NEET Chemistry: Basic Concepts, Atomic Structure and Bonding · Chapter 1

Some Basic Concepts of Chemistry

What to remember

One mole contains 6.022 × 10²³ entities (the Avogadro constant, N_A); the molar mass in g mol⁻¹ is numerically equal to the atomic, molecular or formula mass in u.
Moles are the common currency of every calculation: moles = mass / molar mass = number of particles / N_A = volume at STP / 22.4 L.
The limiting reagent is the reactant that is consumed first; it decides how much product forms.
Molarity (mol L⁻¹) depends on temperature because volume changes; molality (mol kg⁻¹) and mole fraction do not, because they depend only on mass or amount.

Matter and its nature

Matter is anything that has mass and occupies space. It exists in three physical states: solid, liquid and gas.

Solids have a fixed shape and fixed volume. Liquids have a fixed volume but take the shape of the container. Gases have neither fixed shape nor fixed volume.

Matter is classified by composition as shown below.

Class	Meaning	Example
Pure substance: element	One kind of atom only	Hydrogen, iron, oxygen
Pure substance: compound	Two or more elements combined in a fixed ratio by mass	Water, carbon dioxide, sodium chloride
Homogeneous mixture	Uniform composition throughout	Air, sugar solution, brass
Heterogeneous mixture	Non-uniform composition	Sand and water, oil and water

The components of a mixture can be separated by physical methods such as filtration, distillation and crystallisation. The components of a compound can be separated only by chemical methods.

Properties are of two kinds. Physical properties (colour, melting point, density) can be measured without changing the identity of the substance. Chemical properties (acidity, combustibility) involve a change of composition.

Dalton's atomic theory and laws of chemical combination

Dalton's atomic theory states the following.

Matter consists of indivisible atoms.
All atoms of a given element are identical in mass and properties; atoms of different elements differ.
Compounds form when atoms of different elements combine in fixed, simple whole-number ratios.
Chemical reactions involve only the rearrangement of atoms. Atoms are neither created nor destroyed.

The laws of chemical combination are summarised here.

Law	Statement	Example
Conservation of mass	Mass is neither created nor destroyed in a chemical reaction	Mass of reactants = mass of products
Definite proportions	A pure compound always contains the same elements in the same mass ratio	Water always has hydrogen : oxygen = 1 : 8 by mass
Multiple proportions	When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio	CO and CO₂: oxygen combining with 12 g carbon is 16 g and 32 g, ratio 1 : 2
Gay-Lussac's law of gaseous volumes	Gases react in simple whole-number ratios of volumes, at the same temperature and pressure	2 volumes H₂ + 1 volume O₂ → 2 volumes steam
Avogadro's law	Equal volumes of all gases at the same temperature and pressure contain the same number of molecules	1 L H₂ and 1 L N₂ have equal numbers of molecules

Avogadro's idea that gas molecules can be made of more than one atom (such as H₂ and O₂) explained Gay-Lussac's observations.

Atomic and molecular masses

The atomic mass unit (u) is exactly 1/12 of the mass of one carbon-12 atom. 1 u = 1.6605 × 10⁻²⁴ g. The older symbol amu is the same quantity.

The atomic mass of an element is the average over its naturally occurring isotopes, weighted by their abundance.

Average atomic mass = Σ (isotopic mass × fractional abundance).

Worked example 1. Chlorine has two isotopes: chlorine-35 (75%) and chlorine-37 (25%).

Average atomic mass = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 35.5 u.

The molecular mass of a molecule is the sum of the atomic masses of all its atoms. For example, H₂O = 2(1) + 16 = 18 u; glucose C₆H₁₂O₆ = 6(12) + 12(1) + 6(16) = 180 u.

For ionic compounds such as NaCl, which have no discrete molecules, the term formula mass is used: NaCl = 23 + 35.5 = 58.5 u.

