GSEB Class 9 Science Solutions Chapter 3 Atoms and Molecules

Question 1. In a reaction, 5.3 g of sodium carbonate were reacted with 6 g of ethanoic acid solution. The products were 2.2 g of carbon dioxide, 0.9 g of water and 12 g of sodium ethanoate solution. Show that these observations are in agreement with the law of conservation of mass.
Answer: The reaction involves sodium carbonate and ethanoic acid, which produce sodium ethanoate, carbon dioxide, and water.
Sodium carbonate + Ethanoic acid \( \rightarrow \) Sodium ethanoate + Carbon dioxide + Water
Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid
\( = 5.3 + 6.0 = 11.3 \) g
Mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate
\( = 2.2 + 0.9 + 8.2 = 11.3 \) g
Since the total mass of the products is equal to the total mass of the reactants, this observation confirms the law of conservation of mass. This law states that mass is neither created nor destroyed during any chemical change.
In simple words: We added up the weight of things we started with and the weight of things we ended with. Both sums were the same, showing that no matter was lost or gained, which supports the rule of mass conservation.

Exam Tip: To prove the law of conservation of mass, always calculate the total mass of reactants and the total mass of products separately. If they are equal, the law is satisfied.

Question 2. Hydrogen and oxygen combine in the ratio 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer: Hydrogen and oxygen combine in a mass ratio of 1 : 8 to form water. This means that for every 1 gram of hydrogen, 8 grams of oxygen are needed to react.
If 1 g of hydrogen combines with 8 g of oxygen, then 3 g of hydrogen will react with oxygen as follows:
\( = 8 \times 3 \) g
\( = 24 \) g
Therefore, 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.
In simple words: Hydrogen and oxygen always mix in a 1:8 weight ratio to make water. So, if you have 3 grams of hydrogen, you will need 8 times that amount of oxygen, which is 24 grams.

Exam Tip: In ratio-based problems, clearly identify the given ratio and use it as a conversion factor to find the unknown quantity.

Question 3. Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?
Answer: The postulate of Dalton's Atomic theory that helps explain the law of conservation of mass is: 'Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction'. This idea means that during a reaction, atoms are simply rearranged, not made or unmade, so the total mass remains constant.
In simple words: Dalton's theory says atoms can't be made or destroyed. This is why the total mass stays the same in a chemical reaction.

Exam Tip: Remember that the law of conservation of mass is fundamental to stoichiometry and confirms that chemical reactions only involve rearrangement of atoms.

Question 4. Which postulate of Dalton's atomic theory can explain the law of definite proportions?
Answer: The postulate of Dalton's atomic theory that helps explain the law of definite proportions is: 'The relative number and kinds of atoms are constant in a given compound'. This means that a specific compound always has the same elements in the same proportions by count and type of atoms, leading to a fixed proportion by mass.
In simple words: Dalton's idea that each compound has a fixed number and type of atoms helps explain why compounds always have elements in the same proportions.

Exam Tip: The law of definite proportions highlights that a chemical compound always contains the same elements in the same ratio by mass, regardless of its source or preparation.

Question 5. Define the atomic mass unit.
Answer: A mass unit that is exactly one-twelfth the mass of one atom of carbon-12 is called one atomic mass unit (amu). This unit provides a standard reference point for comparing the masses of all other atoms.
In simple words: An atomic mass unit is defined as one-twelfth the mass of a single carbon-12 atom. It's a way to measure how heavy atoms are.

Exam Tip: Clearly state "one-twelfth" and "carbon-12 atom" when defining the atomic mass unit to ensure accuracy.

Question 6. Why is it not possible to see an atom with naked eyes?
Answer: The size of an atom is exceedingly small, making it impossible to observe with the naked eye. Furthermore, atoms of most elements typically do not exist independently; they often combine to form molecules or larger structures. The radius of an atom is generally on the order of \( 10^{-10} \) m, which is far below the resolution limit of the human eye.
In simple words: Atoms are too tiny to see without special tools. Their size is about \( 10^{-10} \) meters, much smaller than what our eyes can detect. Also, most atoms don't float around alone.

Exam Tip: Emphasize both the extremely small size (mentioning the order of magnitude) and the fact that atoms rarely exist independently when explaining why they are invisible to the naked eye.

