Samacheer Kalvi Class 10 Science Solutions Chapter 7 Atoms and Molecules

I. Choose the best answer:

Question 1. Which of the following has the smallest mass?
(a) \( 0.023 \times 10^{23} \) atoms of He
(b) 1 atom of He
(c) 2 g of He
(d) 1-mole atoms of He
Answer: (b) 1 atom of He
In simple words: To find the smallest mass, we compare values in the same units. A single atom of Helium is much lighter than a gram or a mole of atoms. One mole of helium atoms has a mass of 4g, which is a very large number of atoms.

🎯 Exam Tip: Remember that a 'mole' represents a very large number of particles (Avogadro's number), so a single atom will always have a much smaller mass.

Question 2. Which of the following is a triatomic molecule?
(a) Glucose
(b) Helium
(c) Carbon dioxide
(d) Hydrogen.
Answer: (c) Carbon dioxide
In simple words: A triatomic molecule means it has three atoms. Glucose has many atoms, Helium has one atom, and Hydrogen has two atoms. Carbon dioxide (CO2) has one carbon atom and two oxygen atoms, making a total of three atoms.

🎯 Exam Tip: To identify atomicity, simply count the total number of atoms of all elements in the chemical formula (e.g., for CO2, 1 Carbon + 2 Oxygen = 3 atoms).

Question 3. The volume occupied by 4.4 g of CO2 at S.T.P:
(a) 22.4 litre
(b) 2.24 litre
(c) 0.24 litre
(d) 0.1 litre
Answer: (b) 2.24 litre
In simple words: At Standard Temperature and Pressure (STP), one mole of any gas takes up 22.4 liters. The molecular mass of CO2 is 44g, so 44g is one mole. Since we have 4.4g, which is 1/10th of a mole, it will occupy 1/10th of the volume, which is 2.24 liters.

🎯 Exam Tip: Always remember that 1 mole of any ideal gas occupies 22.4 L at STP. This is a fundamental concept for gas calculations.

Question 4. Mass of 1 mole of Nitrogen atom is _____
(a) 28 amu
(b) 14 amu
(c) 28 g
(d) 14 g.
Answer: (d) 14 g
In simple words: The atomic mass of nitrogen is 14.00674. When we talk about 1 mole of nitrogen atoms, its mass is approximately 14 grams. The 'amu' unit is for a single atom, while 'grams' is for a mole of atoms.

🎯 Exam Tip: Understand the difference between atomic mass unit (amu) for a single atom and molar mass (grams) for a mole of atoms; they share the same numerical value.

Question 5. Which of the following represents 1 amu?
(a) Mass of a C – 12 atom
(b) Mass of a hydrogen atom
(c) 1/12 th of the mass of a C – 12 atom
(d) Mass of O – 16 atom
Answer: (c) 1/12 th of the mass of a C – 12 atom
In simple words: One atomic mass unit (amu) is a tiny unit used to measure the mass of atoms. It is defined as exactly one-twelfth of the mass of a single carbon-12 atom. This standard helps keep atomic mass measurements consistent.

🎯 Exam Tip: The definition of 1 amu is based on the carbon-12 isotope because it provides a stable and widely accepted reference point for comparing atomic masses.

Question 6. Which of the following statement is incorrect?
(a) One gram of C – 12 contains Avogadro's number of atoms.
(b) One mole of oxygen gas contains Avogadro's number of molecules.
(c) One mole of hydrogen gas contains Avogadro's number of atoms.
(d) One mole of electrons stands for \( 6.023 \times 10^{23} \) electrons.
Answer: (a) One gram of C – 12 contains Avogadro's number of atoms.
In simple words: One gram of Carbon-12 does not contain Avogadro's number of atoms. Instead, 12 grams of Carbon-12 contains Avogadro's number of atoms. The number of atoms in 1 gram of Carbon-12 is actually \( \frac{6.023 \times 10^{23}}{12} = 5.018 \times 10^{22} \) atoms. This means statement (a) is incorrect.

🎯 Exam Tip: Avogadro's number (approximately \( 6.022 \times 10^{23} \)) is the number of particles (atoms, molecules, ions, etc.) in one mole of any substance.

Question 7. The volume occupied by 1 mole of a diatomic gas at S.T.P is:
(a) 11.2 litre
(b) 5.6 litre
(c) 22.4 litre
(d) 44.8 litre
Answer: (c) 22.4 litre
In simple words: At Standard Temperature and Pressure (STP), one mole of any gas, whether it is made of one atom or two atoms, always takes up 22.4 liters of space. This is a basic rule for gases. This rule holds true for all ideal gases.

🎯 Exam Tip: Remember the molar volume of a gas at STP (0°C and 1 atm pressure) is a constant, 22.4 L, regardless of the gas type.

Question 8. In the nucleus of \( _{20}\mathrm{Ca}^{40} \), there are
(a) 20 protons and 40 neutrons
(b) 20 protons and 20 neutrons
(c) 20 protons and 40 electrons
(d) 40 protons and 20 electrons
Answer: (b) 20 protons and 20 neutrons
In simple words: For an atom like \( _{20}\mathrm{Ca}^{40} \), the bottom number (20) is the atomic number, which tells us how many protons there are. The top number (40) is the mass number, which is the total of protons and neutrons. So, to find the number of neutrons, we subtract the atomic number from the mass number: \( 40 - 20 = 20 \) neutrons.

🎯 Exam Tip: The atomic number (Z) equals the number of protons, and for a neutral atom, also the number of electrons. The mass number (A) equals protons + neutrons.

Question 9. The gram molecular mass of oxygen molecule is _____
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g.
Answer: (c) 32 g
In simple words: An oxygen molecule is written as O2, meaning it has two oxygen atoms. Each oxygen atom has an atomic mass of 16 g/mol. So, for an O2 molecule, the molecular mass is \( 16 \times 2 = 32 \) grams. This is the mass of one mole of oxygen molecules.

🎯 Exam Tip: Always distinguish between atomic mass (for a single atom) and molecular mass (for a molecule, which might consist of multiple atoms).

Question 10. 1 mole of any substance contains _____ molecules.
(a) \( 6.023 \times 10^{23} \)
(b) \( 6.023 \times 10^{-23} \)
(c) \( 3.0115 \times 10^{23} \)
(d) \( 12.046 \times 10^{23} \)
Answer: (a) \( 6.023 \times 10^{23} \)
In simple words: One mole of any substance, whether it's atoms, molecules, or anything else, always contains a specific number of particles. This number is known as Avogadro's number, which is approximately \( 6.023 \times 10^{23} \). This is a foundational constant in chemistry.

🎯 Exam Tip: Avogadro's number is a conversion factor that links the macroscopic world (grams, moles) to the microscopic world (atoms, molecules).

II. Fill in the blanks:

Question 1. Atoms of different elements having _____ mass number, but _____ atomic numbers are called isobars.
Answer: same, different
In simple words: Isobars are like atomic cousins. They come from different elements, meaning they have a different number of protons (atomic number), but they have the same total mass (mass number).

🎯 Exam Tip: Remember "Isobars" share the same mass number (A), while "Isotopes" share the same atomic number (Z) but differ in mass number.

Question 2. Atoms of different elements having same number of _____ are called isotones.
Answer: neutrons
In simple words: Isotones are atoms that have the same number of neutrons, even if they are from different elements and have different numbers of protons. They are defined by their neutron count.

🎯 Exam Tip: To find the number of neutrons, subtract the atomic number (Z) from the mass number (A) for each atom.

Question 3. Atoms of one element can be transmuted into atoms of other element by _____.
Answer: artificial transmutation
In simple words: Artificial transmutation is a process where one element is changed into another by making changes to its atomic nucleus. This usually happens in a laboratory by bombarding the nucleus with particles. This is how new elements are created.

🎯 Exam Tip: Nuclear reactions, unlike chemical reactions, involve changes in the nucleus, leading to the formation of different elements.

Question 4. The sum of the numbers of protons and neutrons of an atom is called its _____.
Answer: mass number
In simple words: The mass number of an atom is a simple count of all the protons and neutrons inside its nucleus. It essentially tells us how heavy an atom is. Each proton and neutron has a mass close to 1 amu.

🎯 Exam Tip: The mass number is crucial for identifying different isotopes of an element, as isotopes have the same number of protons but different numbers of neutrons.

Question 5. Relative atomic mass is otherwise known as _____.
Answer: standard atomic weight
In simple words: Relative atomic mass is just another name for standard atomic weight. It tells us how heavy an atom of an element is compared to a standard, which is usually carbon-12. This average mass considers all the isotopes of an element.

🎯 Exam Tip: The relative atomic mass is a weighted average of the isotopic masses, taking into account their natural abundance.

Question 6. The average atomic mass of hydrogen is _____ amu.
Answer: 1.008
In simple words: The average atomic mass of hydrogen is very close to 1.008 atomic mass units (amu). This value is slightly more than 1 because there are different isotopes of hydrogen, and this is the average mass considering all of them.

🎯 Exam Tip: The slight difference from exactly 1 amu is due to the presence of heavier hydrogen isotopes like deuterium, which contribute to the average.

Question 7. If a molecule is made of similar kind of atoms, then it is called _____ molecule.
Answer: homoatomic
In simple words: A homoatomic molecule is a molecule where all the atoms are of the same type, like an oxygen molecule (O2) or a nitrogen molecule (N2). "Homo" means same, so it's a molecule of the same atoms.

🎯 Exam Tip: Contrast this with a "heteroatomic" molecule, which contains atoms of different elements, like water (H2O) or carbon dioxide (CO2).

Question 8. The number of atoms present in a molecule is called its _____.
Answer: atomicity
In simple words: Atomicity is a simple count of how many atoms are joined together to make one molecule. For example, an oxygen molecule (O2) has an atomicity of 2.

🎯 Exam Tip: Atomicity helps classify molecules (e.g., monoatomic like He, diatomic like O2, triatomic like O3, polyatomic like P4).

Question 9. One mole of any gas occupies _____ ml at S.T.P
Answer: 22,400
In simple words: At Standard Temperature and Pressure (STP), one mole of any gas always fills up 22.4 liters of space. Since 1 liter is 1000 milliliters, this is the same as 22,400 milliliters. This is a standard value for gas volume.

🎯 Exam Tip: Remember to pay attention to the units requested (liters vs. milliliters) when stating the molar volume at STP.

Question 10. Atomicity of phosphorous is _____.
Answer: four
In simple words: Phosphorous usually exists as a molecule with four atoms joined together, written as P4. So, its atomicity is 4. White phosphorus, for example, forms tetrahedral P4 molecules.