Mole concept and molar mass

The mole is the SI unit of amount of substance. One mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of carbon-12. This number is the Avogadro constant, N_A = 6.022 × 10²³ mol⁻¹.

Molar mass is the mass of one mole, in g mol⁻¹. It is numerically equal to the atomic, molecular or formula mass in u.

Quantity	Relation
Number of moles from mass	n = m / M
Number of particles	N = n × N_A
Gas volume at STP	V = n × 22.4 L (0 °C and 1 atm)
Mass of one molecule	M / N_A

Useful reminders:

1 mol of H₂O = 18 g = 6.022 × 10²³ molecules.
1 mol of any gas occupies about 22.4 L at 273 K and 1 atm (STP).
1 mol of CO₂ contains 1 mol of C atoms and 2 mol of O atoms, so 3 mol of atoms in all.

Percentage composition, empirical and molecular formula

Mass percent of an element = (mass of that element in the compound / molar mass of the compound) × 100.

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms. Molecular formula = (empirical formula)ₙ, where n = molar mass / empirical formula mass.

Steps to find the formula:

1. Convert the mass percent of each element to grams (take 100 g).
2. Divide each mass by its atomic mass to get moles.
3. Divide all by the smallest value.
4. Multiply to get whole numbers if needed.
5. Find n from the molar mass.

Worked example 2. A compound has 40.00% C, 6.72% H and 53.28% O, and molar mass 180 g mol⁻¹.

Moles: C = 40.00/12 = 3.33; H = 6.72/1 = 6.72; O = 53.28/16 = 3.33.

Ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O, with mass 30 g mol⁻¹.

n = 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆.

Compound	Molecular formula	Empirical formula
Benzene	C₆H₆	CH
Glucose	C₆H₁₂O₆	CH₂O
Hydrogen peroxide	H₂O₂	HO
Ethane	C₂H₆	CH₃
Water	H₂O	H₂O (same)

Chemical equations, stoichiometry and limiting reagent

A balanced chemical equation has equal numbers of atoms of each element on both sides. The coefficients give the mole ratio of reactants and products. For example, in N₂ + 3H₂ → 2NH₃, 1 mol N₂ reacts with 3 mol H₂ to give 2 mol NH₃. At the same temperature and pressure the volume ratio is also 1 : 3 : 2.

Stoichiometric calculations follow this path: mass (given) → moles (divide by molar mass) → moles of required substance (use coefficients) → mass or volume (multiply by molar mass or 22.4 L).

Mass–volume example. CaCO₃ → CaO + CO₂. Heating 10 g CaCO₃ (100 g mol⁻¹) gives 0.1 mol CO₂, which is 0.1 × 22.4 = 2.24 L at STP.

Limiting reagent. When reactants are not mixed in the exact stoichiometric ratio, the reactant that is used up first is the limiting reagent. It decides the amount of product. The other reactant is in excess.

Worked example 3. 2H₂ + O₂ → 2H₂O. 4 g H₂ and 24 g O₂ are mixed.

Moles: H₂ = 4/2 = 2 mol; O₂ = 24/32 = 0.75 mol.

2 mol H₂ would need 1 mol O₂, but only 0.75 mol is present, so O₂ is the limiting reagent.

Water formed = 2 × 0.75 = 1.5 mol = 1.5 × 18 = 27 g.

H₂ used = 1.5 mol, so 0.5 mol (1 g) H₂ is left over.

Concentration terms

Term	Definition	Unit	Depends on temperature?
Mass percent	(mass of solute / mass of solution) × 100	%	No
Mole fraction (x)	moles of a component / total moles	none	No
Molarity (M)	moles of solute / volume of solution in litres	mol L⁻¹	Yes
Molality (m)	moles of solute / mass of solvent in kg	mol kg⁻¹	No

For a two-component solution, x(solute) + x(solvent) = 1.

Dilution: M₁V₁ = M₂V₂ (moles of solute stay constant).

