10th Standard Syllabus & Materials
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Tamilnadu 10th Standard Social Science GEO - India - Resources and Industries Important Questions And Answers Study Material - QB365 Set A

Published on: 06/12/2019
Atoms and Molecules
Download Tamil Nadu 10th Standard Science question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Science Test1.
Derive the relationship between Relative molecular mass and Vapour density.
2.
Give the salient features of “Modern atomic theory”.
3.
What mass of hydrogen and oxygen will be produced on complete electrolysis of 18 g of water
2 g hydrogen and 32 g oxygen
2 g hydrogen and 16 g oxygen
4 g hydrogen and 32 g oxygen
4 g hydrogen and 14 g oxygen
4.
The total number of electrons present in 16 g of methane gas is
96.352 x 1023
48.176 x 1023
6.023 x 1023
30.11 x 1023
5.
The atomicity of Chlorine is
1
4
8
2
6.
The relative atomic masses of many elements are not whole number because
they are not determined accurately
they exist as isotopes
due to impurities
atoms ionize
7.
An atom of an element has 13 electrons and mass number 27. The nucleus of this atom contains ____________ neutrons.
26
13
14
27
8.
List any two differences between atom and molecules.
9.
State Avogadro's Hypothesis.
10.
Give an example of homo triatomic molecule
11.
What is a hetero atomic molecule? Give two examples
12.
How will you determine the atomicity of gases using Avogadro's hypothesis?
13.
Give the applications of Avogadro's hypothesis.
14.
Calculate the number of molecules in 11g of CO2
15.
Calculate the number of moles in 81g of aluminum.
16.
Calculate the gram molecular mass of carbon dioxide (CO2)
1.
Relative molecular mass:
Relative molecular mass of a gas or vapour is the ratio between the mass of one molecule of gas or vapour to mass of one atom of hydrogen.
Relative molecular mass \(=\frac{\text { Mass of } 1 \text { molecule of a gas (or) vapour at STP }}{\text { Mass of } 1 \text { atom of hydrogen }}\) .......(1)
Vapour density:
It is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
Vapour density (V.D) \(=\frac{\text { Mass of a given volume of gas(or) vapour at S.T.P }}{\text { Mass of the same volume of the hydrogen }}\)
According to Avogadro's law [Let the number of molecules in one volume = 'n']
Vapour density (V.D) \(=\frac{\text { Mass of 'n' molecules of a gas (or) vapour at S.T.P }}{\text { Mass of ' } n \text { 'molecules of a hydrogen }}\)
cancelling 'n' vapour density (V.D) \(=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P }}{\text { Mass of } 1 \text { molecule of a hydrogen }}\)
Hydrogen is diatomic molecule, so
Vapour density \(=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P }}{\text { Mass of } 2 \text { atoms of hydrogen }}\)
Vapour density \(=\frac{\text { mass of } 1 \text { molecule of a gas or vapour at S.T.P }}{2 \times \text { mass of 1atom of hydrogen }}\)
2 x Vapour density \(=\frac{\text { mass of } 1 \text { molecule of a gas (or) vapour at S.T.P }}{\text { mass of 1atom of hydrogen }}\)
From eqn (1)
2 x Vapour density = Relative molecular mass
2.
(i) An atom is no longer indivisible.
(ii) Atoms of the same element may have different atomic masses (isotopes 17Cl35, 17Cl37)
(iii) Atoms of different elements may have the same atomic masses (isobars 18Ar40, 17Ca40)
(iv) Atoms of one element can be transmuted into atoms of other elements. So atom is no longer indestructible. It is called artificial transmutation.
(v) Atoms may not always combine in a simple whole number ratio [Eg: Glucose C6H12O6, sucrose C12H22O11)
(vi) Atom is the smallest particle that takes part in a chemical reaction.
(vii) The mass of an atom can be converted into energy (E = mc2).
3.
(b)
2 g hydrogen and 16 g oxygen
4.
(c)
6.023 x 1023
5.
(d)
2
6.
(b)
they exist as isotopes
7.
(c)
14
8.
|
Atom |
Molecule |
|---|---|
| An atom is the smallest particle of an element. | A molecule is the smallest particle of an element or compound. |
| Atom does not exist in a free state except in a noble gas. | Molecule exist in a free state. |
| Except for noble gas, atoms are highly reactive. | Molecules are less reactive. |
| Atom does not have chemical bond. | Atoms in molecules are held by chemical bonds. |
9.
Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the equal number of molecules"
10.
Ozone O3
11.
The molecules that consists of atoms of different elements are called heteroatomic molecule.
Examples: HCl, H2O
12.
The atomicity of an element can be derived using Avogadro's hypothesis. Let us consider the following equation
| H2(g) | + | CI2(g) | ➝ | 2HCI(g) |
| 1 volume | + | 1 volume | ➝ | 2 volumes (By Gay-Lussac's Law) |
According to Avogadro's law, 1 volume of any gas occupy 'n' number of molecules.
| n molecules |
+ n molecules |
➝ 2n molecules |
if n = 1 then
| 1 molecule | + | 1 molecule | ⟶ | 2 molecules. |
| \(\frac{1}{2}\) molecule | + | \(\frac{1}{2}\) molecule | ➝ | 1 molecule |
(i) 1 molecule of hydrogen chloride gas is made up of \(\frac{1}{2}\) molecule of hydrogen and \(\frac{1}{2}\) molecule of chlorine
(ii) But 1 molecule of hydrogen chloride contains one atom of hydrogen and I atom of chlorine
(iii) Hence \(\frac{1}{2}\) molecule of hydrogen = 1 atom of hydrogen
(iv) Or 1 molecule of hydrogen = 2 atoms of hydrogen
(v) So the atomicity of hydrogen is 2, and molecular formula is H2
Similarly,
\(\frac{1}{2}\) molecule of chlorine = 1 atom of chlorine 1molecule of chlorine = 2 atoms of chlorine So the atomicity of chlorine is 2, and its molecular formula is Cl2
13.
Applications of Avogadro's hypothesis
(i) It explains Gay-Lussac's law.
(ii) It helps in the determination of atomicity of gases.
(iii) Molecular formula of gases can be derived using Avogadro's law.
(iv) It determines the relation between molecular mass and vapour density.
(v) It helps to determine Gram molar volume of all gases (i.e. 22.4 lit at S.T.P)
14.
Gram molecular of CO2 = 44g
Number of molecules present in 44g of CO2
= 6.023 X 1023
∴ Number of molecules present in 11g of CO2
= \(\frac{6.023\times{10}^{23}}{44}\)x 11
= 1.53 x 1023 molecules.
15.
Number of moles = \(\frac{Mass}{Atomic \ mass}\)
= \(\frac{81}{27}\) = 3 moles
16.
Atomic masses of C= 12, O= 16
Gram molecular mass of CO2 = 1(C) + 2(O) = 1(12) + 2(16) = 12 + 32 = 44
Gram molecular mass of CO2 = 44g.
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