Question 7. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer:
(i) Sodium oxide: \( \text{Na}_2\text{O} \)
(ii) Aluminium chloride: \( \text{AlCl}_3 \)
(iii) Sodium sulphide: \( \text{Na}_2\text{S} \)
(iv) Magnesium hydroxide: \( \text{Mg(OH)}_2 \)
In simple words: We write the chemical formulas by knowing the valencies of each element and combining them so the charges balance out, forming a neutral compound.

Exam Tip: When writing chemical formulas, always remember the valencies of the elements involved and criss-cross them to balance the charges, ensuring the compound is electrically neutral.

Question 8. Write down the names of compounds represented by the following formulae:
(i) \( \text{Al}_2(\text{SO}_4)_3 \)
(ii) \( \text{CaCl}_2 \)
(iii) \( \text{K}_2\text{SO}_4 \)
(iv) \( \text{KNO}_3 \)
(v) \( \text{CaCO}_3 \)
Answer:
(i) \( \text{Al}_2(\text{SO}_4)_3 \): Aluminium sulphate
(ii) \( \text{CaCl}_2 \): Calcium chloride
(iii) \( \text{K}_2\text{SO}_4 \): Potassium sulphate
(iv) \( \text{KNO}_3 \): Potassium nitrate
(v) \( \text{CaCO}_3 \): Calcium carbonate
In simple words: We name these compounds by identifying the metal and the non-metal or polyatomic ion and putting their names together.

Exam Tip: Familiarize yourself with common polyatomic ions (like sulfate, nitrate, carbonate) and their charges to accurately name compounds.

Question 9. What is meant by the term chemical formula?
Answer: A chemical formula represents the symbolic depiction of a compound's composition. It shows the number and types of atoms from different elements that make up the compound. For instance, the chemical formula for carbon dioxide (\( \text{CO}_2 \)) indicates that one carbon atom and two oxygen atoms are chemically joined together to form one molecule of carbon dioxide.
In simple words: A chemical formula is like a shorthand way to show what atoms are in a compound and how many of each there are. For example, \( \text{CO}_2 \) tells us there is one carbon atom and two oxygen atoms in carbon dioxide.

Exam Tip: When explaining chemical formula, always mention both the "type" and "number" of atoms present to give a complete definition.

Question 10. How many atoms are present in a
(i) \( \text{H}_2\text{S} \) molecule and
(ii) \( \text{PO}_4^{3-} \) ion?
Answer:
(i) In an \( \text{H}_2\text{S} \) molecule, there are three atoms: two hydrogen atoms and one sulphur atom. Total = \( 2 + 1 = 3 \) atoms.
(ii) In a \( \text{PO}_4^{3-} \) ion, there are five atoms: one phosphorus atom and four oxygen atoms. Total = \( 1 + 4 = 5 \) atoms.
In simple words: Count the small numbers next to each element symbol, and if there is no number, assume one. Add these numbers together to get the total number of atoms.

Exam Tip: Remember to add up the subscripts for each element in the formula or ion to find the total number of atoms.

Question 11. Calculate the molecular masses of \( \text{H}_2 \), \( \text{O}_2 \), \( \text{Cl}_2 \), \( \text{CO}_2 \), \( \text{CH}_4 \), \( \text{C}_2\text{H}_6 \), \( \text{C}_2\text{H}_4 \), \( \text{NH}_3 \), \( \text{CH}_3\text{OH} \).
Answer:
Molecular mass of \( \text{H}_2 = 2 \times \) Atomic mass of H \( = 2 \times 1 = 2 \) u
Molecular mass of \( \text{O}_2 = 2 \times \) Atomic mass of O \( = 2 \times 16 = 32 \) u
Molecular mass of \( \text{Cl}_2 = 2 \times \) Atomic mass of Cl \( = 2 \times 35.5 = 71 \) u
Molecular mass of \( \text{CO}_2 = \) Atomic mass of C \( + 2 \times \) Atomic mass of O \( = 12 + 2 \times 16 = 44 \) u
Molecular mass of \( \text{CH}_4 = \) Atomic mass of C \( + 4 \times \) Atomic mass of H \( = 12 + 4 \times 1 = 16 \) u
Molecular mass of \( \text{C}_2\text{H}_6 = 2 \times \) Atomic mass of C \( + 6 \times \) Atomic mass of H \( = 2 \times 12 + 6 \times 1 = 30 \) u
Molecular mass of \( \text{C}_2\text{H}_4 = 2 \times \) Atomic mass of C \( + 4 \times \) Atomic mass of H \( = 2 \times 12 + 4 \times 1 = 28 \) u
Molecular mass of \( \text{NH}_3 = \) Atomic mass of N \( + 3 \times \) Atomic mass of H \( = 14 + 3 \times 1 = 17 \) u
Molecular mass of \( \text{CH}_3\text{OH} = \) Atomic mass of C \( + 4 \times \) Atomic mass of H \( + \) Atomic mass of O \( = 12 + 4 \times 1 + 16 = 32 \) u
In simple words: To find the molecular mass, add up the atomic masses of all the atoms in the molecule. Multiply each atom's mass by how many of that atom are present before adding them.