🎯 Exam Tip: The common allotrope of phosphorus (white phosphorus) is P4, making it a polyatomic molecule.

III. Match the following:

Question 1. Match the Column I with Column II.

🎯 Exam Tip: To accurately match, always calculate the number of moles using the formula: moles = given mass / molar mass. Pay attention to whether it's an atom or a diatomic molecule.

IV. True or False: (If false give the correct statement)

Question 1. Two elements sometimes can form more than one compound.
Answer: True
In simple words: It is true that two elements can combine in different ways to make more than one compound. For example, carbon and oxygen can form both carbon monoxide (CO) and carbon dioxide (CO2). This is a basic principle in chemistry.

🎯 Exam Tip: This concept is explained by the Law of Multiple Proportions, which states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios.

Question 2. Nobel gases are diatomic.
Answer: False – Noble gases are Monoatomic.
In simple words: Noble gases are special because they exist as single atoms and do not easily combine with other atoms, even themselves. So, they are called monoatomic, not diatomic. This is why they are very stable.

🎯 Exam Tip: Noble gases (like Helium, Neon, Argon) have a full outer electron shell, which makes them very unreactive and prevents them from forming bonds with other atoms.

Question 3. The unit of gram atomic mass of an element has no unit.
Answer: False – The unit of gram atomic mass of an element is gram.
In simple words: The unit for gram atomic mass is indeed grams. When we talk about the mass of one mole of atoms in grams, that unit is explicitly stated. Atomic mass units (amu) are for individual atoms, but gram atomic mass is a bulk measurement.

🎯 Exam Tip: Always include the correct unit (grams) when referring to gram atomic mass or molar mass to avoid confusion.

Question 4. 1 mole of Gold and Silver contain same number of atoms.
Answer: True
In simple words: It is true that one mole of any substance, whether it's gold or silver, will always have the same number of particles. This number is Avogadro's number. Even though they have different masses, they contain the same count of atoms.

🎯 Exam Tip: A mole is a unit of quantity, not mass. Just like a dozen of eggs and a dozen of apples both mean 12 items, one mole of any substance means Avogadro's number of particles.

Question 5. Molar mass of CO2 is 42 g.
Answer: False – Molar mass of CO2 is 44 g.
In simple words: Carbon has an atomic mass of 12 g/mol, and oxygen has an atomic mass of 16 g/mol. Since CO2 has one carbon and two oxygen atoms, its molar mass is \( 12 + (2 \times 16) = 12 + 32 = 44 \) g. So, 42g is not correct.

🎯 Exam Tip: Always calculate molar mass by summing the atomic masses of all atoms present in the chemical formula, considering their subscripts.

V. Assertion and Reason:

Answer the following questions using the data given below:

Question 1. Assertion: Atomic mass of aluminium is 27 Reason: An atom of aluminium is 27 times heavier than 1/12 th of the mass of the C-12 atom.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn't explains Assertion.
Answer: (a) Assertion and Reason are correct, Reason explains the Assertion.
In simple words: The assertion is correct because the atomic mass of aluminium is indeed 27. The reason is also correct, as the atomic mass unit is defined based on carbon-12, and an aluminum atom is 27 times heavier than 1/12th of a carbon-12 atom. The reason directly explains why the atomic mass is 27.

🎯 Exam Tip: For assertion-reason questions, first check if both statements are individually true. If both are true, then check if the reason logically explains the assertion using "because" between them.

Question 2. Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u. Reason: The natural abundance of Chlorine isotopes are not equal.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn't explains Assertion.
Answer: (a) Assertion and Reason are correct, Reason explains the Assertion.
In simple words: Chlorine has an average relative molecular mass of 35.5 amu because it naturally exists as a mix of different isotopes (Cl-35 and Cl-37) that are not present in equal amounts. This difference in abundance causes the average atomic mass to be a decimal number, specifically 35.5.

🎯 Exam Tip: Relative atomic mass values that are not whole numbers indicate the presence of isotopes with different natural abundances.

VI. Short answer questions:

Question 1. Define: Relative atomic mass.
Answer: Relative atomic mass of an element is the ratio between the average mass of its isotopes to \( \frac{1}{12^{th}} \) part of the mass of a carbon-12 atom. It is denoted as Ar.
The formula for relative atomic mass is:
\[ \mathrm{A}_{\mathrm{r}} = \frac{\text{Average mass of the isotopes of the element}}{\frac{1}{12^{\text{th}}} \text{ of the mass of one Carbon - 12 atom}} \]
In simple words: Relative atomic mass compares how heavy an average atom of an element is to a very small piece of a carbon-12 atom (which is \( \frac{1}{12} \) of its mass). It's a way to measure atomic weights relative to a standard.

🎯 Exam Tip: Clearly state the standard reference (carbon-12 atom) when defining relative atomic mass, as this is a key part of the definition.

Question 2. Write the different types of isotopes of oxygen and its percentage abundance.
Answer: Oxygen has three stable isotopes. They are:

🎯 Exam Tip: Isotopic abundance is important because it determines the average atomic mass of an element, which is the value found on the periodic table.

Question 3. Define Atomicity.
Answer: The number of atoms present in the molecule is called its 'Atomicity'.
In simple words: Atomicity is simply a count of how many atoms are bonded together to form a single molecule. For example, a water molecule (H2O) has an atomicity of 3 (two hydrogen atoms and one oxygen atom).

🎯 Exam Tip: Atomicity helps classify molecules as monoatomic (e.g., He), diatomic (e.g., O2), triatomic (e.g., O3), or polyatomic (e.g., S8).

Question 4. Give any two examples for heteroatomic molecules.
Answer: HI, HCl, CO, HBr, HF. (Any two examples from these are acceptable)
In simple words: Heteroatomic molecules are those made up of atoms from different elements. For example, in water (H2O), there are hydrogen and oxygen atoms. HI (hydrogen iodide) and CO (carbon monoxide) are also good examples, as they both have two different types of atoms.

🎯 Exam Tip: The key characteristic of a heteroatomic molecule is that it must contain at least two different chemical elements in its composition.

Question 5. What is Molar volume of a gas?
Answer: One mole of any gas occupies 22.4 litres. (or) 22400 ml at S.T.P. This volume is called as molar volume.
In simple words: Molar volume is the space taken up by one mole of any gas when it's at standard conditions (like a specific temperature and pressure). For most calculations, this is 22.4 liters or 22,400 milliliters. It's a constant value for all ideal gases.

🎯 Exam Tip: Specify the conditions (STP) when defining molar volume, as the volume changes with temperature and pressure.

Question 6. Find the percentage of nitrogen in ammonia.
Answer: Molar mass of \( \mathrm{NH}_{3} = 1(14) + 3(1) = 17 \) g.
\[ \text{Mass \% of Nitrogen} = \frac{\text{Mass of nitrogen in the compound}}{\text{Molar mass of the compound}} \times 100 \]
Mass % of Nitrogen \( = \frac{14}{17} \times 100 = 82.35\% \)
In simple words: Ammonia (NH3) has one nitrogen atom (mass 14) and three hydrogen atoms (mass 1 each), making its total mass 17. To find the percentage of nitrogen, we divide the mass of nitrogen by the total mass of ammonia and multiply by 100, which gives about 82.35%.

🎯 Exam Tip: When calculating percentage composition, ensure you use the molar mass of the element in the compound and the overall molar mass of the compound correctly.

VII. Long answer questions:

Question 1. Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Answer: The molecular mass of water \( (\mathrm{H}_{2}\mathrm{O}) \) is 18.
18 g of water molecule = 1 mole.
0.18 g of water \( = \frac{1}{18} \times 0.18 = 0.01 \) mole.
Since 1 mole of water contains Avogadro's number \( (6.023 \times 10^{23}) \) water molecules.
\( 0.01 \) mole of water contain \( \frac{6.023 \times 10^{23}}{1} \times 0.01 = 6.023 \times 10^{21} \) molecules.
In simple words: First, we find out how many moles are in 0.18 grams of water. Since 1 mole of water weighs 18 grams, 0.18 grams is 0.01 moles. Then, because 1 mole always has Avogadro's number of molecules, we multiply 0.01 by Avogadro's number to get \( 6.023 \times 10^{21} \) molecules in that drop.

🎯 Exam Tip: This problem requires a two-step calculation: first convert mass to moles, then moles to the number of molecules using Avogadro's number.

Question 2. \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \)
(The atomic mass of nitrogen is 14, and that of hydrogen is 1)
1 mole of nitrogen (........g) +
3 moles of hydrogen (.........g) →
2 moles of ammonia (.........g)
Answer:
1 mole of nitrogen (28 g) +
3 moles of hydrogen (6 g) →
2 moles of ammonia (34 g)
In simple words: From the atomic masses, one mole of nitrogen (N2) is \( 2 \times 14 = 28 \) grams. Three moles of hydrogen (3 H2) are \( 3 \times (2 \times 1) = 6 \) grams. When these react, they form two moles of ammonia (2 NH3), which has a total mass of \( 2 \times (14 + 3 \times 1) = 2 \times 17 = 34 \) grams. This shows mass is conserved in the reaction.

🎯 Exam Tip: Always balance the chemical equation first and use the molar masses of the reactants and products to fill in the mass values accurately.

Question 3. Calculate the number of moles in
(i) 27 g of Al;
(ii) \( 1.51 \times 10^{23} \) molecules of \( \mathrm{NH}_{4}\mathrm{Cl} \).
Answer:
(i) 27 g of Al
Given mass atomic mass \( = \frac{\text{Given Mass}}{\text{Atomic Mass}} = \frac{27}{27} \)
\( = 1 \) mole
(ii) \( 1.51 \times 10^{23} \) molecules of \( \mathrm{NH}_{4}\mathrm{Cl} \)
Number of moles \( = \frac{\text{Number of molecules given}}{6.023 \times 10^{23}} \)
\( = \frac{1.51 \times 10^{23}}{6.023 \times 10^{23}} = 0.25 \) moles
In simple words: For aluminum, since 27 grams is its atomic mass, 27 grams means 1 mole. For ammonium chloride molecules, we divide the given number of molecules by Avogadro's number to find the number of moles. This shows us that \( 1.51 \times 10^{23} \) molecules is one-quarter of a mole.

🎯 Exam Tip: Remember two main formulas for moles: moles = mass / molar mass (for grams) and moles = number of particles / Avogadro's number (for counts).