Worked example 4 (conversions). Concentrated H₂SO₄ is 98% by mass with density 1.8 g mL⁻¹.

Take 1 L of solution: mass = 1800 g. H₂SO₄ present = 0.98 × 1800 = 1764 g = 1764/98 = 18 mol.

Molarity = 18 mol / 1 L = 18 M.

Solvent mass = 1800 − 1764 = 36 g = 0.036 kg, so molality = 18 / 0.036 = 500 mol kg⁻¹.

Mole fraction example. 46 g ethanol (46 g mol⁻¹, 1 mol) and 36 g water (2 mol): x(ethanol) = 1/3, x(water) = 2/3.

Significant figures and SI units

Significant figures are the meaningful digits of a measurement, including one uncertain digit.

All non-zero digits are significant (4.56 has three).
Zeros between non-zero digits are significant (2.005 has four).
Leading zeros are not significant (0.0032 has two).
Trailing zeros after a decimal point are significant (2.00 has three).
In multiplication and division the answer has as many significant figures as the least precise factor. In addition and subtraction the answer has as many decimal places as the least precise number.

The seven SI base units are metre (length), kilogram (mass), second (time), ampere (current), kelvin (temperature), mole (amount of substance) and candela (luminous intensity). Prefixes: milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹), pico (10⁻¹²), kilo (10³), mega (10⁶). Density in SI is kg m⁻³. Volume is often expressed in litres: 1 L = 10⁻³ m³ = 1 dm³.

Constants used: N_A = 6.022 × 10²³ mol⁻¹; molar volume at STP = 22.4 L mol⁻¹; 1 u = 1.6605 × 10⁻²⁴ g.

NEET traps

1. Molarity changes with temperature; molality does not. Students often reverse this.
2. 1 u is 1/12 of the mass of a carbon-12 atom, not the whole atom.
3. Gay-Lussac's law is about volumes of gases; the law of conservation of mass is about masses.
4. The empirical formula can be the same as the molecular formula (water), but the molecular formula is never smaller than the empirical formula.
5. Atomic mass is a weighted average of isotopes, so it is rarely a whole number (chlorine 35.5 u).
6. 22.4 L is the volume of 1 mol of gas at STP only. Liquids and solids do not follow this.
7. The limiting reagent is decided by moles divided by coefficient, not by the smaller mass.
8. Mole fraction has no unit; mass percent is not the same as mole fraction.
9. Counting atoms: 1 mol of NH₃ has 4 mol of atoms, not 1 mol.
10. In multiplication of measured values, the answer follows the least number of significant figures, not decimal places.

One-liners

1. 1 u = 1/12 of the mass of one carbon-12 atom.
2. N_A = 6.022 × 10²³ mol⁻¹.
3. Molar volume of a gas at STP is 22.4 L.
4. Molar mass of H₂O is 18 g mol⁻¹; of CO₂ it is 44 g mol⁻¹.
5. Law of definite proportions was given by Proust; conservation of mass by Lavoisier.
6. Law of multiple proportions is a consequence of Dalton's theory.
7. Avogadro's law: equal volumes of gases at the same T and P have equal numbers of molecules.
8. Empirical formula of benzene is CH; of glucose is CH₂O.
9. Molality = moles of solute per kilogram of solvent.
10. Molarity = moles of solute per litre of solution.
11. Dilution formula: M₁V₁ = M₂V₂.
12. Sum of mole fractions of all components is 1.
13. SI unit of amount of substance is the mole; of mass it is the kilogram.
14. 2.00 has three significant figures; 0.0020 has two.
15. The reactant that is used up first is the limiting reagent.

