Class 11 Chemistry Chapter 1 notes, in-text examples, NCERT exercises (1.1 to 1.36), revision formula sheets, and chapter tests.
1.1 Development of Chemistry
Chemistry developed from the search for the Philosopher’s stone (Paras), which was believed to turn base metals into gold, and the elixir of life to grant immortality. In ancient India, chemistry was known as Rasayan Shastra, Rastantra, Ras Kriya, or Rasvidya.
Environment: Developing alternatives to ozone-depleting CFCs (chlorofluorocarbons) and studying greenhouse gases ($CO_2, CH_4$).
1.3 Nature & Classification of Matter
Matter is anything that has mass and occupies space.
Solid: Definite volume and definite shape. Particles are closely packed in an orderly arrangement with minimal freedom of movement.
Liquid: Definite volume, but no definite shape (takes the shape of the container). Particles are close together but can flow.
Gas: Neither definite volume nor definite shape. Particles are far apart and move freely.
Classification of Matter
Pure Substances: Elements vs Compounds
An element contains only one type of particle (atoms or molecules like $O_2$). A compound contains two or more different elements chemically combined in a fixed ratio. The properties of a compound differ completely from those of its constituent elements (for example, hydrogen is combustible and oxygen supports combustion, but their compound water is a fire extinguisher).
1.4 Properties of Matter and Measurement
Physical properties can be measured without changing the chemical identity of the substance (mass, volume, melting point, density). Chemical properties require a chemical change (acidity, combustibility, reactivity).
The International System of Units (SI) uses 7 base units:
Note: $-40^\circ\text{C} = -40^\circ\text{F}$. Negative values are not possible on the Kelvin scale ($0\text{ K}$ is absolute zero).
Mass vs Weight & Density
• Mass: Quantity of matter in a body (constant).
• Weight: Force exerted by gravity ($W = mg$).
• Density: $\frac{\text{Mass}}{\text{Volume}}$ (SI: $kg\text{ m}^{-3}$, common: $g\text{ cm}^{-3}$).
1.5 Uncertainty in Measurement & Significant Figures
Scientific Notation: Numbers are written as $N \times 10^n$, where $1 \le N < 10$ and $n$ is an integer.
Significant Figures Rules
All non-zero digits are significant ($285\text{ cm} \to 3$ sig figs).
Zeros preceding the first non-zero digit are not significant ($0.03\text{ g} \to 1$ sig fig; $0.0052 \to 2$ sig figs).
Zeros between non-zero digits are significant ($2.005\text{ g} \to 4$ sig figs).
Zeros at the end or right of a number are significant if after a decimal point ($0.200\text{ g} \to 3$ sig figs; $100. \to 3$ sig figs).
Exact counts have infinite significant figures ($20\text{ balls} = 20.000...$).
Precision vs Accuracy
• Precision: Closeness of multiple measurements of the same quantity to each other.
• Accuracy: Agreement of a particular value to the true value of the result.
Calculations & Rounding
• Addition/Subtraction: Result cannot have more digits to the right of the decimal point than any of the original numbers ($12.11 + 18.0 + 1.012 = 31.1$).
• Multiplication/Division: Result cannot have more significant figures than the measurement with the fewest significant figures ($2.5 \times 1.25 = 3.1$).
Dimensional Analysis (Factor Label Method)
Ex 1
A piece of metal is 3 inch long. What is its length in cm?
1.6 Laws of Chemical Combinations & Dalton's Atomic Theory
Law
Proposed By
Statement
Example
Conservation of Mass
Antoine Lavoisier (1789)
Matter can neither be created nor destroyed in a chemical reaction.
Combustion experiments showing mass of reactants = mass of products.
Definite Proportions
Joseph Proust (1799)
A given compound always contains exactly the same proportion of elements by weight.
Natural and synthetic cupric carbonate both contain 51.35% Cu, 9.74% C, 38.91% O.
Multiple Proportions
John Dalton (1803)
If two elements can 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.
$H_2 + O_2 \to H_2O$ (16 g O) and $H_2O_2$ (32 g O). Ratio of O masses with 2 g H is $16:32 = 1:2$.
Gay Lussac's Law
Gay Lussac (1808)
Gases combine or are produced in a simple ratio by volume provided all gases are at the same temperature and pressure.
$100\text{ mL } H_2 + 50\text{ mL } O_2 \to 100\text{ mL } H_2O\text{ vapour}$ (Ratio $2:1:2$).
Avogadro's Law
Amedeo Avogadro (1811)
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Distinguished between atoms and diatomic molecules ($H_2, O_2$).
Reciprocal (Equivalent) Proportions
Jeremias Richter (1792)
When two different elements combine separately with a fixed mass of a third element, the ratio in which they do so is either the same or a simple whole-number multiple of the ratio in which they combine with each other.
$C$ combines with $H$ to form $CH_4$ ($12\text{ g C}:4\text{ g H}$ or $3:1$). $O$ combines with $H$ to form $H_2O$ ($16\text{ g O}:2\text{ g H}$ or $32\text{ g O}:4\text{ g H}$). Ratio of $C:O$ combining with fixed $4\text{ g H}$ is $12:32 = 3:8$. When $C$ and $O$ combine directly to form $CO_2$, the ratio is $12:32 = 3:8$ (ratio of ratios is $1:1$).
Dalton's Atomic Theory (1808)
Matter consists of indivisible atoms.
All atoms of a given element have identical mass and chemical properties. Atoms of different elements differ in mass.
Compounds are formed when atoms of different elements combine in fixed ratios.
Chemical reactions involve reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.
1.7 Atomic, Molecular & Formula Masses
Atomic mass is defined relative to the carbon-12 isotope ($^{12}C$), assigned a mass of exactly 12 atomic mass units ($u$):
$$1\text{ u} = \frac{1}{12} \times \text{Mass of one }^{12}C\text{ atom} = 1.66056 \times 10^{-24}\text{ g}$$
Average Atomic Mass
Calculated taking into account the fractional abundance of isotopes:
$$\bar{A} = \sum (A_i \times x_i)$$
Molecular & Formula Mass
• Molecular Mass: Sum of atomic masses of the elements in a molecule ($CH_4 = 12.011 + 4(1.008) = 16.043\text{ u}$).