Question 4. Give the salient features of “Modern atomic theory".
Answer: The salient features of "Modern atomic theory" are:
1. An atom is no longer indivisible.
2. Atoms of the same element may have different atomic mass.
3. Atoms of different elements may have the same atomic masses.
4. Atoms of one element can be transmuted into atoms of other elements. In other words, an atom is no longer indestructible.
5. Atoms may not always combine in a simple whole-number ratio.
6. Atom is the smallest particle that takes part in a chemical reaction.
7. The mass of an atom can be converted into energy \( [\mathrm{E} = \mathrm{mc}^{2}] \).
In simple words: The modern atomic theory updated older ideas by saying that atoms can be broken down into smaller parts, atoms of the same element can have different weights (isotopes), and atoms can even change into other atoms. It also stated that mass can turn into energy. These points changed our understanding of matter.

🎯 Exam Tip: When listing features of a theory, use clear, concise points. Highlighting the key differences from older theories (e.g., Dalton's) demonstrates a deeper understanding.

Question 5. Derive the relationship between Relative molecular mass and Vapour density.
Answer: Relative molecular mass: The relative molecular mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of hydrogen.
Vapour density: Vapour density is the ratio of the mass of 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.
The formula for Vapour density (V.D) is:
\[ \text{Vapour density (V.D)} = \frac{\text{Mass of a given volume of gas or vapour at STP}}{\text{Mass of the same volume of hydrogen}} \]
According to Avogadro's law, equal volumes of all gases contain equal number of molecules.
Let the number of molecules in one volume = n, then
\[ \text{V.D} = \frac{\text{Mass of 'n' molecules of gas or vapour at STP}}{\text{Mass of 'n' molecules of hydrogen}} \]
We can simplify this by removing 'n' because it is common at STP:
\[ \text{V.D} = \frac{\text{Mass of 1 molecule of a gas or vapour at STP}}{\text{Mass of 1 molecule of hydrogen}} \]
Since hydrogen is diatomic \( (\mathrm{H}_{2}) \), one molecule of hydrogen has two atoms:
\[ \text{V.D} = \frac{\text{Mass of 1 molecule of gas or vapour at STP}}{\text{Mass of 2 atoms of hydrogen}} \]
This can be written as:
\[ \text{V.D} = \frac{\text{Mass of 1 molecule of gas or vapour at STP}}{2 \times \text{Mass of 1 atom of hydrogen}} \]
We know that Relative molecular mass \( = \frac{\text{Mass of 1 molecule of gas or vapour at STP}}{\text{Mass of 1 atom of hydrogen}} \).
So,
\[ \text{V.D} = \frac{\text{Relative molecular mass}}{2} \]
\( \implies \) \( 2 \times \) Vapour density = Relative Molecular mass of a gas
[OR]
Relative Molecular Mass = \( 2 \times \) Vapour density
In simple words: Vapour density compares the mass of a gas to the mass of the same volume of hydrogen. We know that if we take the same number of molecules, their mass ratio will be the same as their molecular mass ratio. Since hydrogen is a diatomic molecule (H2), we find that the relative molecular mass of a gas is twice its vapour density. This relationship helps us find the molecular mass of a gas if we know its vapour density.

🎯 Exam Tip: Clearly define both relative molecular mass and vapour density before starting the derivation. The crucial step is applying Avogadro's Law and considering hydrogen's diatomic nature.

VIII. HOT Question:

Question 1. Calcium carbonate is decomposed on heating in the following reaction \( \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2} \)
1. How many moles of Calcium carbonate is involved in this reaction?
2. Calculate the gram molecular mass of calcium carbonate involved in this reaction?
3. How many moles of \( \mathrm{CO}_{2} \) are there in this equation?
Answer:
\( \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2} \)
1. 1 mole of \( \mathrm{CaCO}_{3} \) is involved in this reaction.
2. Gram molecular mass of calcium carbonate
\( \mathrm{CaCO}_{3} = (40 + 12 + 3 \times 16) = 52 + 48 = 100 \) g
3. 1 mole of \( \mathrm{CO}_{2} \) is in this equation.
In simple words: In this reaction, one molecule of calcium carbonate breaks down. This means 1 mole of calcium carbonate is used. The total weight of this 1 mole is 100 grams. From the balanced equation, we can also see that 1 mole of carbon dioxide is made. This is a simple stoichiometric relationship.

🎯 Exam Tip: For stoichiometric questions, always ensure the chemical equation is balanced first. The coefficients in a balanced equation directly give the mole ratios of reactants and products.

IX. Solve the following problems:

Question 1. How many grams are there in the following?
(i) 2 moles of a hydrogen molecule, \( \mathrm{H}_{2} \)
(ii) 3 moles of chlorine molecule, \( \mathrm{Cl}_{2} \)
(iii) 5 moles of sulphur molecule, \( \mathrm{S}_{8} \)
(iv) 4 moles of a phosphorous molecule, \( \mathrm{P}_{4} \)
Answer:
(i) 2 moles of a hydrogen molecule, \( \mathrm{H}_{2} \)
Mass of 1 mole of hydrogen molecule = 2 g
Mass of 2 moles of hydrogen molecule \( = 2 \times 2 = 4 \) g.
(ii) 3 moles of chlorine molecule, \( \mathrm{Cl}_{2} \)
Mass of 1 mole of chlorine molecule = 71 g
Mass of 3 moles of chlorine molecules \( = 71 \times 3 = 213 \) g.
(iii) 5 moles of sulphur molecule, \( \mathrm{S}_{8} \)
Mass of 1 mole of sulphur molecule = 32 g
Mass of 5 moles of sulphur molecules \( = 32 \times 5 = 160 \) g.
(iv) 4 moles of the phosphorous molecule, \( \mathrm{P}_{4} \)
Mass of 1 mole of phosphorous molecule = 30.97 g
Mass of 4 moles of phosphorous molecules \( = 30.97 \times 4 = 123.88 \) g.
In simple words: To find the mass in grams for each substance, we first need to know the mass of one mole of that substance. We multiply the molecular mass (mass of 1 mole) by the number of moles given. For example, since 1 mole of H2 weighs 2g, 2 moles will weigh 4g. We apply this same logic to chlorine, sulfur, and phosphorus molecules to get their total masses.

🎯 Exam Tip: Be careful to use the correct molecular mass for each molecule, especially for polyatomic elements like sulfur (S8) and phosphorus (P4).

Question 2. Calculate the % of each element in calcium carbonate. (Atomic mass: C – 12, O – 16, Ca - 40)
Answer: First, calculate the molar mass of \( \mathrm{CaCO}_{3} \):
\( \mathrm{Ca} = 40 \)
\( \mathrm{C} = 12 \)
\( \mathrm{O}_{3} = 3 \times 16 = 48 \)
Molar mass of \( \mathrm{CaCO}_{3} = 40 + 12 + 48 = 100 \) g/mol.
Now, calculate the percentage of each element:
\[ \text{% of Ca} = \frac{\text{Mass of Ca in the compound}}{\text{Molar mass of CaCO}_{3}} \times 100 = \frac{40}{100} \times 100 = 40\% \]
\[ \text{% of C} = \frac{\text{Mass of C in the compound}}{\text{Molar mass of CaCO}_{3}} \times 100 = \frac{12}{100} \times 100 = 12\% \]
\[ \text{% of O} = \frac{\text{Mass of oxygen in the compound}}{\text{Molar mass of CaCO}_{3}} \times 100 = \frac{3 \times 16}{100} \times 100 = 48\% \]
In simple words: We first find the total weight (molar mass) of calcium carbonate, which is 100 grams. Then, for each element, we take its total weight in the compound, divide it by the total weight of calcium carbonate, and multiply by 100 to get its percentage. Calcium is 40%, Carbon is 12%, and Oxygen is 48%. These percentages sum up to 100.

🎯 Exam Tip: Always calculate the total molar mass of the compound first. Then, for each element, divide the mass contributed by that element by the total molar mass and multiply by 100.

Question 3. Calculate the % of oxygen in \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \).
(Atomic mass: Al – 27, O – 16, S – 32)
Answer: First, calculate the molar mass of \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \).
Molar mass of \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} = [2(\text{Atomic mass of Al}) + 3(\text{Atomic mass of S}) + 12(\text{Atomic mass of O})] \)
\( = 2(27) + 3(32) + 12(16) \)
\( = 54 + 96 + 192 \)
\( = 342 \) g
Now, calculate the percentage of oxygen:
\[ \text{% of Oxygen} = \frac{\text{Mass of oxygen in the compound}}{\text{Molar mass of Al}_{2}(\mathrm{SO}_{4})_{3}} \times 100 \]
Mass of oxygen in \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} = 12 \times 16 = 192 \) g.
\( \text{% of Oxygen} = \frac{192}{342} \times 100 = 56.14\% \).
In simple words: We first find the total weight (molar mass) of aluminum sulfate, which is 342 grams. This compound has 12 oxygen atoms in total. We take the combined mass of these 12 oxygen atoms (192g), divide it by the total mass of the compound, and multiply by 100 to get the percentage. This calculation shows that oxygen makes up about 56.14% of aluminum sulfate.

🎯 Exam Tip: Be careful with subscripts and parentheses in chemical formulas; \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \) means there are 3 sulfur atoms and \( 3 \times 4 = 12 \) oxygen atoms.

Question 4. Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.
Answer: Let the abundance of B – 10 be 'x' and B – 11 be (1 – x).
Average atomic mass of Boron = (Atomic mass of B-10 \( \times \) abundance of B-10) + (Atomic mass of B-11 \( \times \) abundance of B-11)
\( 10.804 = (10 \times x) + (11 \times (1-x)) \)
\( 10.804 = 10x + 11 - 11x \)
\( 10.804 = 11 - x \)
\( x = 11 - 10.804 \)
\( x = 0.196 \)
\( 1 - x = 1 - 0.196 = 0.804 \)
Therefore, % abundance of B – 10 is \( 19.6\% \) and B – 11 is \( 80.4\% \).
[OR - Alternative Method]
Let the % of the isotope B – 10 = x
Then the % of the isotope B – 11 = \( (100 - x) \)
Average atomic mass \( = \frac{(10 \times x) + (11 \times (100-x))}{100} \)
\( 10.804 = \frac{10x + 1100 - 11x}{100} \)
\( 10.804 \times 100 = 1100 - x \)
\( 1080.4 = 1100 - x \)
\( x = 1100 - 1080.4 \)
\( x = 19.6 \)
% abundance of B – 10 = \( 19.6\% \)
% abundance of B – 11 = \( 80.4\% \)
In simple words: We know the average atomic mass of boron and the masses of its two isotopes, B-10 and B-11. We set up an equation where we assume 'x' is the fractional abundance of B-10 and (1-x) for B-11. By solving this equation, we find that B-10 makes up 19.6% of natural boron, and B-11 makes up 80.4%. This is how we determine the natural mix of isotopes.