Exam Tip: Always use the correct atomic mass for each element and carefully count the number of atoms of each element in the molecule before summing them up.

Question 12. Calculate the formula unit masses of ZnO, \( \text{Na}_2\text{O} \), \( \text{K}_2\text{CO}_3 \) given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u.
Answer:
Formula unit mass of ZnO \( = 65 + 16 = 81 \) u
Formula unit mass of \( \text{Na}_2\text{O} = 2 \times 23 + 16 = 62 \) u
Formula unit mass of \( \text{K}_2\text{CO}_3 = 2 \times 39 + 12 \times 1 + 16 \times 3 = 138 \) u
In simple words: To calculate the formula unit mass, you just add up the atomic masses of all the atoms in the given formula. Make sure to multiply the atomic mass by the number of times each atom appears.

Exam Tip: Formula unit mass is calculated similarly to molecular mass; it's simply the sum of the atomic masses of all atoms in a formula unit, especially used for ionic compounds.

Question 13. If one mole of carbon atoms weighs 12 gram, what is the mass (in grams) of 1 atom of carbon?
Answer: One mole of carbon atoms weighs 12 g. This means the mass of 1 mole of carbon atoms is 12 g.
We know that 1 mole contains Avogadro's number of particles, which is \( 6.022 \times 10^{23} \) atoms.
So, the mass of \( 6.022 \times 10^{23} \) carbon atoms \( = 12 \) g.
Therefore, the mass of 1 atom of carbon \( = \frac{12}{6.022 \times 10^{23}} \) g
\( = 1.9926 \times 10^{-23} \) g
In simple words: If a mole of carbon atoms weighs 12 grams, and one mole has \( 6.022 \times 10^{23} \) atoms, then to find the weight of just one atom, you divide 12 grams by that huge number.

Exam Tip: Remember to use Avogadro's number (\( 6.022 \times 10^{23} \)) to convert between moles and the number of individual particles (atoms, molecules, ions).

Question 14. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?
Answer:
For Sodium (Na):
Atomic mass of Na \( = 23 \) u (Given)
Gram atomic mass of Na \( = 23 \) g
We know that 23 g of Na contains \( = 6.022 \times 10^{23} \) number of atoms
Thus, 100 g of Na contains \( = \frac{6.022 \times 10^{23} \times 100}{23} \)
\( = 2.6182 \times 10^{24} \) number of atoms

For Iron (Fe):
Atomic mass of Fe \( = 56 \) u (Given)
Gram atomic mass of Fe \( = 56 \) g
We know that 56 g of Fe contains \( = 6.022 \times 10^{23} \) number of atoms
Thus, 100 g of Fe contains \( = \frac{6.022 \times 10^{23} \times 100}{56} \)
\( = 1.0753 \times 10^{24} \) number of atoms
Therefore, 100 grams of sodium contain a greater number of atoms than 100 grams of iron.
In simple words: To see which has more atoms for the same weight, divide the total weight by the atomic weight of each element. The smaller the atomic weight, the more atoms you get for the same mass. Sodium has a smaller atomic mass than iron, so 100g of sodium has more atoms.

Exam Tip: Remember that a smaller atomic mass for a given total mass will result in a greater number of moles and thus a greater number of atoms.

In-Text Activities Solved

(Textbook Page 1)

Activity 3.1
Answer:
Observation: When the flask is tilted, solution X in an ignition tube and solution Y mix together. A chemical reaction visibly takes place. A cork was placed on the conical flask's mouth to prevent any mass loss during the chemical reaction. Indeed, there is no change in the flask's mass, which is in line with the law of mass conservation.
Conclusion: The law of conservation of mass states that "mass can neither be created nor destroyed during the course of a chemical reaction."
In simple words: When two solutions mixed in a closed flask, a reaction happened, but the total weight of the flask and its contents didn't change. This shows that mass is always conserved, meaning it's neither made nor lost.