• Formula Mass: Used for ionic compounds such as $NaCl$ where discrete molecules do not exist: $Na^+ (23.0) + Cl^- (35.5) = 58.5\text{ u}$.
Equivalent Mass (Equivalent Weight, $E$)
The equivalent mass of a substance is the number of parts by mass of it which combines with or displaces directly or indirectly $1.008\text{ parts by mass of H}$, $8\text{ parts by mass of O}$, or $35.5\text{ parts by mass of Cl}$.
$$\text{Equivalent Mass } (E) = \frac{\text{Atomic Mass or Molar Mass}}{n\text{-factor (Valency Factor)}}$$
For Elements: $E = \frac{\text{Atomic Mass}}{\text{Valency}}$ (e.g., for $Mg$, $E = \frac{24}{2} = 12$; for $Al$, $E = \frac{27}{3} = 9$).
For Acids: $E = \frac{\text{Molar Mass}}{\text{Basicity (number of replaceable } H^+ \text{ ions)}}$.
- For $HCl$: $E = \frac{36.5}{1} = 36.5$
- For $H_2SO_4$: $E = \frac{98}{2} = 49$
- For $H_3PO_4$ (basicity 3): $E = \frac{98}{3} = 32.67$; For $H_3PO_3$ (basicity 2): $E = \frac{82}{2} = 41$; For $H_3PO_2$ (basicity 1): $E = \frac{66}{1} = 66$.
For Salts: $E = \frac{\text{Formula Mass}}{\text{Total positive or negative charge}}$ (e.g., for $Na_2CO_3$, charge $= 2 \implies E = \frac{106}{2} = 53$; for $Al_2(SO_4)_3$, total charge $= 6 \implies E = \frac{M}{6}$).
The density of 3 M solution of NaCl is 1.25 g mL⁻¹. Calculate the molality of the solution.
$3\text{ M} \implies 3\text{ moles } NaCl$ in $1\text{ L solution}$.
Mass of $NaCl = 3 \times 58.5 = 175.5\text{ g}$.
Mass of $1\text{ L solution} = 1000 \times 1.25 = 1250\text{ g}$.
Mass of water $= 1250 - 175.5 = 1074.5\text{ g} = 1.0745\text{ kg}$.
$$\text{Molality} = \frac{3\text{ mol}}{1.0745\text{ kg}} = \mathbf{2.79\text{ m}}$$
Practice Quiz (25 Questions)
Q01
Which state of matter exhibits the highest compressibility and possesses neither a definite volume nor a definite shape?
Incorrect: Solids have constituent particles packed closely with rigid bonds, giving them definite volume and shape with negligible compressibility.
Incorrect: Liquids have definite volume and take the shape of the container; they have very low compressibility.
Correct Answer: Gas particles are far apart with negligible intermolecular forces and high kinetic energy, resulting in maximum compressibility and neither definite volume nor shape.
Incorrect: Ordinary gases at standard room conditions exhibit this behavior without requiring ionization at extreme temperatures.
Q02
Why is pure water (H₂O) classified as a chemical compound rather than a mixture?
Incorrect: Physical state at room temperature does not define whether a substance is an element, compound, or mixture.
Correct Answer: In water, hydrogen and oxygen are chemically combined in a strict 1:8 mass ratio (2:16), producing a substance with unique chemical properties entirely distinct from H₂ and O₂.
Incorrect: Water readily freezes into solid ice at 0 °C (273.15 K).
Incorrect: Water is a pure chemical compound; its constituent elements cannot be separated by physical methods like filtration.
Q03
The ancient Indian philosopher who formulated that all matter consists of eternal, spherical, suprasensible particles called Paramāṇu in 600 BCE was:
Incorrect: Nagarjuna was an eminent alchemist of the 8th9th century CE (author of Rasratnakar), not the 6th century BCE philosopher.
Correct Answer: Acharya Kanada authored the Vaiseshika Sutras in 600 BCE, propounding the concept of eternal, indivisible building blocks called Paramāṇu.
Incorrect: Chakrapani lived in the 11th century CE and is famed for discovering mercury sulphide and soap-making.
Incorrect: Varahamihira (6th century CE) was an astronomer and polymath who authored the Brihat Samhita.
Q04
Which ancient Indian alchemical text (c. 800 CE) first systematically documented furnace construction and identification of metals by their flame test colors?
Incorrect: Charaka Samhita is an ancient Ayurvedic medical treatise detailing therapeutics and Bhasmas, not systematic furnace flame tests.
Incorrect: Sushruta Samhita focuses on surgical methods and instruments.
Correct Answer: Rsarnavam (c. 800 CE) explicitly describes furnace construction, crucibles, and identifying metals by the color of their flames.
Incorrect: Brihat Samhita focuses on architecture, perfumes, and agricultural sciences.
Q05
Which pair of pharmaceuticals represents groundbreaking life-saving drugs used extensively in cancer chemotherapy?
Incorrect: Aspirin and paracetamol are analgesics and antipyretics used for pain and fever relief.
Correct Answer: Cisplatin and Taxol are prominent life-saving chemotherapy agents used to treat cancer.
Incorrect: AZT is an antiretroviral drug for HIV/AIDS and penicillin is an antibacterial antibiotic.
Incorrect: Chloroquine and quinine are antimalarial drugs.
Q06
The SI base units for 'amount of substance' and 'luminous intensity' are respectively:
Incorrect: The SI base unit for mass is the kilogram (not gram), and lumen is the derived unit of luminous flux.
Correct Answer: The mole (mol) is the SI base unit for amount of substance, and candela (cd) is the SI base unit for luminous intensity.
