Atoms and Molecules — NCERT Solutions | Class 9 Science | Academia Aeternum
Ch 3  ·  Q–
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Class 9 Science Exercise NCERT Solutions Olympiad Board Exam
Chapter 3

Atoms and Molecules

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

6 Questions
15–20 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
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.
📘 Concept & Theory Concept Used ›

The percentage composition by mass (or by weight) of an element in a compound tells us what fraction of the total mass of the compound is contributed by that element.

It is calculated using the formula:

\[ \text{Percentage by mass of an element} = \frac{\text{Mass of the element}}{\text{Total mass of compound}} \times 100 \]

Percentage composition is important because it helps chemists determine the composition and formula of compounds and forms the basis of many numerical problems in chemistry.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the total mass of the compound.

  2. Identify the mass of each constituent element.

  3. Apply the percentage composition formula separately for boron and oxygen.

  4. Calculate the percentages.

  5. Verify that the total percentage equals 100%.

📊 Graph / Figure Graph / Figure ›
Percentage Composition Analysis Given Data: Total Sample Mass: 0.24 g Mass of Boron (B): 0.096 g Mass of Oxygen (O): 0.144 g Boron Oxygen Calculations: % of Boron = (0.096 / 0.24) × 100 = 40% % of Oxygen = (0.144 / 0.24) × 100 = 60% Formula: % Composition = (Mass of Element / Total Mass) × 100
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  7 steps
  1. Given
    • Total mass of compound = 0.24 g
    • Mass of boron (B) = 0.096 g
    • Mass of oxygen (O) = 0.144 g
  3. Calculate percentage by weight of boron
  4. \[ \begin{aligned} \text{Percentage of Boron} &= \frac{\text{Mass of Boron}}{\text{Total Mass of Compound}} \times 100 \\ &=\frac{0.096}{0.24} \times 100 \\ &=0.4 \times 100 \end{aligned} \] \[ = 40\% \]
  5. Therefore, the percentage by weight of boron is: \(40\%\)
  6. Calculate percentage by weight of oxygen
  7. \[ \begin{aligned} \text{Percentage of Oxygen} &= \frac{\text{Mass of Oxygen}}{\text{Total Mass of Compound}} \times 100 \\ &= \frac{0.144}{0.24} \times 100 \\ &= 0.6 \times 100 \\ &= 60\% \end{aligned} \]
  8. Therefore, the percentage by weight of oxygen is: \(60\%\)
  9. Verification
  10. \[40\% + 60\% = 100\%\]
  11. Since the total percentage is 100%, the calculation is correct.
💡 Answer Final Answer ›
  • Percentage of Boron = 40%
  • Percentage of Oxygen = 60%
🎯 Exam Significance Exam Significance ›
  • Tests understanding of percentage composition by mass.
  • Frequently asked as a direct numerical problem in school examinations.
  • Develops the skill of applying chemical formulas to real data.
  • Strengthens concepts related to laws of chemical combination and compound composition.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  5 points

  1. Percentage composition gives the mass contribution of each element in a compound.

  2. The formula used is: \[ \text{Percentage by Mass} = \frac{\text{Mass of Element}}{\text{Total Mass of Compound}} \times 100 \]

  3. Boron contributes 40% of the total mass of the compound.

  4. Oxygen contributes 60% of the total mass of the compound.

  5. The sum of percentage compositions of all elements in a compound is always 100%.
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1 / 6  ·  17%
Q2 →
Q2
NUMERIC3 marks
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? Which law of chemical combination will govern your answer?
📘 Concept & Theory Concept Used ›

Chemical compounds are formed when elements combine in a fixed ratio by mass. Carbon dioxide always contains carbon and oxygen in a definite mass ratio.

The formation of carbon dioxide follows the reaction:

\[ C + O_2 \rightarrow CO_2 \]

Even if one reactant is present in excess, the amount of product formed depends upon the reactant that gets completely consumed first. This reactant is called the limiting reactant.

This question is based mainly on the Law of Definite Proportions and also illustrates the Law of Conservation of Mass.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the balanced chemical reaction.