Practice questions

Which law states that a pure chemical compound always contains the same elements combined in the same proportion by mass?
1. Gay-Lussac's law of gaseous volumes
2. Law of conservation of mass
3. Law of multiple proportions
4. Law of definite proportions
Answer
D. Law of definite proportions
Proust's law of definite proportions: composition by mass of a pure compound is fixed.
The number of particles present in one mole of a substance is
1. 6.022 × 10²³
2. 6.022 × 10²²
3. 6.022 × 10²⁴
4. 6.626 × 10²³
Answer
A. 6.022 × 10²³
One mole contains the Avogadro number, 6.022 × 10²³, of entities.
The mass of 0.25 mol of H₂SO₄ (molar mass 98 g mol⁻¹) is
1. 49 g
2. 24.5 g
3. 12.25 g
4. 98 g
Answer
B. 24.5 g
Mass = 0.25 × 98 = 24.5 g.
The total number of atoms in 0.5 mol of NH₃ is
1. 0.5 N_A
2. 4 N_A
3. 2 N_A
4. 1.5 N_A
Answer
C. 2 N_A
Each molecule has 4 atoms: 0.5 × 4 = 2 mol of atoms = 2 N_A.
The volume occupied at STP (22.4 L mol⁻¹) by 11 g of CO₂ (molar mass 44 g mol⁻¹) is
1. 2.8 L
2. 5.6 L
3. 11.2 L
4. 22.4 L
Answer
B. 5.6 L
Moles = 11/44 = 0.25; volume = 0.25 × 22.4 = 5.6 L.
A hydrocarbon contains 80% carbon and 20% hydrogen by mass. Its empirical formula is
1. CH₂
2. C₂H₅
3. CH₃
4. CH₄
Answer
C. CH₃
C = 80/12 = 6.67, H = 20/1 = 20; ratio 1 : 3, so CH₃.
The molality of a solution made by dissolving 4 g of NaOH (molar mass 40 g mol⁻¹) in 500 g of water is
1. 0.8 mol kg⁻¹
2. 0.1 mol kg⁻¹
3. 0.4 mol kg⁻¹
4. 0.2 mol kg⁻¹
Answer
D. 0.2 mol kg⁻¹
Moles = 4/40 = 0.1; solvent = 0.5 kg; molality = 0.1/0.5 = 0.2 mol kg⁻¹.
11.7 g of NaCl (molar mass 58.5 g mol⁻¹) is dissolved in water to make 500 mL of solution. The molarity is
1. 0.4 M
2. 0.1 M
3. 0.2 M
4. 0.8 M
Answer
A. 0.4 M
Moles = 11.7/58.5 = 0.2; molarity = 0.2/0.5 L = 0.4 M.
Which of the following concentration terms does not change with temperature?
1. Volume of the solution
2. Molarity
3. Density of the solution
4. Molality
Answer
D. Molality
Molality uses the mass of solvent, which does not change with temperature; the other quantities depend on volume.
An element has two isotopes of mass 10 u (20% abundance) and 11 u (80% abundance). Its average atomic mass is
1. 10.6 u
2. 10.8 u
3. 10.5 u
4. 10.2 u
Answer
B. 10.8 u
(10 × 20 + 11 × 80)/100 = (200 + 880)/100 = 10.8 u.
The number of significant figures in 0.00250 is
1. 2
2. 4
3. 3
4. 5
Answer
C. 3
Leading zeros are not significant; 2, 5 and the trailing 0 are significant, so 3.
The SI unit of amount of substance is
1. mole
2. kelvin
3. gram
4. litre
Answer
A. mole
The mole is the SI base unit of amount of substance.
The number of moles of oxygen atoms in 3 mol of CaCO₃ is
1. 9
2. 3
3. 6
4. 12
Answer
A. 9
Each formula unit has 3 O atoms: 3 × 3 = 9 mol.
The mass of CO₂ produced on complete combustion of 6 g of carbon (C + O₂ → CO₂) is
1. 33 g
2. 44 g
3. 11 g
4. 22 g
Answer
D. 22 g
Moles of C = 6/12 = 0.5; CO₂ = 0.5 mol × 44 = 22 g.
On heating 10 g of CaCO₃ (molar mass 100 g mol⁻¹) completely, the volume of CO₂ released at STP is
1. 4.48 L
2. 2.24 L
3. 22.4 L
4. 1.12 L
Answer
B. 2.24 L
CaCO₃ → CaO + CO₂; 0.1 mol gives 0.1 mol CO₂ = 0.1 × 22.4 = 2.24 L.
The molecular mass of glucose, C₆H₁₂O₆, is
1. 30 u
2. 162 u
3. 180 u
4. 342 u
Answer
C. 180 u
6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 u.