Incorrect: Dalton is an accepted non-SI unit of mass, and lux is the derived unit of illuminance.
Incorrect: Kilogram is mass and watt is the derived unit of power.
Q07
Under the revised 2019 CGPM SI definitions, the kilogram is anchored to the exact fixed numerical value of:
Incorrect: The Pt-Ir cylinder in Sèvres was retired in the 2019 CGPM revision to eliminate drift.
Correct Answer: In the 2019 CGPM redefinition, the kilogram is anchored to the exact fixed value of the Planck constant h.
Incorrect: The speed of light is used to define the metre, not the kilogram.
Incorrect: Caesium-133 hyperfine transition frequency defines the second.
Q08
What are the correct numerical multiples for the metric prefixes pico (p), femto (f), and micro (μ)?
Incorrect: 10⁻⁹ is nano, 10⁻¹² is pico, and 10⁻³ is milli.
Incorrect: These are micro, nano, and pico respectively.
Incorrect: Femto is 10⁻¹⁵, not 10⁻¹².
Q09
Why does an object's mass remain constant on the Moon while its weight decreases to ~1/6th of its Earth value?
Incorrect: Mass is conserved and cannot be destroyed by changes in location.
Correct Answer: Mass measures the amount of matter and is constant everywhere; weight W = mg depends on the local gravitational acceleration g, which is ~1/6th on the Moon.
Incorrect: The Moon has virtually no atmosphere.
Incorrect: Volume of a solid or liquid remains virtually unchanged in different gravitational fields.
Q10
Normal human body temperature is 37.0 °C. What is its equivalent on the Fahrenheit scale?
Incorrect: 95.0 °F corresponds to (95 - 32) × (5/9) = 35.0 °C.
Incorrect: 100.4 °F corresponds to 38.0 °C (a fever temperature).
Incorrect: 102.0 °F corresponds to 38.9 °C.
Q11
At what numerical value do both Celsius and Fahrenheit thermometer scales coincide?
Incorrect: At 0 °C, °F is (9/5)(0) + 32 = 32 °F.
Correct Answer: Setting x = (9/5)x + 32 gives x - (9/5)x = 32 => -(4/5)x = 32 => x = -40. Thus -40 °C = -40 °F.
Incorrect: At 32 °C, °F is (9/5)(32) + 32 = 89.6 °F.
Incorrect: -273.15 °C is absolute zero, corresponding to -459.67 °F.
Q12
Why is a temperature reading of -10 K physically meaningless?
Incorrect: Mercury freezes at 234.3 K (-38.8 °C), far above absolute zero.
Correct Answer: 0 K is the lowest theoretically possible temperature (absolute zero). No temperature can be below 0 K.
Incorrect: While nitrogen and oxygen solidify at cryogenic temperatures, that does not define the absolute thermodynamic zero.
Incorrect: Specialized cryogenic resistance thermometers work well down into the millikelvin range.
Q13
According to the Atlantic-Pacific Rule, how many significant figures are in 0.005080?
Incorrect: The trailing zero after the decimal point is significant.
Correct Answer: Leading zeros (0.00) are placeholders. In 0.005080, digits 5, 0, 8, 0 are all significant, giving 4 significant figures.
Incorrect: The leading zeros before the digit 5 are not significant.
Incorrect: Includes all zeros regardless of significance rules.
Q14
Report the result of 12.11 + 18.0 + 1.012 to the correct number of significant figures:
Incorrect: Has 3 decimal places; the measurement 18.0 limits the result to 1 decimal place.
Incorrect: Has 2 decimal places instead of 1.
Correct Answer: In addition, the result must have the same number of decimal places as the number with the fewest decimal places (18.0 has 1 decimal place; 31.122 rounds to 31.1).
Incorrect: Rounded to whole number, losing significant precision.
Q15
In an experiment with true mass 2.000 g, Student C reports two trials: 2.001 g and 1.999 g. These results are:
Incorrect: The average is 2.000 g, which matches the true value, so it is accurate.
Incorrect: The two values differ by only 0.002 g, so they are also precise.
Correct Answer: The values 2.001 g and 1.999 g are close to each other (precise) and their average (2.000 g) agrees with the true mass (accurate).
Incorrect: Both criteria for precision and accuracy are satisfied.
Q16
When 12 g of carbon is burnt in 32 g of oxygen in a sealed vessel, exactly 44 g of CO₂ is produced. This verifies:
Incorrect: Multiple proportions applies when two elements form more than one compound.
Correct Answer: Total mass of reactants (12 g + 32 g = 44 g) equals total mass of products (44 g CO₂), verifying conservation of mass.
Incorrect: Gay Lussac's law applies to reacting gaseous volumes.
Incorrect: Avogadro's law relates gas volume to number of molecules.
Q17
Proust demonstrated that natural and synthetic cupric carbonate both contain 51.35% Cu, 9.74% C, and 38.91% O. This proves:
Incorrect: Relates total reactant mass to product mass.
Correct Answer: Proust showed that cupric carbonate has the exact same elemental composition by mass regardless of whether it is natural or synthetic.
Incorrect: Concerns different compounds formed by the same elements.
Incorrect: Concerns ratios of three different elements combining with each other.
Q18
Carbon forms two oxides: CO (12 g C + 16 g O) and CO₂ (12 g C + 32 g O). The mass ratio of oxygen combining with fixed 12 g C is 16:32 = 1:2. This demonstrates:
Incorrect: Does not describe ratios between two distinct oxides.
Correct Answer: The masses of oxygen (16 g and 32 g) combining with fixed 12 g of carbon are in the simple whole-number ratio 1:2.
Incorrect: Concerns a single compound having fixed composition.
Incorrect: Deals with gaseous volumes, not solid/gas mass ratios.
Q19
Gay Lussac found that 100 mL of H₂ gas reacts with 50 mL of O₂ gas to yield 100 mL of water vapour. The volume ratio 2:1:2 illustrates:
Incorrect: Avogadro's law deals with number of molecules in equal gas volumes.