  2. Determine the fixed mass ratio of carbon and oxygen in carbon dioxide.

  3. Calculate the oxygen required for 3.00 g of carbon.

  4. Compare the required oxygen with the available oxygen.

  5. Identify the limiting reactant.

  6. Calculate the mass of carbon dioxide formed.

  7. State the law of chemical combination governing the result.

📊 Graph / Figure Graph / Figure ›
Law of Chemical Combination Challenge Scenario 1 (Observed) Carbon (C) 3.00 g + Oxygen (O₂) 8.00 g → CO₂ Produced Measured Mass: 11.00 g Note the total mass before and after. Scenario 2 (Prediction Required) Carbon (C) 3.00 g + Oxygen (O₂) 50.00 g → CO₂ Formed? Resulting Mass: ? g Excess oxygen is present. Does C react with all of it? Which Law governs this behavior? Constant Proportions? Conservation of Mass? Multiple Proportions?
Physical Chemistry Visualization
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  18 steps
  1. Given
    • 3.00 g carbon + 8.00 g oxygen → 11.00 g carbon dioxide

    In the new situation:

    • Mass of carbon = 3.00 g
    • Mass of oxygen = 50.00 g
  3. Write the chemical reaction
  4. \[C + O_2 \rightarrow CO_2\]
  5. Determine the mass ratio of carbon and oxygen in carbon dioxide

  6. From the given data:

    \[ 3.00 \text{ g carbon} + 8.00 \text{ g oxygen} \rightarrow 11.00 \text{ g carbon dioxide} \]
  7. Therefore, 3.00 g carbon requires exactly 8.00 g oxygen for complete reaction.
  8. Check whether oxygen is sufficient
  9. Oxygen required for 3.00 g carbon:\[8.00 \text{ g}\]
  10. Oxygen available:\[50.00 \text{ g}\]
  11. Since\[50.00 \text{ g} > 8.00 \text{ g}\]
  12. oxygen is present in excess.
  13. Excess oxygen left unreacted:\[50.00 - 8.00 = 42.00 \text{ g}\]
  14. Thus, carbon is the limiting reactant.
  15. Calculate mass of carbon dioxide formed
  16. Since only 8.00 g oxygen can react with 3.00 g carbon, the reaction remains: \[ 3.00 \text{ g carbon} + 8.00 \text{ g oxygen} \rightarrow 11.00 \text{ g carbon dioxide} \]
  17. Therefore,\[\text{Mass of carbon dioxide formed} = 11.00\ \text{g}\]
  18. Verification using Law of Conservation of Mass
  19. \[ \text{Mass of reactants consumed} = 3.00 + 8.00 = 11.00 \text{ g} \]
  20. \[ \text{Mass of products formed} = 11.00 \text{ g} \]
  21. Hence, mass is conserved.
  22. Law Governing the Answer
  23. The answer is primarily governed by the Law of Definite Proportions (Law of Constant Composition).
  24. This law states:
  25. A pure chemical compound always contains the same elements combined in the same fixed proportion by mass.

    Carbon dioxide always contains carbon and oxygen in a fixed mass ratio. Therefore, 3.00 g carbon can react with only 8.00 g oxygen, irrespective of how much extra oxygen is available.

    The result also supports the Law of Conservation of Mass, according to which the total mass of reactants consumed equals the total mass of products formed.

💡 Answer Final Answer ›
\[\text{Mass of carbon dioxide formed} = 11.00\ \text{g}\]

Law Governing the Answer:

  • Law of Definite Proportions (Primary Law)
  • Law of Conservation of Mass (Also Verified)
🎯 Exam Significance Exam Significance ›
  • Directly tests understanding of laws of chemical combination.
  • Frequently asked as a conceptual numerical problem.
  • Develops the concept of fixed composition of compounds.
  • Introduces the idea of limiting and excess reactants.
  • Strengthens understanding of mass relationships in chemical reactions.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. Carbon dioxide is formed in a fixed mass ratio of carbon and oxygen.

  2. 3.00 g carbon requires only 8.00 g oxygen for complete reaction.

  3. When 50.00 g oxygen is supplied, oxygen remains in excess.

  4. Carbon acts as the limiting reactant.

  5. The mass of carbon dioxide formed remains 11.00 g.

  6. The answer is mainly explained by the Law of Definite Proportions.

  7. The Law of Conservation of Mass is also verified because total mass remains unchanged during the reaction.
← Q1
2 / 6  ·  33%
Q3 →
Q3
NUMERIC3 marks
What are polyatomic ions? Give examples.
📘 Concept & Theory Theory / Concept ›

Atoms can gain or lose electrons and become charged particles called ions. Some ions consist of only one atom, such as:

  • Na+ (Sodium ion)
  • Cl− (Chloride ion)
  • Mg2+ (Magnesium ion)

Such ions are known as monatomic ions.