🎯 Exam Tip: When calculating isotopic abundance, set up an algebraic equation where the sum of the abundances (as fractions or percentages) equals 1 or 100, respectively. Always ensure your final percentages add up to 100.

Samacheer Kalvi 10th Science Atoms and Molecules Additional Important Questions and Answers

I. Choose the correct answer:

Question 1. The first scientific theory of an atom was proposed by:
(a) Ruther Ford
(b) Newland
(c) John Dalton
(d) Neils Bohr
Answer: (c) John Dalton
In simple words: John Dalton was the first person to suggest a scientific theory about atoms. He helped us understand that everything is made of tiny particles.

🎯 Exam Tip: Remember key scientists and their contributions to atomic theory, as these are often tested.

Question 2. Identify the pair that indicates isobars among the following
(a) \(_{1}\mathrm{H}^{2},_{1}\mathrm{H}^{3}\)
(b) \(_{17}\mathrm{Cl}^{35},_{17}\mathrm{Cl}^{37}\)
(c) \(_{18}\mathrm{Ar}^{40},_{18}\mathrm{Ca}^{40}\)
(d) \(_{6}\mathrm{C}^{13},_{7}\mathrm{N}^{14}\).
Answer: (c) \(_{18}\mathrm{Ar}^{40},_{18}\mathrm{Ca}^{40}\)
In simple words: Isobars are atoms of different chemical elements that have the same mass number but different atomic numbers. In this choice, both argon and calcium have a mass number of 40, but they are different elements.

🎯 Exam Tip: Remember that isobars have the same mass number but different atomic numbers, while isotopes have the same atomic number but different mass numbers.

Question 3. Which one of the following represents 180 g of water?
(a) 5 moles of water
(b) 90 moles of water
(c) \(6.023 \times 10^{24}\) molecules of water
(d) \(6.023 \times 10^{22}\) molecules of water
Answer: (c) \(6.023 \times 10^{24}\) molecules of water
In simple words: Water has a molar mass of 18 g/mol. So, 180 g of water is 10 moles (180/18). One mole has Avogadro's number (\(6.023 \times 10^{23}\)) of molecules, so 10 moles will have \(10 \times 6.023 \times 10^{23} = 6.023 \times 10^{24}\) molecules.

🎯 Exam Tip: Always remember that 1 mole of any substance contains Avogadro's number of particles and its mass is equal to its molar mass in grams.

Question 4. The isotope of Carbon-12 contains
(a) 6 protons and 7 electrons
(b) 6 protons and 6 neutrons
(c) 12 protons and no neutrons
(d) 12 neutrons and no protons.
Answer: (b) 6 protons and 6 neutrons
In simple words: For Carbon-12, the atomic number is 6, meaning it has 6 protons and 6 electrons. The mass number is 12, so it has 6 neutrons (12 - 6 = 6).

🎯 Exam Tip: For any atom, the number of protons equals the atomic number (Z), and the number of neutrons equals the mass number (A) minus the atomic number (A-Z).

Question 5. Which contains the greatest number of moles of oxygen atoms?
(a) 1 mol of water
(b) 1 mole of NaOH
(c) 1 mole of Na2CO3
(d) 1 mole of CO
Answer: (c) 1 mole of Na2CO3
In simple words: Let's count the oxygen atoms in each option: water (\(\mathrm{H}_2\mathrm{O}\)) has 1, NaOH has 1, \(\mathrm{Na}_2\mathrm{CO}_3\) has 3, and CO has 1. So, 1 mole of \(\mathrm{Na}_2\mathrm{CO}_3\) has the most oxygen atoms.

🎯 Exam Tip: When comparing moles of specific atoms within compounds, always check the chemical formula to see how many of that atom are present in one molecule of the compound.

Question 6. The mass of proton or neutron is approximately ____.
(a) 1 amu
(b) \(1.609 \times 10^{-19}\) g
(c) 1 g
(d) \(6.023 \times 10^{-23}\) g.
Answer: (a) 1 amu
In simple words: Both protons and neutrons are very tiny particles that make up an atom. Their mass is almost exactly 1 atomic mass unit (amu). This unit is useful for comparing the masses of atoms.

🎯 Exam Tip: Remember that 1 amu is defined as one-twelfth of the mass of a carbon-12 atom, making it a convenient unit for atomic and subatomic particle masses.

Question 7. The natural abundance of C-12 and C-13 are 98.90% and 1.10% respectively. The average atomic mass of carbon is:
(a) 12 amu
(b) 12.011 amu
(c) 14 amu
(d) 12.90 amu
Answer: (b) 12.011 amu
In simple words: To find the average atomic mass, you multiply the mass of each isotope by its percentage abundance (as a decimal) and add them up. Here, it's \((12 \times 0.9890) + (13 \times 0.0110)\), which gives 12.011 amu. This average considers how common each type of carbon atom is.

🎯 Exam Tip: Always convert percentages to decimals when calculating average atomic mass, and ensure all isotopic abundances add up to 100%.

Question 8. The relative atomic mass of magnesium-based on C - 12 scale is ____.
(a) 24 g
(b) 24
(c) 24 amu
(d) 24 kg
Answer: (b) 24
In simple words: The relative atomic mass of an element tells us how many times heavier one atom of that element is compared to one-twelfth the mass of a carbon-12 atom. For magnesium, it's simply 24, as it's a ratio and has no units.

🎯 Exam Tip: Relative atomic mass is a dimensionless quantity (a ratio), so it is expressed as a pure number without units like g or amu, unlike atomic mass itself which can be in amu or grams.

Question 9. If 1.5 moles of oxygen combine with Al to form Al2O3, the mass of Al in g (atomic mass of Al = 27) used in the reaction is:
(a) 2.7
(b) 54
(c) 40.5
(d) 81
Answer: (b) 54
In simple words: First, balance the chemical equation for aluminum and oxygen forming aluminum oxide: \(4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3\). This tells us 3 moles of oxygen react with 4 moles of aluminum. If we have 1.5 moles of oxygen, we will need 2 moles of aluminum. Since the atomic mass of aluminum is 27 g/mol, 2 moles of aluminum will weigh \(2 \times 27 = 54\) grams.

🎯 Exam Tip: Always start by writing and balancing the chemical equation when solving stoichiometry problems to get the correct mole ratios.

Question 10. The atomicity of methane is:
(a) 5
(b) 4
(c) 3
(d) 6
Answer: (a) 5
In simple words: Methane has the chemical formula \(\mathrm{CH}_4\). This means it has one carbon atom and four hydrogen atoms. Atomicity is the total number of atoms in one molecule, so \(1 + 4 = 5\).

🎯 Exam Tip: Atomicity is simply the total count of all atoms (of all types) present in a single molecule of a substance.

Question 11. Find the odd one out ____.
(a) \(_{8}\mathrm{O}^{16}\)
(b) \(_{8}\mathrm{O}^{17}\)
(c) \(_{6}\mathrm{O}^{12}\)
(d) \(_{8}\mathrm{O}^{18}\).
Answer: (c) \(_{6}\mathrm{O}^{12}\)
In simple words: Options (a), (b), and (d) all represent different forms (isotopes) of oxygen because they all have an atomic number of 8. Option (c), however, has an atomic number of 6, which means it is a carbon atom (\(_{6}\mathrm{C}^{12}\)), not an oxygen atom. This makes it the different one.

🎯 Exam Tip: Always pay close attention to the subscript (atomic number) and superscript (mass number) when identifying elements and their isotopes. The subscript defines the element.

Question 12. The volume occupied by 3 moles of HCl gas at STP is:
(a) 22.4 L
(b) 44.8 L
(c) 2.24 L
(d) 67.2 L
Answer: (d) 67.2 L
In simple words: At Standard Temperature and Pressure (STP), one mole of any gas always takes up 22.4 liters of space. So, if you have 3 moles of HCl gas, it will occupy three times that volume, which is 67.2 liters.

🎯 Exam Tip: Remember the molar volume of a gas at STP (22.4 L) as a fundamental constant for calculations involving gases.

Question 13. The mass percentage of hydrogen in ethane (\(\mathrm{C}_2\mathrm{H}_6\)) is:
(a) 25%
(b) 75%
(c) 80%
(d) 20%
Answer: (d) 20%
In simple words: Ethane (\(\mathrm{C}_2\mathrm{H}_6\)) has two carbon atoms and six hydrogen atoms. The total mass is \( (2 \times 12) + (6 \times 1) = 24 + 6 = 30 \) units. The hydrogen part is 6 units. So, the percentage of hydrogen is \((6 \div 30) \times 100\%\), which equals 20%.

🎯 Exam Tip: To calculate mass percentage, find the total mass of the element in the compound, divide by the total molar mass of the compound, and multiply by 100.

Question 14. Which one of the following is a homo diatomic molecule?
(a) H2
(b) CO
(c) NO
(d) O3.
Answer: (a) H2
In simple words: A 'homo diatomic' molecule means it is made of two atoms that are exactly the same. Hydrogen gas (\(\mathrm{H}_2\)) fits this description because it has two hydrogen atoms joined together. Carbon monoxide (CO) and Nitric oxide (NO) have two different atoms, and Ozone (\(\mathrm{O}_3\)) has three atoms.

🎯 Exam Tip: "Homo" means same, "hetero" means different, and "diatomic" means two atoms. Combine these prefixes and suffixes to understand molecule types.

Question 15. The percentage of nitrogen in urea is about:
(a) 38.4
(b) 46.6
(c) 59.1
(d) 61.3
Answer: (b) 46.6
In simple words: The formula for urea is \(\mathrm{CO}(\mathrm{NH}_2)_2\). Its total molecular mass is 60 g/mol, and the nitrogen part (two N atoms) has a mass of 28 g. To find the percentage, you divide 28 by 60 and multiply by 100, which gives about 46.6%. Urea is known for its high nitrogen content.

🎯 Exam Tip: For percentage composition problems, always correctly write the chemical formula first, then calculate the total molar mass and the mass contributed by the specific element.

Question 16. Out of the following the largest number of atoms are contained in:
(a) 11 g of CO2
(b) 4 g of H2
(c) 5 g of NH3
(d) 8 g of SO2
Answer: (b) 4 g of H2
In simple words: To find which option has the most atoms, you need to calculate the moles of each substance and then multiply by the number of atoms in each molecule. For hydrogen gas (\(\mathrm{H}_2\)), 4 grams is 2 moles of molecules (\(4 \text{ g} \div 2 \text{ g/mol}\)). Since each molecule has 2 atoms, this gives a total of 4 moles of atoms. This is more than the other options. For example, 11g of CO2 is 0.25 moles of molecules (11g/44g/mol), which is \(0.25 \times 3 = 0.75\) moles of atoms. Hydrogen is very light, so 4g contains a lot of molecules and atoms.