Correct Answer: 2 volumes of H₂ combine with 1 volume of O₂ to produce 2 volumes of water vapour at constant T and P (ratio 2:1:2).
Incorrect: Dalton's theory could not explain Gay Lussac's volumetric observations.
Incorrect: Boyle's law relates pressure and volume of a fixed mass of gas.
Q20
Equal volumes of all gases at the same temperature and pressure contain equal number of molecules. This principle is:
Incorrect: Dalton believed atoms of the same element could not combine with each other (opposing diatomic gases).
Correct Answer: Avogadro proposed in 1811 that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Incorrect: Graham's law relates rates of gas effusion/diffusion.
Incorrect: Charles's law relates volume and absolute temperature at constant pressure.
Q21
One unified atomic mass unit (1 u) is defined as exactly:
Incorrect: Hydrogen was used as an early reference, but replaced by carbon-12 in 1961.
Correct Answer: 1 atomic mass unit (1 u) is defined as exactly 1/12th the mass of an unbound neutral carbon-12 atom.
Incorrect: Electron mass is ~1/1836th of an atomic mass unit.
Incorrect: Oxygen-16 was used prior to 1961 but superseded by carbon-12.
Q22
Which of the following 1.0 g samples contains the greatest number of atoms?
Incorrect: High molar mass yields only 1/197 mol of atoms (~3.06 × 10²¹ atoms).
Incorrect: Gives 1/23 mol (~2.62 × 10²² atoms).
Correct Answer: Lowest atomic mass yields the highest number of moles (1/7 mol) and atoms: (1/7) × 6.022 × 10²³ = 8.60 × 10²² atoms.
Step 2: Calculations
(i) Molar mass of $H_2O = 2(1.008) + 1(16.00) = 2.016 + 16.00 = 18.016\text{ g mol}^{-1} \approx \mathbf{18.02\text{ g mol}^{-1}}$
(ii) Molar mass of $CO_2 = 1(12.011) + 2(16.00) = 12.011 + 32.00 = \mathbf{44.01\text{ g mol}^{-1}}$
(iii) Molar mass of $CH_4 = 1(12.011) + 4(1.008) = 12.011 + 4.032 = \mathbf{16.043\text{ g mol}^{-1}}$
1.2
Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).
Step 1: Molar Mass of Sodium Sulphate
$M(Na_2SO_4) = 2(22.99) + 1(32.06) + 4(16.00) = 45.98 + 32.06 + 64.00 = 142.04\text{ g mol}^{-1}$
Step 2: Mass % Calculation
$$\text{Mass % of an element} = \frac{\text{Total mass of that element in 1 mol}}{\text{Molar mass of compound}} \times 100$$
• $\text{Mass % of } Na = \frac{45.98}{142.04} \times 100 = \mathbf{32.37\%}$
• $\text{Mass % of } S = \frac{32.06}{142.04} \times 100 = \mathbf{22.57\%}$
• $\text{Mass % of } O = \frac{64.00}{142.04} \times 100 = \mathbf{45.06\%}$ Check: $32.37 + 22.57 + 45.06 = 100.0\%$
1.3
Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
Step 1: Convert Mass % to Grams (in 100 g sample)
Mass of $Fe = 69.9\text{ g}$, Mass of $O = 30.1\text{ g}$.
Step 2: Calculate Moles
• $\text{Moles of } Fe = \frac{69.9\text{ g}}{55.85\text{ g mol}^{-1}} = 1.2516\text{ mol}$
• $\text{Moles of } O = \frac{30.1\text{ g}}{16.00\text{ g mol}^{-1}} = 1.8813\text{ mol}$
Calculate the amount of carbon dioxide that could be produced when: (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.
(i) 1 mole C burnt in excess air
Since air is in excess, all $1\text{ mol of } C$ forms $1\text{ mol of } CO_2 = \mathbf{44\text{ g } CO_2}$.
(ii) 1 mole C burnt in 16 g O₂
Moles of $O_2 = \frac{16\text{ g}}{32\text{ g mol}^{-1}} = 0.5\text{ mol}$.
$1\text{ mol } C$ requires $1\text{ mol } O_2$. Here only $0.5\text{ mol } O_2$ is present $\to$ $O_2$ is the limiting reagent.
$\text{Moles of } CO_2 = 0.5\text{ mol} \implies 0.5 \times 44\text{ g} = \mathbf{22\text{ g } CO_2}$.
(iii) 2 moles C burnt in 16 g O₂
Again, $O_2 = 0.5\text{ mol}$ is the limiting reagent (only $0.5\text{ mol } C$ reacts, $1.5\text{ mol } C$ remains unreacted).
$\text{Mass of } CO_2\text{ produced} = 0.5 \times 44\text{ g} = \mathbf{22\text{ g } CO_2}$.
1.5
Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol⁻¹.
Step 1: Given Data
$M = 0.375\text{ mol L}^{-1}$, $V = 500\text{ mL} = 0.500\text{ L}$, $\text{Molar mass } M_w = 82.0245\text{ g mol}^{-1}$.
Step 2: Calculate Required Moles
$$\text{Moles of solute } (n) = M \times V = 0.375\text{ mol L}^{-1} \times 0.500\text{ L} = 0.1875\text{ mol}$$ Step 3: Calculate Mass
$$\text{Mass } (m) = n \times M_w = 0.1875\text{ mol} \times 82.0245\text{ g mol}^{-1} = \mathbf{15.380\text{ g}}$$
Mass required: $\mathbf{15.38\text{ g}}$ of $CH_3COONa$.
1.6
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL⁻¹ and the mass per cent of nitric acid in it being 69%.
Step 1: Understand 69% by Mass
In $100\text{ g}$ of nitric acid solution, there are $69\text{ g}$ of pure $HNO_3$.
Step 2: Molar mass of HNO₃
$M(HNO_3) = 1(1.008) + 14.007 + 3(16.00) = 63.015\text{ g mol}^{-1} \approx 63.02\text{ g mol}^{-1}$.