Exam Tip: For experiments demonstrating conservation of mass, it's crucial to perform the reaction in a closed system to prevent reactants or products from escaping, ensuring an accurate mass measurement.

(Textbook Page 36)

Activity 3.2
Answer:
Calculations are based on the observations from Table 3.4.
(a) The ratio by number of atoms for a water molecule can be found as follows:

(b) The ratio by number of atoms for an ammonia molecule can be found as follows:

(c) The ratio by number of atoms for a carbon dioxide molecule can be found as follows:

Conclusion:
(a) The ratio by number of atoms for H : O \( = 2 : 1 = \text{H}_2\text{O} \)
(b) The ratio by number of atoms for N : H \( = 1 : 3 = \text{NH}_3 \)
(c) The ratio by number of atoms for C : O \( = 1 : 2 = \text{CO}_2 \)
In simple words: By comparing the mass ratios to atomic masses, we can find the simplest whole-number ratio of atoms in a compound, which helps us determine its chemical formula. This shows how elements combine in specific proportions.

Exam Tip: When determining the simplest whole number ratio, always divide the mass ratio by the atomic mass for each element and then find the smallest common divisor to get whole numbers.

Gujarat Board Class 9 Science Atoms and Molecules Textbook Questions and Answers

Question 1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer: The percentage composition of any element in a compound is calculated by the formula:
\( \% \) of any element in a compound \( = \frac{\text{Mass of the element present}}{\text{Mass of the compound}} \times 100 \)

For Boron:
\( \% \) of boron \( = \frac{\text{Mass of boron}}{\text{Mass of the compound}} \times 100 \)
\( = \frac{0.096 \text{ g}}{0.24 \text{ g}} \times 100 = 40\% \)

For Oxygen:
\( \% \) of oxygen \( = \frac{\text{Mass of oxygen}}{\text{Mass of compound}} \times 100 \)
\( = \frac{0.144 \text{ g}}{0.24 \text{ g}} \times 100 = 60\% \)
In simple words: To find out how much of each element is in a compound by weight, we divide the weight of that element by the total weight of the compound, then multiply by 100 to get a percentage. For this compound, it's 40% boron and 60% oxygen.

Exam Tip: Always make sure that the sum of the percentage compositions of all elements in a compound adds up to 100% (or very close to it due to rounding).

Question 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen?
Answer: In the first reaction, 3.0 g of carbon burns in 8.00 g of oxygen to form 11.00 g of carbon dioxide. This indicates that carbon and oxygen combine in a definite mass ratio of 3 : 8 to produce carbon dioxide.
According to the law of definite proportions, elements always combine in fixed proportions by mass to form a specific compound.
Therefore, even when 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be utilized to react with the 3.00 g of carbon. The remaining oxygen (50.00 g - 8.00 g = 42.00 g) will not react.
The mass of carbon dioxide formed will still be 11.00 g, as the reacting masses are fixed.
In simple words: Carbon and oxygen always combine in a 3:8 ratio to make carbon dioxide. So, even if you have a lot of extra oxygen, only the right amount (8g) will react with the carbon (3g) to make 11g of carbon dioxide.

Exam Tip: This problem is an application of the Law of Definite Proportions. Recognize that excess reactants do not change the amount of product formed when the limiting reactant is fixed.

Question 3. What are polyatomic ions? Give examples.
Answer: Polyatomic ions are groups of atoms that carry a net electrical charge, which can be either positive or negative. These atoms are covalently bonded together but behave as a single unit during chemical reactions.
For example:
Ammonium ion (\( \text{NH}_4^+ \))
Hydroxide ion (\( \text{OH}^- \))
Carbonate ion (\( \text{CO}_3^{2-} \))
Sulphate ion (\( \text{SO}_4^{2-} \))
In simple words: Polyatomic ions are like small teams of atoms that stick together and have an overall electrical charge. Examples include ammonium and carbonate.

Exam Tip: When defining polyatomic ions, remember to mention both that they are "groups of atoms" and that they "carry a charge." Provide at least two common examples.

Question 4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride: \( \text{MgCl}_2 \)
(b) Calcium oxide: \( \text{CaO} \)
(c) Copper nitrate: \( \text{Cu(NO}_3)_2 \)
(d) Aluminium chloride: \( \text{AlCl}_3 \)
(e) Calcium carbonate: \( \text{CaCO}_3 \)
In simple words: To write the chemical formulas, balance the charges of the ions involved. For example, magnesium has a \( +2 \) charge and chlorine has a \( -1 \) charge, so you need two chlorines for one magnesium.