🎯 Exam Tip: Remember that hydrogen has a very small molar mass, so a given mass of hydrogen often contains a proportionally larger number of moles and atoms compared to heavier elements or compounds.

Question 17. Which of the following is an example of a homo triatomic molecule?
(a) Phosphorous
(b) Sulphur
(d) Ozone.
Answer: (d) Ozone.
In simple words: A 'homo triatomic' molecule means it is made of three atoms that are all the same. Ozone (\(\mathrm{O}_3\)) fits this because it has three oxygen atoms linked together. Phosphorous usually forms molecules with four atoms (\(\mathrm{P}_4\)) and sulfur forms molecules with eight atoms (\(\mathrm{S}_8\)).

🎯 Exam Tip: Understand the prefixes: 'mono' (1), 'di' (2), 'tri' (3), 'tetra' (4), 'poly' (many) when describing atomicity, and 'homo' (same element) vs. 'hetero' (different elements).

Question 18. For the reaction A + 2B → C, 5 moles of A and 8 moles of B will produce:
(a) 5 moles of C
(b) 4 moles of C
(c) 8 moles of C
(d) 13 moles of C
Answer: (b) 4 moles of C
In simple words: The recipe for this reaction says 1 part A needs 2 parts B to make 1 part C. If you have 5 moles of A, you'd need 10 moles of B. But you only have 8 moles of B, so B is the ingredient that will run out first (the limiting reactant). Since 2 moles of B make 1 mole of C, then 8 moles of B will make 4 moles of C.

🎯 Exam Tip: Always identify the limiting reactant first in stoichiometry problems, as it determines the maximum amount of product that can be formed.

Question 19. The vapour density of a gas is 32. Its relative molecular mass will be:
(a) 32
(b) 16
(c) 64
(d) 96
Answer: (c) 64
In simple words: Vapour density is how many times heavier a gas is compared to hydrogen. The relative molecular mass is double the vapour density. So, if the vapour density is 32, the relative molecular mass will be \(2 \times 32 = 64\). This relationship is a useful shortcut in gas calculations.

🎯 Exam Tip: Remember the relationship: Relative Molecular Mass = 2 \(\times\) Vapour Density. This is a common formula used for gases.

Question 20. Find the odd one out ____.
(a) Silver
(b) Potassium
(c) Iron
(d) Phosphorous.
Answer: (d) Phosphorous.
In simple words: Silver, Potassium, and Iron are all types of metals. Phosphorous, on the other hand, is a non-metal. This makes it different from the other three.

🎯 Exam Tip: Classifying elements as metals, non-metals, or metalloids is a fundamental skill in chemistry, often based on their properties and position in the periodic table.

II. Fill in the blanks:

Question 1. The volume occupied by 16 g of oxygen is __________.
Answer: 11.2 L
In simple words: Oxygen gas (\(\mathrm{O}_2\)) has a molar mass of 32 g/mol. At Standard Temperature and Pressure (STP), one mole (32 g) occupies 22.4 liters. So, 16 grams of oxygen is half a mole (\(16 \text{ g} \div 32 \text{ g/mol} = 0.5 \text{ mol}\)). Half a mole will occupy half of 22.4 liters, which is 11.2 liters.

🎯 Exam Tip: Always use the correct molar mass for diatomic gases like oxygen (\(\mathrm{O}_2\)), which is twice the atomic mass of a single oxygen atom.

Question 2. One mole of a triatomic gas contains __________ atoms.
Answer: \(3 \times 6.023 \times 10^{23}\)
In simple words: A triatomic gas molecule has three atoms. One mole of any substance contains Avogadro's number (\(6.023 \times 10^{23}\)) of molecules. So, one mole of a triatomic gas will contain three times Avogadro's number of atoms.

🎯 Exam Tip: Remember that Avogadro's number refers to the number of *molecules* in a mole, and you need to multiply by the atomicity to find the total number of *atoms*.

Question 3. Equal volume of all gases under the same conditions of temperature and pressure contain equal number of __________.
Answer: molecules
In simple words: This is Avogadro's Law. It means that if you have two different gases in the same-sized containers, at the same temperature and pressure, they will have the same number of gas particles (molecules).

🎯 Exam Tip: Avogadro's Law is crucial for understanding gas behavior and is often used in stoichiometry involving gases.

Question 4. The mass of an atom can be converted into energy by using the formula __________.
Answer: \(E = mc^2\)
In simple words: Albert Einstein's famous equation, \(E = mc^2\), shows that mass and energy are directly related and can be converted into each other. Here, E is energy, m is mass, and c is the speed of light. This idea is fundamental to nuclear energy.

🎯 Exam Tip: Understand that mass-energy equivalence means a small amount of mass can be converted into a very large amount of energy, as seen in nuclear reactions.

Question 5. The percentage composition is useful to determine the __________ formula and __________ formula.
Answer: empirical, molecular
In simple words: Knowing the percentage of each element in a compound helps scientists find its simplest chemical formula (empirical formula) and its actual chemical formula (molecular formula). These formulas are important for understanding what a substance is made of.

🎯 Exam Tip: Empirical formula shows the simplest whole-number ratio of atoms, while molecular formula shows the actual number of atoms in a molecule.

III. Match the following:

Question 1. Match the Column I with Column II.

Answer:
A. (ii)
B. (iii)
C. (i)
D. (v)
E. (iv)
In simple words: This matching exercise tests your understanding of basic mole concepts, molar mass, molar volume, and Avogadro's number. Each item in Column I corresponds to a specific quantity that matches one of the values in Column II. For example, 88g of CO2 is 2 moles, and 1 mole of any gas contains Avogadro's number of molecules.

🎯 Exam Tip: Practice converting between mass, moles, volume (for gases at STP), and number of particles, as these calculations are fundamental to chemistry.

Question 2. Match the Column I with Column II.

Answer:
A. (iii)
B. (i)
C. (iv)
D. (v)
E. (ii)
In simple words: This matching question connects key chemical concepts: Carbon-12 as the atomic mass standard, the large number of particles in a kilogram-mole, Einstein's energy-mass equation, the classification of water as a molecule with three different atoms, and the molar mass of carbon dioxide. Each pair represents a fundamental definition or property.

🎯 Exam Tip: Memorize the definition of atomic mass unit (based on C-12), Avogadro's number, and common classifications of molecules (diatomic, triatomic, homo, hetero).

Question 3. Match the Column I with Column II.

Answer:
A. (iv)
B. (v)
C. (ii)
D. (iii)
E. (i)
In simple words: This matching question combines several concepts: classifying molecules like propane (a complex molecule with different atoms), knowing that noble gases are stable and don't easily react, converting mass to moles (0.012 kg is 12g, which is 1 mole of carbon), converting gas volume at STP to moles, and using the formula to find relative molecular mass from vapour density.

🎯 Exam Tip: Pay attention to unit conversions (e.g., kg to g) and conditions (e.g., STP) as they are crucial for accurate calculations in matching questions.

Question 4. Match the Column I with Column II.

Answer:
A. (v)
B. (iii)
C. (iv)
D. (i)
E. (ii)
In simple words: This matching set clarifies key terms about atoms and molecules. Isobars have the same mass but different atomic numbers. Isotopes have the same atomic number but different masses. Isotones have the same number of neutrons. Atomicity refers to the number of atoms in a molecule, which can be found by dividing its molecular mass by the atomic mass. Finally, \(2 \times 16\) relates to the molar mass of diatomic oxygen.

🎯 Exam Tip: Ensure you can clearly define isobars, isotopes, and isotones, as their distinctions are a common source of confusion in exams.

IV. True or False: (If false give the correct statement)

Question 1. Atoms always combine in a simple whole number ratio.
Answer: False – Atoms may not combine always in a simple whole number ratio.
In simple words: While many compounds show simple whole-number ratios (like water, H2O), complex organic molecules or non-stoichiometric compounds might not always have such simple ratios. So, saying "always" makes the statement false.

🎯 Exam Tip: Be careful with absolute terms like "always" or "never" in scientific statements, as there are often exceptions, especially in chemistry.

Question 2. \(2 \times \mathrm{RMM} = \mathrm{VD}\)
Answer: False – \(2 \times \mathrm{VD} = \mathrm{RMM}\)
In simple words: The correct relationship is that the relative molecular mass (RMM) is twice the vapour density (VD), not the other way around. The formula helps us find the molecular mass of a gas from its density compared to hydrogen.

🎯 Exam Tip: Always remember the formula: Relative Molecular Mass = 2 \(\times\) Vapour Density. Getting this formula wrong is a common mistake.

Question 3. The average atomic mass of Beryllium is 9.012 because of the presence of isotopes.
Answer: True
In simple words: Beryllium's average atomic mass is 9.012 because it has different isotopes. The average mass takes into account the mass of each isotope and how common each one is.

🎯 Exam Tip: The average atomic mass listed on the periodic table is a weighted average of all naturally occurring isotopes of an element.

Question 4. The noble gases are diatomic.
Answer: False – The noble gases are mono atomic.
In simple words: Noble gases like Helium (He) and Neon (Ne) exist as single atoms, not as pairs of atoms. They are very stable and don't usually react with other atoms to form molecules.

🎯 Exam Tip: Noble gases are known for their inertness and stable electron configurations, which means they exist as individual atoms (monoatomic) rather than diatomic molecules.

Question 5. The number of atoms present in one mole of phosphorus(\(\mathrm{P}_4\)) is \(4 \times 6.023 \times 10^{23}\)
Answer: True
In simple words: One molecule of phosphorus is \(\mathrm{P}_4\), which means it has 4 phosphorus atoms. One mole of \(\mathrm{P}_4\) molecules contains Avogadro's number (\(6.023 \times 10^{23}\)) of molecules. So, the total number of atoms will be 4 times Avogadro's number.

🎯 Exam Tip: For polyatomic elements like phosphorus (\(\mathrm{P}_4\)) or sulfur (\(\mathrm{S}_8\)), remember to multiply Avogadro's number by the atomicity to find the total number of atoms in a mole of that element.