$\text{Moles of } HNO_3 = \frac{69\text{ g}}{63.02\text{ g mol}^{-1}} = 1.0949\text{ mol}$.
Step 3: Volume of 100 g Solution
$$\text{Volume } V = \frac{\text{Mass}}{\text{Density}} = \frac{100\text{ g}}{1.41\text{ g mL}^{-1}} = 70.922\text{ mL} = 0.07092\text{ L}$$ Step 4: Molarity
$$\text{Molarity } (M) = \frac{\text{Moles}}{\text{Volume in L}} = \frac{1.0949\text{ mol}}{0.07092\text{ L}} = \mathbf{15.44\text{ M}}$$
How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Step 1: Molar Mass of CuSO₄
$M(CuSO_4) = 63.55 + 32.06 + 4(16.00) = 63.55 + 32.06 + 64.00 = 159.61\text{ g mol}^{-1}$.
Step 2: Proportional Copper Content
$1\text{ mole of } CuSO_4 (159.61\text{ g})$ contains $1\text{ mole of } Cu (63.55\text{ g})$.
$$\text{Mass of } Cu\text{ in } 100\text{ g } CuSO_4 = \frac{63.55\text{ g}}{159.61\text{ g}} \times 100\text{ g} = \mathbf{39.81\text{ g}}$$
Amount of Copper obtained $= \mathbf{39.81\text{ g}}$
1.8
Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. (Molar mass = 159.69 g mol⁻¹)
Step 1: From Problem 1.3
Empirical formula was determined to be $\mathbf{Fe_2O_3}$.
Step 2: Empirical Formula Mass
$\text{Empirical mass of } Fe_2O_3 = 2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70\text{ g mol}^{-1}$.
Step 3: Factor n
$$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{159.69}{159.70} \approx 1$$
Molecular Formula $= (Fe_2O_3)_1 = \mathbf{Fe_2O_3}$
1.9
Calculate the atomic mass (average) of chlorine using the following data: ³⁵Cl: 75.77% (34.9689 u), ³⁷Cl: 24.23% (36.9659 u).
Average atomic mass of Chlorine $= \mathbf{35.45\text{ u}}$
1.10
In three moles of ethane (C₂H₆), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane.
Step-by-Step Analysis
$1\text{ mole of } C_2H_6$ contains $2\text{ moles of } C$ atoms and $6\text{ moles of } H$ atoms.
(i) $\text{Moles of } C = 3 \times 2 = \mathbf{6\text{ moles of Carbon}}$
(ii) $\text{Moles of } H = 3 \times 6 = \mathbf{18\text{ moles of Hydrogen}}$
(iii) $\text{Molecules of ethane} = 3 \times N_A = 3 \times 6.022 \times 10^{23} = \mathbf{1.807 \times 10^{24}\text{ molecules}}$
1.11
What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L⁻¹ if its 20 g are dissolved in enough water to make a final volume up to 2 L?
Step 1: Molar Mass of Cane Sugar (C₁₂H₂₂O₁₁)
$M = 12(12.011) + 22(1.008) + 11(16.00) = 144.13 + 22.18 + 176.00 = 342.31\text{ g mol}^{-1}$.
Step 2: Moles of Sugar
$$n = \frac{20\text{ g}}{342.31\text{ g mol}^{-1}} = 0.0584\text{ mol}$$ Step 3: Molarity in 2 L
$$\text{Molarity } (M) = \frac{0.0584\text{ mol}}{2\text{ L}} = \mathbf{0.0292\text{ mol L}^{-1}} = \mathbf{0.0292\text{ M}}$$
1.12
If the density of methanol is 0.793 kg L⁻¹, what is its volume needed for making 2.5 L of its 0.25 M solution?
Step 1: Molar Mass of Methanol (CH₃OH)
$M(CH_3OH) = 12.011 + 4(1.008) + 16.00 = 32.04\text{ g mol}^{-1}$.
Step 2: Moles Required for 2.5 L of 0.25 M Solution
$$n = M \times V = 0.25\text{ mol L}^{-1} \times 2.5\text{ L} = 0.625\text{ mol}$$ Step 3: Mass of Methanol Required
$$m = 0.625\text{ mol} \times 32.04\text{ g mol}^{-1} = 20.025\text{ g} = 0.020025\text{ kg}$$ Step 4: Volume Needed (V = m / d)
$$V = \frac{0.020025\text{ kg}}{0.793\text{ kg L}^{-1}} = 0.02525\text{ L} = \mathbf{25.25\text{ mL}}$$
1.13
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is: 1 Pa = 1 N m⁻². If mass of air at sea level is 1034 g cm⁻², calculate the pressure in pascal.
Step 1: Conversion of Units to SI
Mass per unit area $= 1034\text{ g cm}^{-2} = \frac{1034 \times 10^{-3}\text{ kg}}{10^{-4}\text{ m}^2} = 10340\text{ kg m}^{-2}$.
Step 2: Force = Mass × Acceleration due to gravity (g = 9.8 m s⁻²)
$$\text{Force per } m^2 = m \times g = 10340\text{ kg m}^{-2} \times 9.8\text{ m s}^{-2} = 101332\text{ N m}^{-2}$$
SI Unit: The SI base unit of mass is the kilogram (symbol: $kg$).
Official 2019 CGPM Definition:
The kilogram is defined by taking the fixed numerical value of the Planck constant $h$ to be $6.62607015 \times 10^{-34}$ when expressed in the unit $\text{J s}$, which is equal to $\text{kg m}^2\text{ s}^{-1}$, where the metre and the second are defined in terms of $c$ (speed of light) and $\Delta \nu_{Cs}$ (hyperfine transition frequency of Caesium-133).
1.15
Match the following prefixes with their multiples: (i) micro, (ii) deca, (iii) mega, (iv) giga, (v) femto.
