mass = 48 / 32 = 1.5 mol
(b) Molecular weight of CO = 12 + 2 × 16 = 44 g
44 g of CO = 1 mol
∴ 22 g of CO =
1/44 x 22 = 0.5 mol
Number of moles = given mass / molar mass = 22 / 44 = 0.5 mol
Question. Two students of class IX, Kaveri and Nalini, were asked to take 5.3 g of sodium carbonate and 6 g of ethanoic acid to make 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Kaveri followed the instructions but Nalini took the chemicals without measuring their amounts.
Read the above passage and answer the following questions
Answer. (a) whose activity do you think will be in agreement with the law of conservation of mass?
(b) State the law of conservation of mass.
(c) Whose method do you like and why?
(a) Kaveri took the chemicals after measuring them so she is following the law of conservation of mass.
(b) Accoring to law of conservation of mass , “Mass can neither be created nor be destroyed in a chemical reaction.”
(c) I liked the method of Kaveri, as she performed the experiment following all the instructions.
Question. (i) What do the following symbols / formulae stand for :
(a) 2O (b) O2 (c) O2 (d) H2O
(ii) Give the chemical formula of the following compounds :
(a) Potassium carbonate (b) Calcium chloride
(iii) Calculate the formula unit mass of Al2(SO4) .
Answer. (Given : Atomic mass of Al = 27 u, S = 32 u, O = 16 u)
(i) (a) 2O = 2 atoms of oxygen
(b) O2 = Diatomic oxygen
(c) O3 = Triatomic oxygen / ozone
(d) H2O=Two atoms of hydrogen and one atom of oxygen construct one molecule of water (H2O)
(ii) Potassium carbonate – K2CO3
Calcium chloride – CaCl2
(iii) Formula unit mass of any compound is the sum of atomic masses of all atoms in the formula of that compound.
Formula unit mass of Al2(SO4)3
Atomic mass of Al = 27 × 2 = 54 u
Atomic mass of S = 32 × 3 = 96 u
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