V. Assertion and Reason:

Question 1. Assertion: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1/12th of the mass of the C-12 atom.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn't explains Assertion.
Answer: (a) Assertion and Reason are correct, Reason explains the Assertion.
In simple words: The statement that aluminum's atomic mass is 27 is true. The reason correctly explains this by defining atomic mass. It states that an aluminum atom is 27 times heavier than the standard unit for measuring atomic mass, which is 1/12th of a carbon-12 atom.

🎯 Exam Tip: For Assertion-Reason questions, first check if both statements are individually true. If both are true, then check if the Reason actually explains the Assertion.

Question 2. Assertion: The volume occupied by 44 g of CO2 is 22.4 L
Reason: The volume occupied by one mole of any gas is 22.4 L
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn't explains Assertion.
Answer: (a) Assertion and Reason are correct, Reason explains the Assertion.
In simple words: Carbon dioxide (\(\mathrm{CO}_2\)) has a molar mass of 44 g/mol. This means 44 grams of \(\mathrm{CO}_2\) is exactly one mole. The reason states that one mole of any gas takes up 22.4 liters at standard conditions, which is also true. Since 44 grams of \(\mathrm{CO}_2\) is one mole, it will indeed occupy 22.4 liters, so the reason clearly explains the assertion.

🎯 Exam Tip: This question tests your knowledge of both molar mass calculations and the molar volume of gases at STP, which are linked by the concept of the mole.

VI. Short answer questions:

Question 1. Define vapour density.
Answer: Vapour density is a measure that compares the mass of a certain volume of a gas or vapor to the mass of an equal volume of hydrogen gas (\(\mathrm{H}_2\)). Both volumes are measured under the same conditions of temperature and pressure. It essentially tells us how much heavier a gas is compared to hydrogen.
\[ \text{V.D. = } \frac{\text{Mass of a certain volume of gas or vapour at STP}}{\text{Mass of the same volume of H2 at STP}} \]In simple words: Vapour density is just a way to see how much heavier a gas is than hydrogen gas when you compare the same amount of both gases under the same conditions.

🎯 Exam Tip: Remember that vapour density is a ratio, so it does not have units. It is often used to determine the molecular mass of an unknown gas.

Question 2. What are isobars? Give an example.
Answer: Isobars are atoms that belong to different chemical elements but have the same mass number. This means they have a different number of protons (atomic number) but the total count of protons and neutrons is identical. A good example of isobars is Argon-40 and Calcium-40.
For example, \(_{18}\mathrm{Ar}^{40}\) and \(_{20}\mathrm{Ca}^{40}\).In simple words: Isobars are like twin atoms from different families; they have the same weight (mass number) but are different elements because they have a different number of protons.

🎯 Exam Tip: Always provide a clear definition and a correct chemical example when asked about terms like isobars, isotopes, or isotones.

Question 3. Write the differences between an atom and a molecule.

Answer: Atoms are the most basic units of an element and participate in chemical reactions, but they might not always exist independently. For example, a single oxygen atom is very reactive. Molecules, however, are formed when two or more atoms bond together, and these combined units can exist freely. This bonding makes them more stable.
In simple words: An atom is a single, tiny building block, like a single Lego piece. A molecule is when two or more of these blocks join together, like a small Lego structure, and it can exist on its own.

🎯 Exam Tip: Focus on key distinctions like participation in reactions versus free existence, and single units versus bonded units, when differentiating between atoms and molecules.

Question 4. What is artificial transmutation?
Answer: Artificial transmutation is a process where one chemical element is changed into another element, meaning atoms are transformed. This shows that atoms are not indestructible, as was once thought. Scientists achieve this by bombarding atomic nuclei with high-energy particles.
In simple words: Artificial transmutation is when humans change one type of atom into another type. It means atoms can be broken apart and changed, not just rearranged.

🎯 Exam Tip: Contrast artificial transmutation with natural radioactivity to highlight that one is human-induced, while the other occurs spontaneously in unstable isotopes.

Question 5. Classify the following based on atomicity.
(i) Bromine
(ii) Argon
(iii) Ozone
(iv) Sulphur
Answer:
(i) Bromine: Bromine exists as \(\mathrm{Br}_2\), so its atomicity is 2. It is a diatomic molecule.
(ii) Argon: Argon is a noble gas and exists as single atoms (Ar), so its atomicity is 1. It is a monoatomic molecule.
(iii) Ozone: Ozone exists as \(\mathrm{O}_3\), so its atomicity is 3. It is a triatomic molecule.
(iv) Sulphur: Sulphur commonly exists as \(\mathrm{S}_8\) (a ring structure), so its atomicity is 8. It is a polyatomic molecule.
In simple words: Atomicity tells us how many atoms are in one molecule. Bromine has 2 atoms, Argon has 1, Ozone has 3, and Sulphur has 8.

🎯 Exam Tip: Knowing the common molecular forms of elements (e.g., \(\mathrm{H}_2\), \(\mathrm{O}_2\), \(\mathrm{S}_8\)) is essential for correctly determining their atomicity.

Question 6. Define atomic mass unit.
Answer: The atomic mass unit (amu) is a standard unit used to measure the mass of atoms and molecules. It is precisely defined as one-twelfth (\(\frac{1}{12}\)) of the mass of a single carbon-12 atom. The carbon-12 isotope is specifically chosen because it has exactly 6 protons and 6 neutrons, providing a stable reference point. It is often simply referred to as 'amu'.
In simple words: An atomic mass unit (amu) is a very tiny unit of weight. It's exactly the same as one-twelfth of the mass of a carbon-12 atom, which is used as a standard to measure how heavy other atoms are.

🎯 Exam Tip: Understand that the carbon-12 isotope is the chosen standard because of its stability and the convenience of its mass. This definition is a core concept in atomic structure.

VII. Long answer questions:

Question 1. Explain how Avogadro hypothesis is used to derive the value of atomicity.
Answer:
(i) Avogadro's Law states that if you have different gases, and they are all at the same temperature, pressure, and volume, then they will all contain the same number of molecules. This is a very important principle for understanding gases.
(ii) Let's look at the reaction where hydrogen gas and chlorine gas combine to form hydrogen chloride gas:
\( \mathrm{H}_2 \text{ (g)} + \mathrm{Cl}_2 \text{ (g)} \rightarrow 2\mathrm{HCl} \text{ (g)} \)
This means: 1 volume of hydrogen + 1 volume of chlorine
\( \implies \) 2 volumes of hydrogen chloride.
(iii) Now, if we use Avogadro's Law, we can say that 'n' number of molecules occupy one volume of any gas. So, for the reaction above:
n molecules of hydrogen + n molecules of chlorine
\( \implies \) 2n molecules of hydrogen chloride.
If we consider the simplest case where 'n' is 1, then:
1 molecule of hydrogen + 1 molecule of chlorine
\( \implies \) 2 molecules of hydrogen chloride.
This also means that half a molecule of hydrogen and half a molecule of chlorine combine to make one molecule of hydrogen chloride.
(iv) From this, we understand that one molecule of hydrogen chloride is made from half a molecule of hydrogen and half a molecule of chlorine.
(v) If half a molecule of hydrogen contains 1 atom, then one whole molecule of hydrogen must contain 2 atoms. Therefore, the atomicity of hydrogen is 2. Similarly, by applying the same logic, the atomicity of chlorine is also found to be 2. This proves that both hydrogen and chlorine exist as diatomic molecules, meaning \(\mathrm{H}_2\) and \(\mathrm{Cl}_2\).
In simple words: Avogadro's hypothesis helps us figure out how many atoms are in a gas molecule. By looking at how gases react and combine in simple volume ratios, we can deduce that molecules like hydrogen and chlorine are made of two atoms each.

🎯 Exam Tip: When deriving atomicity using Avogadro's Law, remember that the smallest unit that reacts (a molecule) can be broken down into atoms during the reaction, providing insight into its composition.

Question 2. Write a note on the following,
(i) Isotopes
(ii) Isobars
(iii) Relative atomic mass.
Answer:
(i) Isotopes: Isotopes are atoms of the same chemical element. This means they have the same number of protons (and thus the same atomic number), but they have different numbers of neutrons. Because of this, their mass numbers are different. For example, two common isotopes of chlorine are Chlorine-35 and Chlorine-37, represented as \(_{17}\mathrm{Cl}^{35}\) and \(_{17}\mathrm{Cl}^{37}\).
(ii) Isobars: Isobars are atoms of different chemical elements. They have different atomic numbers (different numbers of protons) but possess the same mass number (same total number of protons and neutrons). A classic example includes Argon-40 and Calcium-40, which are written as \(_{18}\mathrm{Ar}^{40}\) and \(_{20}\mathrm{Ca}^{40}\).
In simple words: Isotopes are atoms of the same element that have different "weights" due to different numbers of neutrons. Isobars are atoms of different elements that happen to have the same "weight".

🎯 Exam Tip: Clearly distinguish between isotopes and isobars by focusing on what is the same (atomic number for isotopes, mass number for isobars) and what is different (mass number for isotopes, atomic number for isobars).

Question 3. Sodim bicarbonate breaks down on heating as follows:
\( 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
(Atomic mass of Na = 23, H = 1, C = 12, O = 16)
(i) How many moles of \( \text{NaHCO}_3 \) are there in the equation?
(ii) What is the mass of \( \text{CO}_2 \) produced in the equation?
(iii) How many moles of water molecules are produced in the equation?
(iv) What is the mass of \( \text{NaHCO}_3 \) used in this equation?
(v) What is the volume occupied by \( \text{CO}_2 \) in this equation?
Answer:
(i) Based on the balanced chemical equation, there are 2 moles of sodium bicarbonate \( (\text{NaHCO}_3) \) involved in the reaction.
(ii) The molar mass of \( \text{CO}_2 \) is \( 12 + 2(16) = 44 \) g. From the equation, one mole of \( \text{CO}_2 \) is produced, so its mass is 44 g.
(iii) The balanced equation shows that 1 mole of water \( (\text{H}_2\text{O}) \) molecules is produced. Water is an essential product in many decomposition reactions.
(iv) First, calculate the molar mass of \( \text{NaHCO}_3 \):
\( = \text{Na} + \text{H} + \text{C} + 3(\text{O}) \)
\( = 23 + 1 + 12 + 3(16) \)
\( = 23 + 1 + 12 + 48 \)
\( = 84 \text{ g/mol} \)
Since 2 moles of \( \text{NaHCO}_3 \) are used, the total mass is:
\( = 2 \times 84 \text{ g} \)
\( = 168 \text{ g} \)
(v) At standard temperature and pressure (STP), one mole of any gas occupies 22.4 litres. Since 1 mole of \( \text{CO}_2 \) gas is produced, it will occupy 22.4 litres.
In simple words: Look at the numbers in front of each chemical in the balanced equation to find moles. To get mass, use atomic weights to find the total weight of one molecule, then multiply by the number of moles. For gases at standard conditions, one mole always fills 22.4 litres.