Prefix
Symbol
Multiple
(i) micro
$\mu$
$10^{-6}$
(ii) deca
$da$
$10^1$ (or 10)
(iii) mega
$M$
$10^6$
(iv) giga
$G$
$10^9$
(v) femto
$f$
$10^{-15}$
1.16
What do you mean by significant figures?
Definition: Significant figures are the total number of digits in a measured quantity that are known with certainty plus one last digit that is estimated or uncertain.
For instance, in a measurement reported as $11.2\text{ mL}$, the digits '11' are certain while '2' is uncertain with an inherent uncertainty of $\pm 1$ in the last digit.
1.17
A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.
(i) Percent by Mass
$15\text{ ppm} = 15\text{ parts per } 10^6\text{ parts by mass}$.
$$\text{Mass \%} = \frac{15}{10^6} \times 100 = 15 \times 10^{-4}\% = \mathbf{1.5 \times 10^{-3}\%}$$ (ii) Molality Calculation
In $10^6\text{ g}$ of solution, mass of solute $CHCl_3 = 15\text{ g}$.
Mass of solvent $\approx 10^6\text{ g} = 1000\text{ kg}$.
Molar mass of $CHCl_3 = 12.01 + 1.008 + 3(35.45) = 119.36\text{ g mol}^{-1}$.
$$\text{Moles of } CHCl_3 = \frac{15\text{ g}}{119.36\text{ g mol}^{-1}} = 0.1257\text{ mol}$$
$$\text{Molality } (m) = \frac{0.1257\text{ mol}}{1000\text{ kg}} = \mathbf{1.26 \times 10^{-4}\text{ m}}$$
1.18
Express the following in scientific notation: (i) 0.0048, (ii) 234,000, (iii) 8008, (iv) 500.0, (v) 6.0012.
How many significant figures are present in the following? (i) 0.0025, (ii) 208, (iii) 5005, (iv) 126,000, (v) 500.0, (vi) 2.0034.
(i) $0.0025 \to$ 2 (leading zeros non-significant)
(ii) $208 \to$ 3 (captive zero significant)
(iii) $5005 \to$ 4 (captive zeros significant)
(iv) $126,000 \to$ 3 (trailing zeros without decimal are non-significant)
(v) $500.0 \to$ 4 (trailing zeros after decimal are significant)
(vi) $2.0034 \to$ 5 (all digits between non-zeros significant)
1.20
Round up the following upto three significant figures: (i) 34.216, (ii) 10.4107, (iii) 0.04597, (iv) 2808.
(i) $34.216 \to \mathbf{34.2}$ (next digit 1 is $<5$)
(ii) $10.4107 \to \mathbf{10.4}$ (next digit 1 is $<5$)
(iii) $0.04597 \to \mathbf{0.0460}$ (next digit 7 is $>5$, 9 rounds up to 10)
(iv) $2808 \to \mathbf{2810}$ or $\mathbf{2.81 \times 10^3}$ (next digit 8 is $>5$)
1.21
The following data are obtained when dinitrogen and dioxygen react together: (i) 14 g N₂ + 16 g O₂, (ii) 14 g N₂ + 32 g O₂, (iii) 28 g N₂ + 32 g O₂, (iv) 28 g N₂ + 80 g O₂. (a) Which law is obeyed? (b) Fill in the conversion blanks.
(a) Law of Chemical Combination
Fix the mass of dinitrogen to $28\text{ g}$ across all cases:
• Case (i): $14\text{ g } N_2 + 16\text{ g } O_2 \implies 28\text{ g } N_2$ combines with $32\text{ g } O_2$
• Case (ii): $14\text{ g } N_2 + 32\text{ g } O_2 \implies 28\text{ g } N_2$ combines with $64\text{ g } O_2$
• Case (iii): $28\text{ g } N_2$ combines with $32\text{ g } O_2$
• Case (iv): $28\text{ g } N_2$ combines with $80\text{ g } O_2$
Masses of oxygen combining with fixed $28\text{ g } N_2$ are: $32 : 64 : 32 : 80$, which simplifies to $2 : 4 : 2 : 5$ (or $1 : 2 : 1 : 2.5$).
These are small whole numbers $\to$ obeys the Law of Multiple Proportions.
If the speed of light is 3.0 × 10⁸ m s⁻¹, calculate the distance covered by light in 2.00 ns.
Calculation
Time $t = 2.00\text{ ns} = 2.00 \times 10^{-9}\text{ s}$.
Speed $c = 3.0 \times 10^8\text{ m s}^{-1}$.
$$\text{Distance } d = c \times t = (3.0 \times 10^8\text{ m s}^{-1}) \times (2.00 \times 10^{-9}\text{ s}) = \mathbf{0.600\text{ m}} = \mathbf{6.00 \times 10^{-1}\text{ m}}$$
1.23
In a reaction: A + B₂ → AB₂, identify the limiting reagent, if any: (i) 300 atoms A + 200 molecules B₂, (ii) 2 mol A + 3 mol B₂, (iii) 100 atoms A + 100 molecules B₂, (iv) 5 mol A + 2.5 mol B₂, (v) 2.5 mol A + 5 mol B₂.
According to balanced reaction: $1\text{ atom } A$ reacts with $1\text{ molecule } B_2$.
(i) 300 atoms A + 200 molecules B₂ $\to$ B₂ is limiting (only 200 atoms of A will react, 100 remain).
(ii) 2 mol A + 3 mol B₂ $\to$ A is limiting (needs 2 mol B₂, 1 mol B₂ in excess).
(iii) 100 atoms A + 100 molecules B₂ $\to$ No limiting reagent (stoichiometrically equal).
(iv) 5 mol A + 2.5 mol B₂ $\to$ B₂ is limiting (needs 5 mol B₂, only 2.5 mol available).
(v) 2.5 mol A + 5 mol B₂ $\to$ A is limiting (needs 2.5 mol B₂, 2.5 mol B₂ in excess).