Exam Tip: Always balance the charges of the cations and anions to make the compound electrically neutral. Use parentheses for polyatomic ions when more than one is needed.

Question 5. Give the names of the elements present in the following compounds.
(a) Quicklime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
(a) Quicklime (\( \text{CaO} \)): Calcium and oxygen
(b) Hydrogen bromide (\( \text{HBr} \)): Hydrogen and bromine
(c) Baking powder (commonly sodium bicarbonate, \( \text{NaHCO}_3 \)): Sodium, hydrogen, carbon, and oxygen
(d) Potassium sulphate (\( \text{K}_2\text{SO}_4 \)): Potassium, sulphur, and oxygen
In simple words: We list the elements that make up each compound based on its chemical formula. For baking powder, it typically contains sodium, hydrogen, carbon, and oxygen.

Exam Tip: For common names like "Quicklime" or "Baking powder," know their chemical names/formulas to accurately identify the constituent elements.

Question 6. Calculate the molar mass of the following substances.
(a) Ethyne, \( \text{C}_2\text{H}_2 \)
(b) Sulphur molecule, \( \text{S}_8 \)
(c) Phosphorus molecule, \( \text{P}_4 \) (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, \( \text{HNO}_3 \)
Answer:
(a) Ethyne, \( \text{C}_2\text{H}_2 = (2 \times 12) + (2 \times 1) = 24 + 2 = 26 \) g/mol
(b) Sulphur molecule, \( \text{S}_8 = 8 \times 32 = 256 \) g/mol
(c) Phosphorus molecule, \( \text{P}_4 = 4 \times 31 = 124 \) g/mol
(d) Hydrochloric acid, HCl \( = (1 \times 1) + (1 \times 35.5) = 1 + 35.5 = 36.5 \) g/mol
(e) Nitric acid, \( \text{HNO}_3 = (1 \times 1) + (1 \times 14) + (3 \times 16) = 1 + 14 + 48 = 63 \) g/mol
In simple words: To find the molar mass, add up the atomic masses of all the atoms present in one molecule of the substance. Make sure to multiply each atomic mass by the number of times that atom appears in the formula.

Exam Tip: Always refer to the atomic masses from the periodic table (or provided in the question) and multiply by the number of atoms of each element before summing them for the molar mass.

Question 7. What is the mass of:
(a) 1 mole of nitrogen atom?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (\( \text{Na}_2\text{SO}_3 \))?
Answer:
(a) The mass of 1 mole of nitrogen atoms is equal to its atomic mass, which is 14 g.
(b) The mass of 4 moles of aluminium atoms \( = 4 \times \) Atomic mass of aluminium \( = 4 \times 27 \) g \( = 108 \) g.
(c) First, calculate the molar mass of sodium sulphite (\( \text{Na}_2\text{SO}_3 \)):
Molar mass of \( \text{Na}_2\text{SO}_3 = (2 \times 23) + 32 + (3 \times 16) = 46 + 32 + 48 = 126 \) g/mol.
Then, the mass of 10 moles of sodium sulphite \( = 10 \times 126 \) g \( = 1260 \) g.
In simple words: To find the mass of a certain number of moles, multiply the number of moles by the molar mass of the substance. The molar mass of an element is its atomic mass in grams, and for a compound, it's the sum of atomic masses in grams.

Exam Tip: Remember the relationship: mass = moles \( \times \) molar mass. Always calculate the molar mass of the compound correctly before multiplying by the number of moles.

Question 8. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer: The number of moles (n) is calculated using the formula:
\( \text{n} = \frac{\text{Given mass (m)}}{\text{Molar mass (M)}} \)

(a) For 12 g of oxygen gas (\( \text{O}_2 \)):
Molar mass of \( \text{O}_2 = 2 \times 16 = 32 \) g/mol
Number of moles \( = \frac{12 \text{ g}}{32 \text{ g/mol}} = 0.375 \) mole

(b) For 20 g of water (\( \text{H}_2\text{O} \)):
Molar mass of \( \text{H}_2\text{O} = (2 \times 1.0) + 16 = 2 + 16 = 18 \) g/mol
Number of moles \( = \frac{20 \text{ g}}{18 \text{ g/mol}} \approx 1.11 \) mole

(c) For 22 g of carbon dioxide (\( \text{CO}_2 \)):
Molar mass of \( \text{CO}_2 = 12 + (2 \times 16) = 12 + 32 = 44 \) g/mol
Number of moles \( = \frac{22 \text{ g}}{44 \text{ g/mol}} = 0.5 \) mole
In simple words: To change a given mass of a substance into moles, divide its mass by its molar mass. The molar mass is the weight of one mole of that substance in grams.