Given below are two statements: Statement I: One mole of a substance contains as many entities as there are atoms in exactly 12 g of carbon-12. Statement II: The molar mass of water is 18 g mol⁻¹. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are standard: the mole definition, and 2(1) + 16 = 18 g mol⁻¹.
Given below are two statements: Statement I: Molarity of a solution changes with temperature. Statement II: Molality of a solution changes with temperature because the mass of solvent changes. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Molarity depends on volume and so changes; molality depends on mass and does not change.
Given below are two statements: Statement I: The empirical formula of a compound is always different from its molecular formula. Statement II: The molecular formula is a whole-number multiple of the empirical formula. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
In water both are H₂O (n = 1), so I is wrong; II is correct.
Given below are two statements: Statement I: Gay-Lussac's law relates the masses of reacting gases. Statement II: One atomic mass unit is the mass of one carbon-12 atom. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Gay-Lussac's law relates volumes; 1 u is 1/12 of the mass of a carbon-12 atom.
Given below are two statements: Statement I: In the reaction 2H₂ + O₂ → 2H₂O, two volumes of hydrogen react with one volume of oxygen at the same temperature and pressure. Statement II: This observation is an example of Gay-Lussac's law of gaseous volumes. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
The volumes combine in the simple ratio 2 : 1, which is Gay-Lussac's law.
Which of the following statements are correct? 1. In CO and CO₂, the masses of oxygen combining with a fixed mass of carbon are in a simple whole-number ratio. 2. The percentage composition of a pure compound depends on the size of the sample. 3. The limiting reagent decides the amount of product formed.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 3 only
1 is the law of multiple proportions (true); 2 is false, composition is independent of sample size; 3 is true.
Given below are two statements: Statement I: Dalton proposed that atoms of different elements combine in fixed whole-number ratios to form compounds. Statement II: Dalton proposed that atoms can be created during a chemical reaction. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Dalton's theory says atoms are neither created nor destroyed in a reaction.
Given below are two statements: Statement I: One mole of H₂O and one mole of CO₂ have the same mass. Statement II: One mole of H₂O and one mole of CO₂ contain the same number of molecules. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Masses are 18 g and 44 g; both contain N_A molecules.
Which of the following statements about significant figures are correct? 1. Zeros between non-zero digits are significant. 2. Leading zeros are significant. 3. 2.00 has three significant figures.
1. 1 and 2 only
2. 2 and 3 only
3. 1, 2 and 3
4. 1 and 3 only
Answer
D. 1 and 3 only
Leading zeros are not significant, so 2 is false; 1 and 3 are true.
Given below are two statements: Statement I: The sum of the mole fractions of all components of a solution is 1. Statement II: Mole fraction is a dimensionless quantity. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Mole fraction = moles of component / total moles; it has no unit and the sum is 1.