However, in many compounds, a group of two or more atoms remains bonded together and carries an overall positive or negative charge. This charged group behaves as a single ion during chemical reactions.

Such ions are called polyatomic ions.

The atoms within a polyatomic ion are held together by covalent bonds, but the entire group possesses a net electrical charge.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Understand the meaning of an ion.

  2. Recognize that some ions contain more than one atom.

  3. Define polyatomic ions.

  4. Give suitable examples with their formulae and charges.

📊 Graph / Figure Graph / Figure ›
Understanding Polyatomic Ions Definition: A cluster of atoms that are covalently bonded together and carry a net electric charge. Hydroxide Ion (OH⁻) O H - Simplified unit Ammonium Ion (NH₄⁺) N H H H + Group of atoms acts as a single ion Note: They participate in reactions as stable, single units.
Chemistry Fundamentals
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  2 steps

  1. A polyatomic ion is a group of two or more atoms that are chemically bonded together and carry an overall positive or negative charge.

    The entire group behaves as a single ion in chemical reactions.

    Unlike simple ions such as Na+ or Cl−, polyatomic ions contain multiple atoms joined together by covalent bonds.

  2. Examples of polyatomic ions are:
  3. Name of Ion Formula Charge
    Ammonium NH4+ +1
    Hydroxide OH− −1
    Nitrate NO3− −1
    Carbonate CO32− −2
    Sulphate SO42− −2
    Phosphate PO43− −3
    Hydrogen Carbonate (Bicarbonate) HCO3− −1
    Acetate CH3COO− −1
💡 Answer Final Answer ›

Polyatomic ions are charged groups of two or more atoms that are covalently bonded together and behave as a single ion during chemical reactions.

Examples:

  • Ammonium ion, NH4+
  • Hydroxide ion, OH−
  • Nitrate ion, NO3−
  • Carbonate ion, CO32−
  • Sulphate ion, SO42−
🎯 Exam Significance Exam Significance ›
  • Polyatomic ions are frequently asked in definitions, short-answer, and objective-type questions.
  • Knowledge of common polyatomic ions is essential for writing correct chemical formulae.
  • Helps in understanding the composition of acids, bases, and salts.
  • Forms the basis for balancing chemical equations and naming compounds.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points

  1. A polyatomic ion consists of more than one atom.

  2. The atoms are held together by covalent bonds.

  3. The entire group carries a net positive or negative charge.

  4. Polyatomic ions behave as a single unit during chemical reactions.

  5. Important examples include NH4+, OH−, NO3−, CO32−, and SO42−.

  6. Knowledge of polyatomic ions is essential for writing chemical formulae correctly.
← Q2
3 / 6  ·  50%
Q4 →
Q4
NUMERIC3 marks
Write the chemical formulae of the following.
  1. Magnesium chloride
  2. Calcium oxide
  3. Copper nitrate
  4. Aluminium chloride
  5. Calcium carbonate
📘 Concept & Theory Theory / Concept ›

A chemical formula represents the composition of a compound using symbols of elements and ions. For ionic compounds, the formula is written in such a way that the total positive charge of the cation equals the total negative charge of the anion.

To write a chemical formula correctly, we must:

  1. Identify the positive ion (cation).
  2. Identify the negative ion (anion).
  3. Write their valencies (charges).
  4. Balance the charges using the criss-cross method.
  5. Reduce the ratio to the simplest whole-number form if possible.

Some compounds in this question contain polyatomic ions such as nitrate \[(NO_3^-)\] and carbonate \[(CO_3^{2-})\].

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the symbols of the cation and anion.