Exam Tip: Always ensure you use the correct molar mass for the substance (e.g., \( \text{O}_2 \) for oxygen gas, not just O) in the denominator of the moles calculation.

Question 9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) For 0.2 mole of oxygen atoms (O):
Mass of one mole of oxygen atoms \( = 16 \) g
Then, the mass of 0.2 mole of oxygen atoms \( = 0.2 \times 16 \) g \( = 3.2 \) g.

(b) For 0.5 mole of water molecules (\( \text{H}_2\text{O} \)):
Mass of one mole of water molecules \( = 18 \) g
Then, the mass of 0.5 mole of water molecules \( = 0.5 \times 18 \) g \( = 9 \) g.
In simple words: To find the mass from moles, you multiply the number of moles by the molar mass of the substance. For oxygen atoms, the molar mass is 16g. For water molecules, it's 18g.

Exam Tip: Differentiate between atoms and molecules. For oxygen atoms, use atomic mass (O=16u). For water molecules, use molecular mass (\( \text{H}_2\text{O}=18 \)u).

Question 10. Calculate the number of molecules of sulphur (\( \text{S}_8 \)) present in 16 g of solid sulphur.
Answer:
First, find the molar mass of \( \text{S}_8 \):
1 mole of \( \text{S}_8 = 8 \times 32 = 256 \) g
We know that 1 mole of any substance contains Avogadro's number of molecules.
So, 1 mole of \( \text{S}_8 = 6.023 \times 10^{23} \) molecules
Therefore, 256 g of \( \text{S}_8 \) has \( = 6.023 \times 10^{23} \) \( \text{S}_8 \) molecules.
Now, to find the number of molecules in 16 g of \( \text{S}_8 \):
Number of molecules \( = \frac{6.023 \times 10^{23}}{256 \text{ g}} \times 16 \text{ g} \)
\( = 3.76 \times 10^{22} \) molecules
In simple words: To find how many molecules are in 16 grams of sulphur (\( \text{S}_8 \)), first calculate the weight of one mole of \( \text{S}_8 \). Then, use Avogadro's number to find how many molecules are in 16 grams based on that molar mass.

Exam Tip: Remember the two-step process: convert mass to moles, then convert moles to the number of particles using Avogadro's number.

Question 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)
Answer:
First, calculate the molar mass of aluminium oxide (\( \text{Al}_2\text{O}_3 \)):
Molar mass of \( \text{Al}_2\text{O}_3 = (2 \times 27) + (3 \times 16) = 54 + 48 = 102 \) u \( = 102 \) g/mol
We know that 1 mole of \( \text{Al}_2\text{O}_3 \) contains Avogadro's number of molecules.
So, 102 g of \( \text{Al}_2\text{O}_3 \) has \( = 6.023 \times 10^{23} \) \( \text{Al}_2\text{O}_3 \) molecules.
Now, calculate the number of \( \text{Al}_2\text{O}_3 \) molecules in 0.051 g:
Number of molecules \( = \frac{6.023 \times 10^{23}}{102 \text{ g}} \times 0.051 \text{ g} \)
\( = 3.01 \times 10^{20} \) \( \text{Al}_2\text{O}_3 \) molecules
From the formula \( \text{Al}_2\text{O}_3 \), we see that 1 molecule of \( \text{Al}_2\text{O}_3 \) contains 2 aluminium ions (\( \text{Al}^{3+} \)).
Hence, 0.051 g of \( \text{Al}_2\text{O}_3 \) contains \( = 2 \times (3.01 \times 10^{20}) \) \( \text{Al}^{3+} \) ions
\( = 6.023 \times 10^{20} \) aluminium ions.
In simple words: First, find the molar mass of aluminium oxide. Then, use that and Avogadro's number to figure out how many molecules are in 0.051 grams. Since each molecule of aluminium oxide has two aluminium ions, multiply the number of molecules by two to get the total number of ions.

Exam Tip: Pay close attention to the stoichiometry: if each molecule/formula unit contains multiple ions of interest, remember to multiply the total number of molecules by that factor to get the number of ions.