1.24
Dinitrogen and dihydrogen react according to: N₂(g) + 3H₂(g) → 2NH₃(g). (i) Calculate mass of NH₃ produced from 2.00 × 10³ g N₂ and 1.00 × 10³ g H₂. (ii) Will any reactant remain unreacted? (iii) If yes, which one and what mass?
Step 1: Calculate Moles of Reactants
• $\text{Moles of } N_2 = \frac{2000\text{ g}}{28.02\text{ g mol}^{-1}} = 71.38\text{ mol}$
• $\text{Moles of } H_2 = \frac{1000\text{ g}}{2.016\text{ g mol}^{-1}} = 496.03\text{ mol}$
Step 2: 5-Second Limiting Reagent Test
• $\text{Index}(N_2) = \frac{71.38}{1} = 71.38$
• $\text{Index}(H_2) = \frac{496.03}{3} = 165.34$
Since $71.38 < 165.34$, $N_2$ is the LIMITING REAGENT!
Step 3: Mass of NH₃ Formed
$1\text{ mol } N_2 \to 2\text{ mol } NH_3$.
$\text{Moles of } NH_3 = 2 \times 71.38 = 142.76\text{ mol}$.
$$\text{Mass of } NH_3 = 142.76\text{ mol} \times 17.03\text{ g mol}^{-1} = \mathbf{2431\text{ g}} = \mathbf{2.43 \times 10^3\text{ g}}$$ Step 4: Unreacted Reactant
$H_2$ remains unreacted.
Moles of $H_2$ consumed $= 3 \times 71.38 = 214.14\text{ mol}$.
Moles of $H_2$ remaining $= 496.03 - 214.14 = 281.89\text{ mol}$.
$$\text{Mass of } H_2\text{ remaining} = 281.89\text{ mol} \times 2.016\text{ g mol}^{-1} = \mathbf{568.3\text{ g}} \approx \mathbf{568\text{ g}}$$
1.25
How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
0.50 mol Na₂CO₃
Represents an amount of substance (mass).
Molar mass of $Na_2CO_3 = 106\text{ g mol}^{-1}$.
$$\text{Mass} = 0.50\text{ mol} \times 106\text{ g mol}^{-1} = \mathbf{53\text{ g}}$$
0.50 M Na₂CO₃
Represents a molar concentration.
It means $0.50\text{ mol}$ ($53\text{ g}$) of $Na_2CO_3$ is dissolved in $1\text{ Litre of solution}$.
1.26
If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Gay Lussac's Volume Ratios
$$2H_2(g) + O_2(g) \to 2H_2O(g)$$
$$2\text{ volumes } H_2 + 1\text{ volume } O_2 \to 2\text{ volumes } H_2O(g)$$
Given: $10\text{ volumes } H_2$ and $5\text{ volumes } O_2$.
Ratio of given volumes is $10:5 = 2:1$, which matches the stoichiometric ratio exactly.
Volume of water vapour produced $= \mathbf{10\text{ volumes}}$
1.27
Convert the following into basic units: (i) 28.7 pm, (ii) 15.15 pm, (iii) 25365 mg.
Basic SI units are metre ($m$) for length and kilogram ($kg$) for mass.
Which one of the following will have the largest number of atoms? (i) 1 g Au(s), (ii) 1 g Na(s), (iii) 1 g Li(s), (iv) 1 g of Cl₂(g).
Formula: Number of atoms = (m / M) × N_A × atomicity
(i) $1\text{ g } Au = \frac{1}{197} \times N_A = 0.00507\, N_A$
(ii) $1\text{ g } Na = \frac{1}{23} \times N_A = 0.04348\, N_A$
(iii) $1\text{ g } Li = \frac{1}{7} \times N_A = \mathbf{0.14286\, N_A}$
(iv) $1\text{ g } Cl_2 = \frac{1}{71} \times 2 \times N_A = \frac{1}{35.5} \times N_A = 0.02817\, N_A$
$$\mathbf{1\text{ g of } Li(s)}$$ has the largest number of atoms because Lithium has the smallest atomic mass!
1.29
Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be 1.0 g mL⁻¹).
Step 1: Understand Mole Fraction
$\chi_{\text{ethanol}} = 0.040 \implies \chi_{\text{water}} = 1 - 0.040 = 0.960$.
In a mixture of $1\text{ mole}$, there is $0.040\text{ mol}$ ethanol and $0.960\text{ mol}$ water.
Step 2: Volume of Solution
Mass of $0.960\text{ mol } H_2O = 0.960 \times 18.02\text{ g} = 17.30\text{ g}$.
With density of water $= 1.0\text{ g mL}^{-1}$, Volume $\approx 17.30\text{ mL} = 0.01730\text{ L}$.
Step 3: Molarity Calculation
$$M = \frac{\text{Moles of ethanol}}{\text{Volume in L}} = \frac{0.040\text{ mol}}{0.01730\text{ L}} = \mathbf{2.31\text{ M}}$$
Molarity $= \mathbf{2.31\text{ mol L}^{-1}}$
1.30
What will be the mass of one ¹²C atom in g?
Calculation
$1\text{ mole of }{}^{12}C\text{ atoms} = 12\text{ g}$ and contains $6.022 \times 10^{23}\text{ atoms}$.
$$\text{Mass of 1 atom} = \frac{12\text{ g}}{6.02214 \times 10^{23}} = \mathbf{1.992648 \times 10^{-23}\text{ g}}$$
1.31
How many significant figures should be present in the answer of the following calculations? (i) (0.02856 × 298.15 × 0.112) / 0.5785, (ii) 5 × 5.364, (iii) 0.0125 + 0.7864 + 0.0215.
(i) In multiplication/division, the term with least sig figs governs the result. $0.112$ has 3 sig figs $\to$ 3 significant figures.
(ii) 5 is an exact counting number (infinite sig figs). $5.364$ has 4 sig figs $\to$ 4 significant figures.
(iii) In addition, decimal places govern. All three numbers have 4 decimal places $\to$ 4 decimal places (4 significant figures).