  2. Determine their charges (valencies).

  3. Apply the criss-cross method.

  4. Check that the overall charge becomes zero.

  5. Write the final chemical formula.

📊 Graph / Figure Graph / Figure ›
Constructing Chemical Formulae Applying the Criss-Cross Method Magnesium Chloride Mg 2+ Cl 1- MgCl2 Calcium Oxide Ca 2+ O 2- CaO (Simplified 1:1) Copper(II) Nitrate Cu 2+ NO3 1- Cu(NO3)2 Aluminium Chloride Al 3+ Cl 1- AlCl3 Calcium Carbonate Ca 2+ CO3 2- CaCO3 (Simplified 1:1) Step-by-Step Method: 1. Write symbols of constituent ions. 2. Write valencies/charges above them. 3. Criss-cross the valencies. 4. Simplify the ratio if necessary.
Chemical Nomenclature Visual Guide
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  22 steps
  1. (a) Magnesium chloride
  2. Magnesium ion: \[ Mg^{2+} \]
  3. Magnesium ion: \[ Mg^{2+} \]
  4. Balancing charges:\[Mg^{2+} + 2Cl^{-}\]
  5. Therefore, the chemical formula is:\[MgCl_2\]
  6. (b) Calcium oxide
  7. Calcium ion: \[ Ca^{2+} \]
  8. Oxide ion: \[ O^{2-} \]
  9. Since the charges are equal and opposite:\[Ca^{2+}+O^{2-}\]
  10. Ratio:\[1:1\]
  11. Therefore, the chemical formula is:\[CaO\]
  12. (c) Copper nitrate
  13. Copper ion: \[ Cu^{2+} \]
  14. Nitrate ion: \[ NO_3^{-} \]
  15. Balancing charges:\[Cu^{2+}+2(NO_3^-)\]
  16. Therefore, the chemical formula is:\[Cu(NO_3)_2\] Parentheses are used because nitrate is a polyatomic ion.
  17. (d) Aluminium chloride
  18. Aluminium ion: \[ Al^{3+} \]
  19. Chloride ion: \[ Cl^{-} \]
  20. Balancing charges:\[Al^{3+}+3Cl^{-}\]
  21. Therefore, the chemical formula is:\[AlCl_3\]
  22. (e) Calcium carbonate
  23. Calcium ion: \[ Ca^{2+} \]
  24. Carbonate ion: \[ CO_3^{2-} \]
  25. Since the charges are equal and opposite:\[Ca^{2+}+CO_3^{2-}\]
  26. Ratio: \[1:1\]
  27. Therefore, the chemical formula is:\[CaCO_3\]
💡 Answer Final Answer ›
Compound Chemical Formula
Magnesium chloride \(MgCl_2\)
Calcium oxide \(CaO\)
Copper nitrate \(Cu(NO_3)_2\)
Aluminium chloride \(AlCl_3\)
Calcium carbonate \(CaCO_3\)
🎯 Exam Significance Exam Significance ›
  • Writing chemical formulae is one of the most frequently asked topics in examinations.
  • Tests understanding of valency and ionic charges.
  • Helps students learn correct representation of compounds.
  • Forms the basis for writing chemical equations.
  • Important for objective, short-answer, and long-answer questions.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points

  1. Chemical formulae are written by balancing positive and negative charges.

  2. The net charge of a compound must always be zero.

  3. Polyatomic ions such as nitrate \((NO_3^-)\) and carbonate \((CO_3^{2-})\) should be treated as single units.

  4. Parentheses are used when more than one polyatomic ion is present in a formula.

  5. Knowledge of valencies is essential for writing correct chemical formulae.

  6. Correct formula writing forms the foundation of all higher-level chemistry calculations.
← Q3
4 / 6  ·  67%
Q5 →
Q5
NUMERIC3 marks
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
📘 Concept & Theory Theory / Concept ›

A compound is a pure substance formed when two or more elements combine chemically in a fixed proportion.

To identify the elements present in a compound, we first write its chemical formula and then identify the symbols of the constituent elements.

Some common compounds are known by their traditional names, such as quick lime and baking powder. Therefore, it is important to know both their names and chemical formulae.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the chemical formula of each compound.

