Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

Class 9 Science by VP Classes

Class 9 : Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

The document Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Science by VP Classes.
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Q1. (a) How do atoms exist?
 (b) What is atomicity?
 (c) What are polyatomic ions?
 Ans.
(a) Atoms of some elements are not able to exist independently. For such elements atoms form molecules and ions. In case of metals and inert gases atoms can exist independently.
Atoms of metals and inert gases: E.g.,

Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

Non-metals: E.g., H2, Cl2, P4, S8
Exceptional non-metal C

(b) The number of atoms constituting a molecule is known as its atomicity.  E.g., O3 → atomicity is 3O2 → atomicity is 2

(c) Polyatomic ions: When more than two atoms combine together and act like an atom with a charge on it is called polyatomic ion.
E.g., OH, NO3, NH4
 

Q2. Calculate (a) the mass of one atom of oxygen
 (b) the mass of one molecule of oxygen
 (c) the mass of one mole of oxygen gas
 (d) the mass of one ion of oxygen
 (e) the number of atoms in 1 mole of oxygen molecule
 Ans. 
(a) Mass of one atom of oxygen 1 mole of oxygen atom = 16gm = 6.022 × 1023 atoms.
∴ Mass of one atom of oxygen 

Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

(b) Mass of one molecule of oxygen 1 molecule of oxygen = O2 = 2 × 16
= 32 u

(c) Mass of one mole of oxygen gas 1 mole of oxygen gas is O2 = 32 u or 32 g

(d) Mass of one ion of oxygen One mole of oxygen = 6.022 × 1023 atoms = 16g.

Mass of one ion of oxygen  Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

e) Number of atoms in one mole of oxygen molecule 1 mole of oxygen molecule i.e.,
O= 6.022 × 1023 molecules.
1 molecule of O2 = 2 atoms.
∴ Number of atoms in 1 mole of oxygen molecule = 6.022 × 1023 × 2 atoms
= 1.2044 × 1024 atoms
 

Q3. What is meant by atomic mass, gram atomic mass of an element? Why is the mass have different expressions i.e., ‘u’ and ‘g’?
 Ans. 
The atoms are very tiny and their individual mass cannot be calculated as it is negligible. Hence the mass of atoms is expressed in units with respect to a fixed standard. Initially hydrogen atom with mass 1 was taken as standard unit by Dalton. Later, it was replaced by oxygen atom (O=16). But due to the isotopes the masses were found in fractions instead of whole number. Hence, carbon (C=12) isotope was taken as standard unit and was universally accepted.
The atomic mass unit is equal to one twelfth (1/12) the mass of an atom of carbon-12, its unit is u.

Gram atomic mass: When the atomic mass of an element is expressed in grams, it is called the gram atomic mass of the element.
The mass of atoms, molecules is expressed in ‘u’ and the mass of moles i.e., molar mass is expressed in g.
 

Q4. Define a mole. Give the significance of the mole.
 Ans.
Mole–One mole of any species (atoms, molecules, ions or particles) is that quantity or number having a mass equal to its atomic or molecular mass in grams.
1 mole = 6.022 × 1023 in number (atoms, molecules, ions or particles)
 

Significance of the mole

1. A mole gives the number of entities present i.e, 6.022 × 1023 particles of the substance.

2. Mass of 1 mole is expressed as M grams.

3. Mass of 1 mole = mass of 6.022 × 1023 atoms of the element.

E.g., 1 mole of O= 6.022 × 1023 atoms
1 mole of O2 = 2 × 16 = 32 g 6.022 × 2 × 1023 = 1.2044 × 104 atoms  
1 mole of (compound) HCl = 6.022 × 1023 atoms of H and Cl atoms(1 + 35.5 = 36.5 g) (6.022 × 1023 molecules of HCl)
 

Q5. Barium carbonate decomposes when heated BaCO(s)  → BaO(s) + CO2(g)
 (a) A student heated a 10.0 g sample of barium carbonate until it was fully decomposed.
 (i) Calculate the number of moles of barium carbonate the student used.
 (ii) Calculate the volume of carbon dioxide gas produced at room temperature and pressure. Give your answer in dm3.
 (b) The student added 2.00 g of the barium oxide produced to water.
 BaO + H2O → Ba(OH)2 Calculate the mass of barium hydroxide that can be made from 2.00 g of barium oxide. The molecular mass of Ba(OH)2 is 171.
 Ans.
(a) (i) Relative formula mass BaCO= 197, moles of barium carbonate = (10.0/197 = 0.0508 mol.
(ii) Volume of carbon dioxide = 1.22 dm3
(b) Mass of barium hydroxide = 2.24 g


Q6. Magnesium sulphate crystals are hydrated. A student heated some hydrated magnesium sulphate crystals in a crucible and obtained the following results.
 Mass of hydrated magnesium sulfate crystals = 4.92 g Mass of water removed = 2.52 g
 (i) Calculate the number of moles of water removed.
 (ii) Calculate the number of moles of anhydrous magnesium sulfate remaining in the crucible. The molecular mass of anhydrous magnesium sulfate is 120.
 (iii) Calculate the ratio of moles of anhydrous magnesium sulfate : moles of water.
 Ans. 
(i) Moles of water = 2.52/18 = 0.14. moles.
(ii) Moles of anhydrous magnesium sulfate = 0.02 moles.
(iii) Ratio = 0.02/0.02 : 0.14/0.02 = ratio in whole numbers is 1 : 7.
 

Q7. Calculate the mass of each element in potassium carbonate, K2CO3.
 Ans. 
l First calculate the formula mass for K2CO3. Find the atomic mass of each element from the periodic table. Multiply it by the number of times it appears in the formula and add up the total

Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

  • To find the percent of each element divide the part of the formula mass that pertains to that element with the total formula mass

Percent of Potassium Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

Percent of Carbon Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

Percent of Oxygen Long Answers - Atoms and Molecules, Science, Class 9 Class 9 Notes | EduRev

Q8. For each of the following calculate the empirical formula:
 (a) When 2.20 g of a hydrocarbon, D, is burnt in excess oxygen, 6.90 g of CO2 and 2.83 g of water are produced.
 (b) When 1.52 g of compound E, which contains carbon, hydrogen and oxygen only, is burnt in excess oxygen, 3.04 g CO2 and 1.24 g H2O are produced.
 Ans.
(a) Moles of CO= 6.90 ÷ 44.01 = 0.157 mol
Moles of H2O = 2.83 ÷ 18.02 = 0.157 mol
Moles of C = 0.157 mol
moles of H = 2 ÷ 0.157 = 0.304 mol
Empirical formula = CH2

(b) Mass of C in CO2 = 12.01 ÷ 44.01 3.04 = 0.830 g
Mass of H in H2O = 2.02 ÷ 18.02 1.24 = 0.139 g
Mass of O = 1.52 – (0.830 + 0.139) = 0.551 g
Ratio of moles C : H : O = 0.0691 : 0.139 : 0.0344
Whole number ratio = 2.01 : 4.04 : 1
Empirical formula: C2H4O

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