🎯 Exam Tip: Always balance the chemical equation first to ensure the correct mole ratios are used for all calculations. Also, remember the molar volume of a gas at STP (22.4 L) for volume-related questions.

VIII. Hot Questions

Question 1. Why do we take an atomic mass of Carbon - 12 as standard?
Answer: Carbon-12 is used as the standard for measuring atomic masses. This is because, apart from Carbon-12, no other naturally occurring atoms (nuclides) have exactly whole-number masses on this scale. The small differences in the masses of protons and neutrons, along with their different ratios in other atoms compared to Carbon-12, cause their total masses to deviate from whole numbers. Using Carbon-12 helps create a consistent and precise relative scale. It also helps to ensure atomic masses are reported uniformly.
In simple words: We use Carbon-12 as the main reference for measuring how heavy atoms are. This is because it's very stable and its mass is a perfect whole number (12). This helps us compare the weights of all other atoms easily and accurately.

🎯 Exam Tip: When defining atomic mass standards, emphasize the whole-number mass of C-12 and its role in establishing a relative scale for all other elements.

Question 2. The cost of common salt (NaCl) is Rs 18 per kg. Calculate the cost of the salt per mole.
Answer: First, we need to find the molar mass of \( \text{NaCl} \):
Atomic mass of Na = 23
Atomic mass of Cl = 35.5
Molar mass of \( \text{NaCl} = 23 + 35.5 = 58.5 \text{ g/mol} \)
Next, we are given that 1 kg (which is 1000 g) of \( \text{NaCl} \) costs Rs 18. We need to find the cost of 58.5 g.
Cost of 1000 g of \( \text{NaCl} = \text{Rs } 18 \)
Cost of 1 g of \( \text{NaCl} = \frac{\text{Rs } 18}{1000} \)
Cost of 58.5 g (1 mole) of \( \text{NaCl} = \frac{\text{Rs } 18}{1000} \times 58.5 \)
\( = \text{Rs } 0.018 \times 58.5 \)
\( = \text{Rs } 1.053 \)
So, the cost of one mole of \( \text{NaCl} \) is Rs 1.053. This calculation connects everyday cost with the chemical quantity.
In simple words: First, find how much one molecule of salt weighs (molar mass). Then, if you know the price for a big amount of salt, you can figure out the price for just that one molar mass quantity.

🎯 Exam Tip: Remember to convert kilograms to grams to match the units of molar mass before performing calculations. Clearly show each step of finding the molar mass and then the cost per mole.

Question 3. What will be the mass of one \( ^{12}\text{C} \) atom in g?
Answer: We know that 1 mole of \( ^{12}\text{C} \) atoms is equal to \( 6.022 \times 10^{23} \) atoms (Avogadro's number).
We also know that 1 mole of \( ^{12}\text{C} \) atoms has a mass of 12 g.
So, \( 6.022 \times 10^{23} \) atoms of \( ^{12}\text{C} \) have a mass of 12 g.
To find the mass of just one \( ^{12}\text{C} \) atom, we divide the total mass by the number of atoms:
Mass of 1 atom of \( ^{12}\text{C} = \frac{12 \text{ g}}{6.022 \times 10^{23}} \)
\( \implies = 1.9927 \times 10^{-23} \text{ g} \)
This incredibly small mass shows why we use atomic mass units for convenience. Its tiny size makes direct measurement impossible without advanced tools.
In simple words: One mole of Carbon-12 atoms weighs 12 grams. A mole contains a huge number of atoms. So, to find the weight of just one atom, you divide the total weight of the mole by that huge number.

🎯 Exam Tip: Clearly state Avogadro's number and the molar mass of carbon-12. The key is to correctly set up the division to find the mass of a single atom.

IX. Solve the following problems:

Question 1. Calculate the average atomic mass of carbon, if the natural abundance of C - 12 and C - 13 are 98.90% and 1.10% respectively.
Answer: The average atomic mass of an element is calculated by taking a weighted average of the masses of its isotopes, based on their natural abundance.
Given:
Atomic mass of C-12 = 12 amu, Abundance = 98.90%
Atomic mass of C-13 = 13 amu, Abundance = 1.10%
Average atomic mass of carbon \( = \left(12 \times \frac{98.9}{100}\right) + \left(13 \times \frac{1.1}{100}\right) \)
\( \implies = (12 \times 0.989) + (13 \times 0.011) \)
\( \implies = 11.868 + 0.143 \)
\( \implies = 12.011 \text{ amu} \)
This average value explains why the atomic mass on the periodic table is not a whole number. This method is fundamental for understanding how the atomic mass of elements is determined from isotopic data.
In simple words: To find the average weight of carbon, you multiply the weight of each carbon type (isotope) by how much of it naturally exists. Then you add these numbers up to get the final average weight.

🎯 Exam Tip: Remember to convert percentages to decimal fractions (divide by 100) before multiplying by the isotopic mass. Ensure you sum up all weighted contributions.

Question 2. Find how many moles are there in
(a) 98 g of \( \text{H}_2\text{SO}_4 \)
(b) \( 18.069 \times 10^{23} \) atoms of calcium
(c) 4.48 L of \( \text{CO}_2 \)
Answer:
(a) To find the number of moles in 98 g of \( \text{H}_2\text{SO}_4 \), we first need its Gram Molar Mass (GMM).
Atomic masses: H=1, S=32, O=16
GMM of \( \text{H}_2\text{SO}_4 = 2(1) + 32 + 4(16) \)
\( = 2 + 32 + 64 \)
\( = 98 \text{ g/mol} \)
Number of moles \( = \frac{\text{Given Mass}}{\text{Mol. Mass}} \)
\( = \frac{98 \text{ g}}{98 \text{ g/mol}} \)
\( = 1 \text{ mole} \)
This means that 98 grams of sulfuric acid contains exactly one Avogadro's number of molecules.
(b) To find the number of moles from a given number of atoms, we use Avogadro's number \( (6.023 \times 10^{23} \text{ atoms/mol}) \).
Number of moles \( = \frac{\text{Given number of atoms}}{\text{Avogadro number}} \)
\( = \frac{18.069 \times 10^{23}}{6.023 \times 10^{23}} \)
\( = 3 \text{ moles} \)
This shows a direct relationship between the count of particles and the mole concept.
(c) To find the number of moles for a gas at STP, we use the molar volume of a gas.
At STP, 1 mole of any gas occupies 22.4 L.
Number of moles \( = \frac{\text{Given volume}}{\text{Molar volume at STP}} \)
\( = \frac{4.48 \text{ L}}{22.4 \text{ L/mol}} \)
\( = 0.2 \text{ moles} \)
This calculation is useful for gases and helps relate volume to chemical quantity.
In simple words: To find moles, either divide the given weight by the weight of one molecule, or divide the count of atoms/molecules by Avogadro's number. For gases, divide the volume by 22.4 litres if it's at standard conditions.

🎯 Exam Tip: Be careful to use the correct formula for calculating moles based on whether you are given mass, number of particles, or volume (for gases at STP).

Question 3. Calculate the number of moles in
(i) \( 12.046 \times 10^{23} \) atoms of copper
(ii) 27.95 g of iron
(iii) \( 1.51 \times 10^{23} \) molecules of \( \text{CO}_2 \)
Answer:
(i) To find the number of moles from a given number of atoms, we use Avogadro's number \( (6.023 \times 10^{23} \text{ atoms/mol}) \).
Number of moles \( = \frac{\text{Given number of atoms}}{\text{Avogadro number}} \)
\( = \frac{12.046 \times 10^{23}}{6.023 \times 10^{23}} \)
\( = 2 \text{ moles of copper} \)
This shows that if you have twice Avogadro's number of atoms, you have two moles.
(ii) To find the number of moles in 27.95 g of iron, we need the atomic mass of iron (Fe). The atomic mass of iron is approximately 55.9 g/mol.
Number of moles \( = \frac{\text{Given Mass}}{\text{Atomic Mass}} \)
\( = \frac{27.95 \text{ g}}{55.9 \text{ g/mol}} \)
\( = 0.5 \text{ mole of iron} \)
This calculation helps convert mass directly into a count of moles.
(iii) To find the number of moles from a given number of molecules, we use Avogadro's number \( (6.023 \times 10^{23} \text{ molecules/mol}) \).
Number of moles \( = \frac{\text{No. of molecules}}{\text{Avogadro number}} \)
\( = \frac{1.51 \times 10^{23}}{6.023 \times 10^{23}} \)
\( = 0.25 \text{ mole of CO}_2 \)
This directly links the count of molecules to the mole unit.
In simple words: To get the number of moles, divide the given number of atoms or molecules by the special Avogadro's number. If you have grams, divide by the weight of one mole (molar mass) of that substance.

🎯 Exam Tip: Pay attention to the units given (atoms, grams, or molecules) and choose the appropriate formula involving Avogadro's number or molar mass.

Question 4. Calculate the number of atoms of mercury present in 1 kg of Mercury. [Atomic mass of Hg = 200.6]
Answer: First, convert the given mass from kg to g:
1 kg = 1000 g
We know the atomic mass of mercury (Hg) is 200.6 g/mol.
This means that 200.6 g of Hg contains Avogadro's number of atoms, which is \( 6.023 \times 10^{23} \) Hg atoms.
Number of atoms in 1 g of Hg \( = \frac{6.023 \times 10^{23}}{200.6} \)
Number of atoms in 1000 g (1 kg) of Hg \( = \frac{6.023 \times 10^{23}}{200.6} \times 1000 \)
\( \implies = 30.0249 \times 10^{23} \text{ atoms} \)
\( \implies = 3.0025 \times 10^{24} \text{ Hg atoms} \)
This calculation shows how many individual mercury atoms are in a macroscopic amount like a kilogram. This helps visualize the scale of atoms.
In simple words: To find how many atoms are in a kilogram of mercury, first change kilograms to grams. Then, use the atomic weight of mercury and Avogadro's number to figure out how many atoms are in that many grams.

🎯 Exam Tip: Ensure consistent units (grams) throughout the calculation. Clearly show the conversion from mass to moles, and then from moles to number of atoms using Avogadro's number.