  2. Identify the symbols present in the formula.

  3. Convert the symbols into element names.

  4. List all constituent elements.

📊 Graph / Figure Graph / Figure ›
Elements in Common Compounds Quick Lime Calcium Oxide (CaO) Calcium (Ca) Oxygen (O) Derived from heating limestone. Hydrogen Bromide (HBr) Hydrogen (H) Bromine (Br) A colorless gas at room temp. Baking Powder Sodium Hydrogen Carbonate (NaHCO3) Sodium (Na) Hydrogen (H) Carbon (C) Oxygen (O) Also known as Sodium Bicarbonate. Potassium Sulphate (K2SO4) Potassium (K) Sulphur (S) Oxygen (O) Used commonly in fertilizers.
Chemical Composition Guide
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. (a) Quick lime
  2. Chemical formula:\[CaO\]
  3. Elements present:
    • Ca = Calcium
    • O = Oxygen
  4. Therefore, quick lime contains:\[\text{Calcium and Oxygen}\]
  5. (b) Hydrogen bromide
  6. Chemical formula:\[HBr\]
  7. Elements present:
    • H = Hydrogen
    • Br = Bromine
  8. Therefore, hydrogen bromide contains:\[\text{Hydrogen and Bromine}\]
  9. (c) Baking powder
  10. In Class 9 NCERT, baking powder is generally considered to contain sodium hydrogen carbonate as its main constituent.
  11. Chemical formula of sodium hydrogen carbonate:\[NaHCO_3\]
  12. Elements present:
    • Na = Sodium
    • H = Hydrogen
    • C = Carbon
    • O = Oxygen
  13. Therefore, baking powder contains:\[\text{Sodium, Hydrogen, Carbon and Oxygen}\]
  14. (d) Potassium sulphate
  15. Chemical formula:\[K_2SO_4\]
  16. Elements present:
    • K = Potassium
    • S = Sulphur
    • O = Oxygen
  17. Therefore, potassium sulphate contains:\[\text{Potassium, Sulphur and Oxygen}\]
💡 Answer Final Answer ›
Compound Chemical Formula Elements Present
Quick lime \(CaO\) Calcium, Oxygen
Hydrogen bromide \(HBr\) Hydrogen, Bromine
Baking powder* \(NaHCO_3\) Sodium, Hydrogen, Carbon, Oxygen
Potassium sulphate \(K_2SO_4\) Potassium, Sulphur, Oxygen

*For NCERT Class 9 exercises, baking powder is generally treated through its active ingredient, sodium hydrogen carbonate (\(NaHCO_3\)).

🎯 Exam Significance Exam Significance ›
  • Tests knowledge of chemical symbols and formulae.
  • Strengthens understanding of the composition of compounds.
  • Frequently appears in objective and short-answer questions.
  • Helps students connect common names of substances with their chemical formulae.
  • Develops familiarity with important compounds used in everyday life.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  7 points

  1. A compound consists of two or more elements chemically combined in a fixed ratio.

  2. Elements present in a compound can be identified from its chemical formula.

  3. Quick lime (\(CaO\)) contains calcium and oxygen.

  4. Hydrogen bromide (\(HBr\)) contains hydrogen and bromine.

  5. Sodium hydrogen carbonate (\(NaHCO_3\)) contains sodium, hydrogen, carbon, and oxygen.

  6. Potassium sulphate (\(K_2SO_4\)) contains potassium, sulphur, and oxygen.

  7. Knowledge of symbols and formulae is essential for learning chemistry effectively.
← Q4
5 / 6  ·  83%
Q6 →
Q6
NUMERIC3 marks
Calculate the molar mass of the following substances.
(a) Ethyne, \(\mathrm{C_2H_2}\)
(b) Sulphur molecule, \(\mathrm{S_8}\)
(c) Phosphorus molecule, \(\mathrm{P_4}\)
(Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, \(\mathrm{HCl}\)
(e) Nitric acid, \(\mathrm{HNO_3}\)
📘 Concept & Theory Theory / Concept ›

Molar mass is the mass of one mole of a substance. It is expressed in \[ \text{g mol}^{-1} \]

The molar mass of a compound is obtained by adding the atomic masses of all the atoms present in its chemical formula.

Mathematically,

\[ \text{Molar Mass} = \sum (\text{Number of atoms} \times \text{Atomic Mass}) \]

Atomic masses used:

  • \(C = 12\)
  • \(H = 1\)
  • \(S = 32\)
  • \(P = 31\)
  • \(Cl = 35.5\)
  • \(N = 14\)
  • \(O = 16\)
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Write the chemical formula of the substance.