Assertion (A): The law of conservation of mass is obeyed in all chemical reactions. Reason (R): Atoms are neither created nor destroyed in a chemical reaction. Choose the correct option.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Rearrangement of atoms with no loss keeps the total mass unchanged, which explains A.
Assertion (A): One mole of H₂ and one mole of O₂ contain the same number of molecules. Reason (R): One mole of H₂ and one mole of O₂ have the same mass. Choose the correct option.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Both contain N_A molecules (A true), but the masses are 2 g and 32 g (R false).
Assertion (A): The molecular mass of a compound can be lower than its empirical formula mass. Reason (R): The molecular formula is a whole-number multiple of the empirical formula. Choose the correct option.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Molecular mass = n × empirical formula mass with n ≥ 1, so A is false; R is true.
Assertion (A): The mass percent of a solute in a solution does not change with temperature. Reason (R): The molarity of a solution depends on its volume. Choose the correct option.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true: mass percent uses masses only; R is true but is not the reason for A.
Assertion (A): The average atomic mass of chlorine is 35.5 u. Reason (R): Natural chlorine is a mixture of two isotopes, of mass 35 u and 37 u, in the ratio of about 3 : 1. Choose the correct option.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The weighted average of 35 (75%) and 37 (25%) is 35.5 u, so R explains A.
Assertion (A): Gay-Lussac's law deals with the masses of gases that react. Reason (R): Gases react in simple whole-number ratios of their volumes. Choose the correct option.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
The law deals with volumes (A false); R is its correct statement.
Match List I with List II. List I: a. Law of conservation of mass b. Law of definite proportions c. Atomic theory of matter d. Law of gaseous volumes. List II: i. Lavoisier ii. Proust iii. Dalton iv. Gay-Lussac
1. a-i, b-iii, c-ii, d-iv
2. a-i, b-ii, c-iii, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
B. a-i, b-ii, c-iii, d-iv
Lavoisier: mass; Proust: definite proportions; Dalton: atomic theory; Gay-Lussac: gaseous volumes.
Match List I with List II. List I: a. Molarity b. Molality c. Molar mass d. Density (SI). List II: i. mol kg⁻¹ ii. mol L⁻¹ iii. g mol⁻¹ iv. kg m⁻³
1. a-ii, b-i, c-iv, d-iii
2. a-iii, b-i, c-ii, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-i, b-ii, c-iii, d-iv
Answer
C. a-ii, b-i, c-iii, d-iv
Molarity is mol per litre, molality is mol per kg of solvent, molar mass is g per mol and SI density is kg m⁻³.
Match List I with List II. List I: a. 2 mol of H₂O b. 0.5 mol of CO₂ c. 0.1 mol of NaOH d. 3 mol of O₂. List II: i. 4 g ii. 22 g iii. 36 g iv. 96 g
1. a-ii, b-iii, c-i, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-iv, b-ii, c-i, d-iii
Answer
B. a-iii, b-ii, c-i, d-iv
2 × 18 = 36 g; 0.5 × 44 = 22 g; 0.1 × 40 = 4 g; 3 × 32 = 96 g.
Match List I with List II. List I: a. 0.0450 b. 4.500 c. 0.0007 d. 100.05. List II (number of significant figures): i. 1 ii. 3 iii. 4 iv. 5
1. a-ii, b-iii, c-iv, d-i
2. a-iii, b-ii, c-i, d-iv
3. a-ii, b-iii, c-i, d-iv
4. a-i, b-iii, c-ii, d-iv
Answer
C. a-ii, b-iii, c-i, d-iv
0.0450 has 3; 4.500 has 4; 0.0007 has 1; 100.05 has 5 significant figures.
Match List I (molecular formula) with List II (empirical formula). List I: a. C₆H₆ b. C₆H₁₂O₆ c. H₂O₂ d. C₂H₆. List II: i. CH₂O ii. CH iii. CH₃ iv. HO
1. a-i, b-ii, c-iv, d-iii
2. a-iv, b-i, c-ii, d-iii
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
D. a-ii, b-i, c-iv, d-iii
Divide each formula by the highest common factor: CH, CH₂O, HO, CH₃.
Match List I with List II. List I: a. 12 g of carbon-12 b. 9 g of water c. 5.6 L of a gas at STP d. 6.022 × 10²² molecules. List II (amount): i. 0.1 mol ii. 0.25 mol iii. 0.5 mol iv. 1 mol
1. a-iv, b-iii, c-i, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iv, b-iii, c-ii, d-i
4. a-iv, b-ii, c-iii, d-i
Answer
C. a-iv, b-iii, c-ii, d-i
12 g C = 1 mol; 9/18 = 0.5 mol; 5.6/22.4 = 0.25 mol; 6.022 × 10²²/6.022 × 10²³ = 0.1 mol.
10 mL of 1 M HCl is diluted with water to a total volume of 100 mL. The molarity of the diluted solution is
1. 0.1 M
2. 0.5 M
3. 0.01 M
4. 1 M
Answer
A. 0.1 M
M₁V₁ = M₂V₂: 1 × 10 = M₂ × 100, so M₂ = 0.1 M.
2.3 g of sodium (23 g mol⁻¹) reacts completely with excess water: 2Na + 2H₂O → 2NaOH + H₂. The volume of H₂ liberated at STP is
1. 0.56 L
2. 1.12 L
3. 2.24 L
4. 11.2 L
Answer
B. 1.12 L
Moles of Na = 0.1; H₂ = 0.05 mol; volume = 0.05 × 22.4 = 1.12 L.
3 mol of H₂ and 2 mol of N₂ react as N₂ + 3H₂ → 2NH₃. The maximum amount of NH₃ formed is
1. 2 mol
2. 6 mol
3. 4 mol
4. 1 mol
Answer
A. 2 mol
3 mol H₂ needs only 1 mol N₂, so H₂ is limiting; NH₃ = 3 × 2/3 = 2 mol.
The mass of Fe₂O₃ (160 g mol⁻¹) formed when 5.6 g of iron (56 g mol⁻¹) burns completely in oxygen (4Fe + 3O₂ → 2Fe₂O₃) is
1. 11.2 g
2. 8 g
3. 4 g
4. 16 g
Answer
B. 8 g
Fe = 0.1 mol; Fe₂O₃ = 0.05 mol; mass = 0.05 × 160 = 8 g.
A hydrochloric acid solution is 36.5% HCl by mass and has density 1.2 g mL⁻¹ (molar mass of HCl 36.5 g mol⁻¹). Its molarity is
1. 10 M
2. 36.5 M
3. 1.2 M
4. 12 M
Answer
D. 12 M
1 L = 1200 g; HCl = 0.365 × 1200 = 438 g = 12 mol; molarity = 12 M.
A solution contains 46 g of ethanol (46 g mol⁻¹) and 90 g of water (18 g mol⁻¹). The mole fraction of ethanol is
1. 1/2
2. 5/6
3. 1/6
4. 1/5
Answer
C. 1/6
Ethanol = 1 mol, water = 5 mol; x = 1/(1 + 5) = 1/6.
A compound has empirical formula CH₂ and molar mass 70 g mol⁻¹. Its molecular formula is
1. C₅H₁₀
2. C₂H₄
3. C₆H₁₂
4. C₄H₈
Answer
A. C₅H₁₀
Empirical mass = 14; n = 70/14 = 5; formula C₅H₁₀.

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1.Sunil made a paste of rice flour and added a drop of iodine solution to the paste. What will he see?
2.The process used for removing gangue from the ore works on the basis of difference of their chemical or physical properties and is called ______ of ores.
3.Which of the following is an example of a third-class lever?
4.Which of the following methods is NOT used for preventing iron from rusting?
5.Which of the following methods is most commonly used nowadays to determine nitrogen and protein content in foods?
6.Which of the following is the reverse of electroplating?
7.Which process is used to get petrol, petroleum gas, diesel, kerosene and fuel oil from petroleum?
8.Which technique is useful for the separation of colloidal particles?
9.Which of the following procedures would you use to separate ammonium chloride from sodium chloride?
10.Which of the following parts are cleaned using ultrasound?

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