Question 5. How many molecules are present in \( 7 \times 10^{-3} \text{ m}^3 \) of \( \text{NH}_3 \) at STP?
Answer: We need to find the number of molecules in a given volume of gas at STP.
First, let's establish the molar volume. At STP, the molar volume of any gas is 22.4 dm\(^3\) (litres) or \( 2.24 \times 10^{-2} \text{ m}^3 \).
We know that \( 2.24 \times 10^{-2} \text{ m}^3 \) of \( \text{NH}_3 \) at STP contains \( 6.023 \times 10^{23} \) \( \text{NH}_3 \) molecules (Avogadro's number).
Now, calculate how many molecules are in the given volume of \( 7 \times 10^{-3} \text{ m}^3 \):
Number of molecules \( = \frac{\text{Given volume}}{\text{Molar volume at STP}} \times \text{Avogadro number} \)
\( = \frac{7 \times 10^{-3} \text{ m}^3}{2.24 \times 10^{-2} \text{ m}^3/\text{mol}} \times 6.023 \times 10^{23} \text{ molecules/mol} \)
\( \implies = \frac{7}{2.24} \times \frac{10^{-3}}{10^{-2}} \times 6.023 \times 10^{23} \)
\( \implies = 3.125 \times 10^{-1} \times 6.023 \times 10^{23} \)
\( \implies = 3.125 \times 6.023 \times 10^{22} \)
\( \implies = 18.82 \times 10^{22} \text{ NH}_3 \text{ molecules} \)
This calculation demonstrates the relationship between volume, moles, and number of molecules for gases. It's important to use consistent units for volume throughout.
In simple words: To find how many ammonia molecules are in a specific volume at standard conditions, compare that volume to the volume one mole of gas takes up. Then, multiply by Avogadro's number to get the total count of molecules.

🎯 Exam Tip: Ensure you use the molar volume in \( \text{m}^3 \) to match the given volume units. Remember that Avogadro's number is key to converting moles or volume to particle count.

Question 6. What is the mass in grams of the following?
(a) 3 moles of \( \text{NaOH} \)
(b) \( 6.023 \times 10^{22} \) atoms of Ca
(c) 224 L of \( \text{CO}_2 \)
Answer:
(a) To find the mass of 3 moles of \( \text{NaOH} \), we first need its Gram Molar Mass (GMM).
Atomic masses: Na=23, O=16, H=1
GMM of \( \text{NaOH} = 23 + 16 + 1 = 40 \text{ g/mol} \)
Mass of 3 moles of \( \text{NaOH} = \text{Number of moles} \times \text{Molar mass} \)
\( = 3 \text{ mol} \times 40 \text{ g/mol} \)
\( = 120 \text{ g} \)
This directly shows how molar mass converts moles to grams. This conversion is crucial in stoichiometry.
(b) To find the mass of \( 6.023 \times 10^{22} \) atoms of Ca, we relate it to Avogadro's number and the atomic mass of Calcium (Ca).
Atomic mass of Ca = 40 g/mol
Avogadro's number \( = 6.023 \times 10^{23} \text{ atoms/mol} \)
Number of moles of Ca \( = \frac{\text{Given number of atoms}}{\text{Avogadro number}} \)
\( = \frac{6.023 \times 10^{22}}{6.023 \times 10^{23}} \)
\( = 0.1 \text{ mole} \)
Mass of Ca \( = \text{Number of moles} \times \text{Atomic mass} \)
\( = 0.1 \text{ mol} \times 40 \text{ g/mol} \)
\( = 4 \text{ g} \)
This calculation highlights the link between the number of particles and their mass.
(c) To find the mass of 224 L of \( \text{CO}_2 \) at STP, we use the molar volume and molar mass.
At STP, 1 mole of any gas occupies 22.4 L.
First, find the number of moles:
Number of moles \( = \frac{\text{Given volume}}{\text{Molar volume at STP}} \)
\( = \frac{224 \text{ L}}{22.4 \text{ L/mol}} \)
\( = 10 \text{ moles} \)
Next, find the molar mass of \( \text{CO}_2 \):
Atomic masses: C=12, O=16
Molar mass of \( \text{CO}_2 = 12 + 2(16) = 12 + 32 = 44 \text{ g/mol} \)
Mass of 224 L of \( \text{CO}_2 = \text{Number of moles} \times \text{Molar mass} \)
\( = 10 \text{ mol} \times 44 \text{ g/mol} \)
\( = 440 \text{ g} \)
This calculation shows how gas volume, molar mass, and mass are interconnected. This conversion is crucial for gas calculations.
In simple words: To get the mass in grams: if you have moles, multiply by the weight of one mole. If you have atoms, first turn atoms into moles using Avogadro's number, then convert to mass. If you have gas volume at standard conditions, turn volume into moles, then convert to mass.

🎯 Exam Tip: Always identify what information is given (moles, number of particles, or volume) and choose the appropriate conversion factor (molar mass, Avogadro's number, or molar volume at STP).

Question 7. How many grams are therein:
(i) 5 moles of water
(ii) 2 moles of Ammonia
(iii) 2 moles of Glucose
Answer:
To find the mass in grams, we use the formula: Mass = Number of moles \( \times \) Molar mass.
(i) For water \( (\text{H}_2\text{O}) \):
Atomic masses: H=1, O=16
Molar mass of \( \text{H}_2\text{O} = 2(1) + 16 = 18 \text{ g/mol} \)
Mass of 5 moles of \( \text{H}_2\text{O} = 5 \text{ mol} \times 18 \text{ g/mol} = 90 \text{ g} \)
Water is a very common molecule, and this calculation shows its mass relationship. This is a basic conversion from moles to grams.
(ii) For ammonia \( (\text{NH}_3) \):
Atomic masses: N=14, H=1
Molar mass of \( \text{NH}_3 = 14 + 3(1) = 17 \text{ g/mol} \)
Mass of 2 moles of \( \text{NH}_3 = 2 \text{ mol} \times 17 \text{ g/mol} = 34 \text{ g} \)
Ammonia is an important chemical, and understanding its molar mass is key to chemical reactions. This illustrates the mole-to-mass conversion.
(iii) For glucose \( (\text{C}_6\text{H}_{12}\text{O}_6) \):
Atomic masses: C=12, H=1, O=16
Molar mass of \( \text{C}_6\text{H}_{12}\text{O}_6 = 6(12) + 12(1) + 6(16) \)
\( = 72 + 12 + 96 \)
\( = 180 \text{ g/mol} \)
Mass of 2 moles of \( \text{C}_6\text{H}_{12}\text{O}_6 = 2 \text{ mol} \times 180 \text{ g/mol} = 360 \text{ g} \)
Glucose is a vital sugar, and its larger molar mass results in a higher mass for the same number of moles. This reinforces the concept of molar mass for more complex compounds.
In simple words: To find the mass in grams, simply multiply the number of moles by the molar mass (the weight of one mole) of that substance. Each substance has its own unique molar mass.

🎯 Exam Tip: Always calculate the molar mass correctly by summing the atomic masses of all atoms in the chemical formula. Then, multiply by the number of moles given.

Question 8. Calculate the molar mass of the following compounds.
(a) Urea \( (\text{NH}_2\text{CONH}_2) \)
(b) Ethanol \( (\text{C}_2\text{H}_5\text{OH}) \)
(c) Boric acid \( (\text{H}_3\text{BO}_3) \)
[Atomic mass of N = 14, H = 1, C = 12, B = 11, O = 16]
Answer: Molar mass is the sum of the atomic masses of all atoms in a molecule.
(a) For Urea \( (\text{NH}_2\text{CONH}_2) \):
Molar mass \( = 2(\text{N}) + 4(\text{H}) + 1(\text{C}) + 1(\text{O}) \)
\( = 2(14) + 4(1) + 1(12) + 1(16) \)
\( = 28 + 4 + 12 + 16 \)
\( = 60 \text{ g/mol} \)
Urea is a key compound in fertilizers, and its molar mass helps in calculating amounts needed. This shows how to calculate molar mass for organic molecules.
(b) For Ethanol \( (\text{C}_2\text{H}_5\text{OH}) \):
Molar mass \( = 2(\text{C}) + 6(\text{H}) + 1(\text{O}) \)
\( = 2(12) + 6(1) + 1(16) \)
\( = 24 + 6 + 16 \)
\( = 46 \text{ g/mol} \)
Ethanol is a common alcohol, and its molar mass is important in various chemical industries. This applies the molar mass calculation to another organic compound.
(c) For Boric acid \( (\text{H}_3\text{BO}_3) \):
Molar mass \( = 3(\text{H}) + 1(\text{B}) + 3(\text{O}) \)
\( = 3(1) + 1(11) + 3(16) \)
\( = 3 + 11 + 48 \)
\( = 62 \text{ g/mol} \)
Boric acid is used as an antiseptic, and knowing its molar mass is vital for proper usage and calculations. This completes the set of molar mass calculations.
In simple words: To find the molar mass of a compound, simply add up the atomic weights of all the atoms in its chemical formula. Remember to count each atom multiple times if it appears more than once.

🎯 Exam Tip: Always double-check the number of each type of atom in the chemical formula. Use the provided atomic masses accurately for each element to avoid calculation errors.

Question 9. Mass of one atom of an element is \( 6.645 \times 10^{-23} \text{ g} \). How many moles of element are there in 0.320 kg.
Answer: We are given the mass of one atom of the element as \( 6.645 \times 10^{-23} \text{ g} \).
First, calculate the molar mass of the element. Molar mass is the mass of Avogadro's number of atoms.
Molar mass \( = \text{Mass of one atom} \times \text{Avogadro's number} \)
\( = 6.645 \times 10^{-23} \text{ g/atom} \times 6.023 \times 10^{23} \text{ atoms/mol} \)
\( = 40 \text{ g/mol} \)
So, the molar mass of the element is 40 g/mol. This step converts the mass of a single atom to a macroscopic molar mass.
Next, convert the given total mass from kilograms to grams:
0.320 kg \( = 0.320 \times 1000 \text{ g} = 320 \text{ g} \)
Now, calculate the number of moles in 320 g of the element:
Number of moles \( = \frac{\text{Given mass in g}}{\text{Atomic mass (molar mass)}} \)
\( = \frac{320 \text{ g}}{40 \text{ g/mol}} \)
\( = 8 \text{ moles} \)
This calculation shows how to work backwards from the mass of a single atom to find the molar mass, then apply it to a large sample. It connects the microscopic and macroscopic world of chemistry.
In simple words: First, use the weight of one atom and Avogadro's number to find the weight of one mole. Then, change the given total weight from kilograms to grams. Finally, divide the total grams by the weight of one mole to find out how many moles are present.

🎯 Exam Tip: Remember to convert the mass of a single atom to molar mass using Avogadro's number. Also, be careful with unit conversions, especially from kilograms to grams, before calculating moles.