  2. Identify the number of atoms of each element.

  3. Multiply the number of atoms by their respective atomic masses.

  4. Add all contributions.

  5. Express the result in g mol−1.

📊 Graph / Figure Graph / Figure ›
Molar Mass Analysis Atomic Masses (u): C = 12 | H = 1 | S = 32 | P = 31 | Cl = 35.5 | N = 14 | O = 16 (a) Ethyne C2H2 Molar Mass = 2 × (Mass of C) + 2 × (Mass of H) (b) Sulphur Molecule S8 Molar Mass = 8 × (Mass of S) (c) Phosphorus Molecule P4 Molar Mass = 4 × (Mass of P) (d) Hydrochloric Acid HCl Molar Mass = (Mass of H) + (Mass of Cl) (e) Nitric Acid HNO3 Molar Mass = (Mass of H) + (Mass of N) + 3 × (Mass of O) Note: Multiply the atomic mass of Oxygen by its subscript. Calculation Guide Sum the masses of all atoms present in the chemical formula. Unit: g/mol (Grams per mole) Concept: 1 Mole = Molecular Mass expressed in grams
Stoichiometry Visualization
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  15 steps
  1. (a) Ethyne, \(\mathrm{C_2H_2}\)
  2. Number of carbon atoms = 2\[2 \times 12 = 24\]
  3. Number of hydrogen atoms = 2\[2 \times 1 = 2\]
  4. Therefore, \[ \begin{aligned} \text{Molar Mass of } C_2H_2 &=24 + 2\\ &=26 \end{aligned} \]
  5. 26\ \text{g mol}^{-1}
  6. (b) Sulphur molecule, \(\mathrm{S_8}\)
  7. Number of sulphur atoms = 8 \[\begin{aligned}\text{Molar Mass of } S_8&=8 \times 32\\&= 256\ \text{g mol}^{-1}\end{aligned}\]
  8. (c) Phosphorus molecule, \(\mathrm{P_4}\)
  9. Atomic mass of phosphorus = 31
  10. Number of phosphorus atoms = 4
  11. Therefore, \[ \begin{aligned} \text{Molar Mass of } P_4 &= 4 \times 31\\ &= 124\ \text{g mol}^{-1} \end{aligned} \]
  12. (d) Hydrochloric acid, \(\mathrm{HCl}\)
  13. Hydrogen contribution:\[1 \times 1 = 1\]
  14. Chlorine contribution:\[1 \times 35.5 = 35.5\]
  15. Therefore, \[ \begin{aligned} \text{Molar Mass of HCl} &=1 + 35.5\\ &= 36.5\ \text{g mol}^{-1} \end{aligned} \]
  16. (e) Nitric acid, \(\mathrm{HNO_3}\)
  17. Hydrogen contribution:\[1 \times 1 = 1\]
  18. Nitrogen contribution:\[1 \times 14 = 14\]
  19. Oxygen contribution:\[3 \times 16 = 48\]
  20. Therefore, \[ \begin{aligned} \text{Molar Mass of HNO_3} &=1 + 14 + 48\\ &= 63\ \text{g mol}^{-1} \end{aligned} \]
💡 Answer Final Answer ›
Substance Formula Molar Mass (g mol−1)
Ethyne \(C_2H_2\) 26
Sulphur molecule \(S_8\) 256
Phosphorus molecule \(P_4\) 124
Hydrochloric acid \(HCl\) 36.5
Nitric acid \(HNO_3\) 63
🎯 Exam Significance Exam Significance ›
  • Molar mass calculation is one of the most important numerical concepts in chemistry.
  • Frequently asked in direct, short-answer, and objective-type questions.
  • Tests understanding of chemical formulae and atomic masses.
  • Builds the foundation for mole concept calculations in higher classes.
  • Develops accuracy in handling chemical data and numerical problems.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  8 points

  1. Molar mass is the mass of one mole of a substance.

  2. It is calculated by adding the atomic masses of all atoms present in the formula.

  3. \(C_2H_2\) has a molar mass of 26 g mol−1.

  4. \(S_8\) has a molar mass of 256 g mol−1.

  5. \(P_4\) has a molar mass of 124 g mol−1.

  6. \(HCl\) has a molar mass of 36.5 g mol−1.

  7. \(HNO_3\) has a molar mass of 63 g mol−1.

  8. Molar mass serves as a bridge between microscopic particles and measurable mass.
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