1.32
Use the data given in the table to calculate the molar mass of naturally occurring argon: ³⁶Ar: 35.96755 (0.337%), ³⁸Ar: 37.96272 (0.063%), ⁴⁰Ar: 39.9624 (99.600%).
Molar mass of Argon $= \mathbf{39.95\text{ g mol}^{-1}}$
1.33
Calculate the number of atoms in each of the following: (i) 52 moles of Ar, (ii) 52 u of He, (iii) 52 g of He.
(i) $52\text{ moles of } Ar = 52 \times 6.022 \times 10^{23} = \mathbf{3.131 \times 10^{25}\text{ atoms of Ar}}$.
(ii) $1\text{ atom of } He = 4\text{ u}$.
$$\text{Number of atoms} = \frac{52\text{ u}}{4\text{ u/atom}} = \mathbf{13\text{ atoms of He}}$$
(iii) $1\text{ mole of } He = 4\text{ g}$.
$$\text{Moles of } He = \frac{52\text{ g}}{4\text{ g mol}^{-1}} = 13\text{ mol}$$
$$\text{Number of atoms} = 13 \times 6.022 \times 10^{23} = \mathbf{7.828 \times 10^{24}\text{ atoms of He}}$$
1.34
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it gives 3.38 g CO₂, 0.690 g H₂O. A volume of 10.0 L (measured at STP) weighs 11.6 g. Calculate: (i) empirical formula, (ii) molar mass, and (iii) molecular formula.
Step 1: Find Masses of C and H
• $\text{Mass of } C = 3.38\text{ g } CO_2 \times \frac{12.011}{44.01} = 0.9224\text{ g } C$
• $\text{Mass of } H = 0.690\text{ g } H_2O \times \frac{2.016}{18.02} = 0.0772\text{ g } H$
Total mass of sample $= 0.9224 + 0.0772 = 0.9996\text{ g} \approx 1.00\text{ g}$.
Step 2: Moles of C and H
• $\text{Moles of } C = \frac{0.9224}{12.011} = 0.0768\text{ mol}$
• $\text{Moles of } H = \frac{0.0772}{1.008} = 0.0766\text{ mol}$
Molar ratio $C : H = 1 : 1$.
(i) Empirical Formula: $\mathbf{CH}$ (Empirical mass $= 13.02\text{ g mol}^{-1}$)
Step 3: Molar Mass from STP Data
$10.0\text{ L}$ at STP weighs $11.6\text{ g}$.
$$\text{Molar mass } (22.4\text{ L}) = \frac{11.6\text{ g}}{10.0\text{ L}} \times 22.4\text{ L} = \mathbf{25.98\text{ g mol}^{-1}} \approx \mathbf{26.0\text{ g mol}^{-1}}$$ Step 4: Molecular Formula
$$n = \frac{\text{Molar mass}}{\text{Empirical mass}} = \frac{26.0}{13.02} = 2$$
Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Step 1: Calculate Moles of HCl
$$n(HCl) = M \times V = 0.75\text{ mol L}^{-1} \times 0.025\text{ L} = 0.01875\text{ mol}$$ Step 2: Stoichiometric Ratio
$2\text{ moles of } HCl$ react with $1\text{ mole of } CaCO_3$.
$$\text{Moles of } CaCO_3\text{ required} = \frac{0.01875}{2} = 0.009375\text{ mol}$$ Step 3: Mass of CaCO₃
$M(CaCO_3) = 40.08 + 12.01 + 3(16.00) = 100.09\text{ g mol}^{-1}$.
$$\text{Mass} = 0.009375\text{ mol} \times 100.09\text{ g mol}^{-1} = \mathbf{0.938\text{ g}}$$
Mass of $CaCO_3$ required $= \mathbf{0.94\text{ g}}$ (or $0.938\text{ g}$)
1.36
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid: 4HCl(aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g). How many grams of HCl react with 5.0 g of manganese dioxide?
Step 2: Stoichiometry from Balanced Equation
$1\text{ mole of } MnO_2 (86.94\text{ g})$ reacts with $4\text{ moles of } HCl (4 \times 36.46 = 145.84\text{ g})$.
Step 3: Calculation for 5.0 g MnO₂
$$\text{Mass of } HCl = 5.0\text{ g } MnO_2 \times \frac{145.84\text{ g } HCl}{86.94\text{ g } MnO_2} = \mathbf{8.387\text{ g}}$$
L3.1
Naturally occurring boron consists of two isotopes: ¹⁰B (atomic mass = 10.01 u) and ¹¹B (atomic mass = 11.01 u). If the average atomic mass of boron is 10.81 u, the percentage abundance of ¹⁰B is:
L3.2
A 3.0 M aqueous solution of NaCl has a density of 1.25 g mL⁻¹. The molality of the solution is approximately:
Incorrect: Underestimated solvent mass.
Correct Answer: 1 L solution mass = 1250 g. Solute mass = 3 × 58.5 = 175.5 g. Solvent mass = 1250 - 175.5 = 1074.5 g = 1.0745 kg. Molality = 3 / 1.0745 = 2.79 m.
Incorrect: Molality equals molarity only when density is ~1 g/mL with negligible solute volume.
Incorrect: Overestimated molality.
L3.3
Equal masses of methane (CH₄) and oxygen (O₂) are mixed in an empty container at 25 °C. The fraction of total pressure exerted by oxygen is:
Incorrect: Assumes equal moles, but molar masses differ (16 vs 32).
Correct Answer: Let mass = w. Moles CH₄ = w/16 = 2w/32. Moles O₂ = w/32. Total moles = 3w/32. Mole fraction O₂ = (w/32) / (3w/32) = 1/3. Hence PO₂ = (1/3) Ptotal.
Incorrect: 2/3 is the fraction of pressure exerted by methane.
Incorrect: Ratio of masses was mistaken.
L3.4
If Avogadro's number NA is changed from 6.022 × 10²³ mol⁻¹ to 6.022 × 10²⁰ mol⁻¹, this would change: