Class 9 Science Chapter 3 Atoms and Molecules

Key Concept: Law of Definite Proportions

c) 32 g

[Solution Description] According to the Law of Definite Proportions, elements in a compound are always in a fixed ratio by mass. The first sample has a ratio of 12 g C to 16 g O. Keeping this ratio constant, for 24 g of C, we use the proportion $\frac{12}{16} = \frac{24}{x}$, where $x$ is the mass of oxygen. Solving for $x$, $x = \frac{24 \times 16}{12} = 32$. Hence, 32 g of oxygen would combine with 24 g of carbon.

Key Concept: Comparison of atomic sizes

b) Molecule of water

[Solution Description] To solve this, we look at the given relative sizes: an atom of hydrogen is $10^{-10}$ m, a molecule of water is $10^{-9}$ m, a molecule of haemoglobin is $10^{-8}$ m, a grain of sand is $10^{-4}$ m, an ant is $10^{-3}$ m, and an apple is $10^{-1}$ m. The one closest to $10^{-9}$ m is a molecule of water.

Key Concept: Polyatomic Ions

c) $KNO_3$

[Solution Description] Potassium has a valency of +1 and the nitrate ion $(NO_3)^-$ has a valency of -1. Since their valencies are equal and opposite, they combine in a 1:1 ratio. Hence, the chemical formula for potassium nitrate is $KNO_3$.

Key Concept: Dalton’s Atomic Theory

d) Different masses and different chemical properties

[Solution Description] Dalton’s atomic theory states that atoms of different elements possess different masses and chemical properties. This helps differentiate one element from another based on their unique properties and explains why chemical reactions involve specific changes in substances.

Key Concept: Molecules composed of single atoms

c) Helium

[Solution Description] Monoatomic molecules are those that consist of a single atom. Noble gases, such as helium and argon, are typical examples of elements existing as monoatomic molecules due to their full valence electron shells making them chemically inert. Hence, both helium and argon are monoatomic, but helium is included in the options.

Key Concept: Formula Unit Mass Calculation

c) 208 u

[Solution Description] The formula unit of barium chloride (BaCl$_2$) consists of one atom of barium and two atoms of chlorine. Thus, the formula unit mass is calculated as:

Formula unit mass of BaCl$_2$ = $(1 \times \text{atomic mass of Ba}) + (2 \times \text{atomic mass of Cl})$

Substituting in the given atomic masses:

$\text{Formula unit mass of BaCl}_2 = 1 \times 137\, \text{u} + 2 \times 35.5\, \text{u} = 137\, \text{u} + 71\, \text{u} = 208\, \text{u}$

Hence, the formula unit mass of BaCl$_2$ is 208 u.

Key Concept: Chemical Formula Derivation, Valency, Ionic Compound Analysis

b) 95 $u$

[Solution Description]

The chemical formula of $MgCl_2$ is determined by balancing the charge of Magnesium

($+2$) with two Chlorine ions (each $-1$):

Thus, the formula becomes $MgCl_2$.

Calculate the formula unit mass:

$\text{Formula Unit Mass of } MgCl_2 = (\text{Atomic Mass of } Mg) + 2 \times (\text{Atomic Mass of } Cl)$

$= 24 + 2 \times 35.5 = 24 + 71 = 95 \, u$

This confirms the charge balance and correctness of the chemical formula.

Key Concept: Concept of Atomic Mass

c) 24

[Solution Description]

The relative atomic mass of an element in terms of carbon-12 is calculated by comparing it to $\frac{1}{12}$th the mass of a carbon-12 atom. For magnesium, which consists of isotopes such as $^{24}$Mg, whose atomic mass is approximately 24 u, we can directly use this number as its atomic mass since these values are standardized against carbon-12.

Key Concept: Law of Constant Proportions

c) 9 g

[Solution Description]

Given the mass ratio of nitrogen to hydrogen in ammonia is 14:3, for every 14 g of nitrogen, there are 3 g of hydrogen.

Using the proportionality:

$\frac{42}{y} = \frac{14}{3}$

Solving for $y$, the mass of hydrogen:

$y = \frac{42 \times 3}{14} = 9 \, \text{g}$

Thus, 9 g of hydrogen would be present along with 42 g of nitrogen.

Key Concept: Determining formula with simplified ratios

c) $Ca(NO_3)_2$

Calcium has a valency of 2, shown as $Ca^{2+}$, and nitrate has a valency of 1, shown as $NO_3^-$. Balancing the charges requires two nitrate ions for each calcium ion. Therefore, the formula is $Ca(NO_3)_2$.

Key Concept: Conceptual evolution, historical context

b) They provided a logical basis that encouraged experimentation to classify substances.

[Solution Description] The philosophical ideas proposed by Leucippus and Democritus laid the foundation for thinking about matter at a microscopic level, even though it was devoid of empirical evidence. This groundwork inspired later scientists to investigate the composition of substances. By the eighteenth century, scientific advancements enabled chemists to experimentally distinguish between pure substances (elements) and combinations thereof (compounds), influenced by the notion that matter could be broken down into basic units or atoms.

Key Concept: Molecular Mass Calculation

c) 111 u

[Solution Description] The formula unit mass of $CaCl_2$ can be calculated by adding the atomic masses of its constituent elements. Calcium chloride has one calcium atom and two chlorine atoms. Thus, the calculation is:

$\text{Formula Unit Mass} = (\text{Atomic Mass of Ca}) + 2 \times (\text{Atomic Mass of Cl})$

Substituting the given atomic masses:

$40 \, \text{u} + 2 \times 35.5 \, \text{u} = 40 \, \text{u} + 71 \, \text{u} = 111 \, \text{u}$

Key Concept: Formation of ions

d) Sodium is positive and chloride is negative

[Solution Description]

In a compound like sodium chloride ($NaCl$), the sodium ion ($Na^+$) is positively charged, termed as a ‘cation’, while the chloride ion ($Cl^-$) is negatively charged, termed as an ‘anion’. This charge separation contributes to the electrical neutrality of the compound.

Key Concept: Writing Formula from Valencies

c) $Al_2S_3$

Aluminum ($Al$) has a valency of +3, while sulfur ($S$) has a valency of -2. By using the crossover method, we need two $Al^{3+}$ ions to balance with three $S^{2-}$ ions to neutralize the compound. Therefore, the chemical formula for aluminum sulfide is $Al_2S_3$.

Key Concept: Multi-step Solutions, Law of Constant Proportions with Ancient Views

c) Water maintains a consistent mass ratio of hydrogen to oxygen

[Solution Description] From ancient views, indivisible particles combine to form substances. Using the law of constant proportions, water is always composed of hydrogen and oxygen in a specific mass ratio (1:8 by mass). This implies that regardless of the source, each molecule of water will have the same consistent ratio of hydrogen atoms combining with oxygen atoms, reflecting both ancient concepts of fixed particle combinations and modern chemical understanding.

Key Concept: Non-Straightforward Path, Numerical and Analytical Difficulty

b) 232 pm

To find the distance between the outermost oxygen atoms in a linear CO$_2$ molecule, you add the bond lengths on both sides of carbon. Given each C-O bond is 116 pm, the distance across the entire molecule is $116 \text{ pm} + 116 \text{ pm} = 232 \text{ pm}$.

Key Concept: Comparison of sizes

b) Molecule of water

[Solution Description] The size of a molecule of haemoglobin is $10^{-8} \text{ m}$. Comparing this to other given entities:

– Atom of hydrogen: $10^{-10} \text{ m}$

– Molecule of water: $10^{-9} \text{ m}$

– Grain of sand: $10^{-4} \text{ m}$

The molecule of haemoglobin, being $10^{-8} \text{ m}$, is closest in size to the molecule of water which is $10^{-9} \text{ m}$.

Key Concept: Historical Development of Atomic Theory, Law of Conservation of Mass, Dalton’s Postulates

d) The mass will remain unchanged from the initial mass of reactants

[Solution Description] According to the Law of Conservation of Mass, in a chemical reaction, the total mass of reactants must equal the total mass of products. Since we start with 10 g of hydrogen and excess oxygen, after the formation of water, the mass of all substances within the closed system (vessel) remains constant. Therefore, the total mass inside the vessel should still be the original mass of hydrogen plus the mass of any unreacted oxygen.

Key Concept: Molecular Mass, Complex Molecule

b) 180 u

[Solution Description]

The molecular formula of glucose is $C_6H_{12}O_6$, which means it contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. To find the molecular mass, we sum up the atomic masses of all the atoms:

– Atomic mass of carbon = 12 u

– Atomic mass of hydrogen = 1 u

– Atomic mass of oxygen = 16 u

Now, calculate the total mass:

$\text{Molecular mass of } C_6H_{12}O_6 = (6 \times 12) + (12 \times 1) + (6 \times 16)$

$ = 72 + 12 + 96 = 180 \, \text{u}$

Thus, the molecular mass of glucose is 180 u.

Key Concept: Understanding scale of measurement

d) $10^6$

[Solution Description] Given that the thickness of a sheet of paper is approximately $0.1 \text{ mm}$, which is equal to $0.1 \times 10^{-3} \text{ m} = 10^{-4} \text{ m}$, and an atom of hydrogen measures $10^{-10} \text{ m}$, the number of hydrogen atoms stacked together would be:

$\frac{10^{-4} \text{ m}}{10^{-10} \text{ m}} = 10^{6}$

Hence, approximately $10^6$ or one million hydrogen atoms are needed to match the thickness of a sheet of paper.

Key Concept: Criss-cross method for aluminium chloride

d) AlCl$_3$

[Solution Description] Aluminium has a charge of $\text{Al}^{3+}$ and chlorine has a charge of $\text{Cl}^-$. Using the criss-cross method, the formula becomes: $\text{AlCl}_3$.

Key Concept: Modern Day Symbols

c) K

[Solution Description] Potassium’s symbol comes from its Latin name “kalium.” According to IUPAC naming conventions, the symbol for potassium uses one letter from its Latin name and has been approved as ‘K’.

Key Concept: Molecular Mass Concept Application

c) 98 u

[Solution Description] The molecular mass of $H_2SO_4$ can be found by adding up the atomic masses of all atoms in its chemical formula. The compound contains two hydrogen atoms, one sulfur atom, and four oxygen atoms. Thus, the calculation is:

$\text{Molecular mass of } H_2SO_4 = 2 \times \text{atomic mass of H} + \text{atomic mass of S} + 4 \times \text{atomic mass of O}$

$2 \times 1 + 32 + 4 \times 16 = 2 + 32 + 64 = 98 \, \text{u}$

Key Concept: Conceptual Complexity, Molecular Mass, Formula Unit Mass, Numerical Difficulty

d) 32 u

[Solution Description]

We compute the molecular mass for the structure $CH_3OH$ using the given atomic masses:

$\text{Molecular mass of } CH_3OH = (1 \times 12) + (4 \times 1) + (1 \times 16)$

Break down each part:

$1 \times 12 = 12$

$4 \times 1 = 4$

$1 \times 16 = 16$

Sum these calculations:

$12 + 4 + 16 = 32 \text{ u}$

Thus, the molecular mass of $CH_3OH$ is $32 \text{ u}$.

Key Concept: Molecular Mass, Multi-step Solutions, Real-World Application

c) 180 u

[Solution Description]

In order to compute the molecular mass of aspirin, $C_9H_8O_4$, count the total atomic mass of each constituent atom:

– Carbon ($C$): There are 9 carbon atoms.

– Hydrogen ($H$): There are 8 hydrogen atoms.

– Oxygen ($O$): There are 4 oxygen atoms.

Proceeding with the calculations:

$\text{Molecular mass of } C_9H_8O_4 = (9 \times 12) + (8 \times 1) + (4 \times 16)$

Continuing with simplification:

$ = 108 + 8 + 64$

$ = 180 \text{ u}$

Thus, the molecular mass of acetylsalicylic acid is 180 u.

Key Concept: Significance of Laws in Atomic Theory Development

a) It proposed that atoms are indivisible and combine in simple whole-number ratios.

[Solution Description] Dalton’s atomic theory supported the laws of chemical combination by explaining how atoms interact during chemical reactions. According to his theory, all matter is composed of tiny, indivisible atoms, and each element has identical atoms. This idea reinforced the Law of Conservation of Mass, as atoms rearrange but are never created or destroyed. His theory also explained the Law of Definite Proportions, stating that compounds always contain elements in fixed ratios due to the combination of specific atoms. Additionally, the Law of Multiple Proportions was supported by Dalton’s idea that atoms combine in whole-number ratios, forming different compounds.

Key Concept: Molecular Mass, Calculation Using Unit Mass, Understanding Brackets

a) 149 u

[Solution Description]

In $(NH_4)_3PO_4$, calculate the individual components:

– For $(NH_4)_3$: 3 ammonium ions, each containing 1 nitrogen and 4 hydrogen atoms:

$3 \times (14 + 4 \times 1) = 3 \times 18 = 54 \, u$

– For $PO_4$: 1 phosphorus and 4 oxygen atoms:

$31 + 4 \times 16 = 31 + 64 = 95 \, u$

Therefore, total molecular mass:

$54 \, u + 95 \, u = 149 \, u$

The molecular mass of $(NH_4)_3PO_4$ is $149 \, u$.

Key Concept: Molecular Mass calculation, Real-World Application

b) 180 u

[Solution Description] The molecular mass of glucose can be calculated by adding up the atomic masses of all atoms present in one molecule of glucose. We have:

– Carbon ($C$): 6 atoms, each with an atomic mass of approximately 12 u

– Hydrogen ($H$): 12 atoms, each with an atomic mass of approximately 1 u

– Oxygen ($O$): 6 atoms, each with an atomic mass of approximately 16 u

Thus, the total molecular mass is:

$\text{Molecular mass of } C_6H_{12}O_6 = (6 \times 12) + (12 \times 1) + (6 \times 16)$

Simplifying this:

$ = 72 + 12 + 96$

$ = 180 \, \text{u}$

Therefore, the molecular mass of glucose is 180 u.

Key Concept: Law of Conservation of Mass, Chemical Reaction Observations

c) 50 g since mass is conserved

[Solution Description] According to the law of conservation of mass, the total mass before and after a chemical reaction remains constant as mass can neither be created nor destroyed. Therefore, even though the substances undergo chemical change, their masses are simply rearranged in new compounds or molecules. In this case, the initial mass of the reactants is given as 50 g. Since the system is closed, no mass is lost to the surroundings. Thus, the mass of the products must also be 50 g.

Key Concept: Balancing Charges in Ionic Compounds

b) $Na_3PO_4$

[Solution Description] Sodium ion has a charge of +1 ($Na^+$), while phosphate ion has a charge of -3 ($PO_4^{3-}$). To balance the charges, three sodium ions are needed for each phosphate ion. Hence, the formula is $Na_3PO_4$.

Key Concept: Law of Conservation of Mass

d) 27 g

[Solution Description] According to the law of conservation of mass, the mass of reactants should equal the mass of products in a chemical reaction. Here, the reactants are calcium and chlorine.

Given:

Mass of calcium = 10 g

Mass of chlorine = 17 g

Total mass of reactants = Mass of calcium + Mass of chlorine

$= 10 \text{ g} + 17 \text{ g}$

$= 27 \text{ g}$

Therefore, the total mass of calcium chloride formed should also be 27 g.

Key Concept: Relative Size Comparison

d) Grain of sand

[Solution Description] According to the syllabus, the relative sizes are given as follows:

– Atom of hydrogen: $10^{-10} \text{ m}$

– Molecule of water: $10^{-9} \text{ m}$

– Molecule of haemoglobin: $10^{-8} \text{ m}$

– Grain of sand: $10^{-4} \text{ m}$

Comparing these values, a grain of sand with a size of $10^{-4} \text{ m}$ is the largest among the given options.

Key Concept: Conceptual Complexity, Advanced Reasoning

c) A group of atoms with a net charge

[Solution Description] A polyatomic ion is a charged species consisting of two or more atoms bonded together, which collectively carry a net charge. Unlike simple ions, which are formed from single atoms gaining or losing electrons, polyatomic ions result from the collective charge of multiple atoms. For instance, $SO_4^{2-}$ is a common polyatomic ion where sulfur and oxygen atoms bond to form an ion with a net negative charge.

Key Concept: Identifying the Compound from its Formula

b) Magnesium hydroxide

[Solution Description] The formula $Mg(OH)_2$ represents magnesium hydroxide. Magnesium has a valency of +2 which balances with two hydroxide ions (each having a valency of -1), forming magnesium hydroxide.

Key Concept: Laws of Chemical Combination

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

The assertion is about the Law of Conservation of Mass, which indicates that during any chemical reaction, the total mass of reactants is equal to the total mass of products. This law was established by experimental verification showing that mass remains constant before and after a reaction. The reason is tied to Dalton’s Atomic Theory, particularly the postulate stating that atoms are indivisible and cannot be created or destroyed during chemical reactions. Thus, since the number of each type of atom remains the same throughout a reaction, their combined mass also remains unchanged, directly supporting the Law of Conservation of Mass.