Atomic Foundations of Matter Class 9 CBSE Important Questions with Answers

Answer: (b) Joseph Proust.

Proust stated that a pure compound always contains the same elements combined in the same proportion by mass. Lavoisier established the law of conservation of mass, and Dalton explained both laws with his atomic theory.

Answer: (b) Law of constant proportions.

A pure compound always contains the same elements in the same fixed proportion by mass, whatever its source — here hydrogen to oxygen is always $1:8$1:8 in water.

Answer: (c) $100$100 u.

$\text{CaCO}_3=40+12+(3\times16)=40+12+48=100$CaCO₃=40+12+(3×16)=40+12+48=100 u. Calcium carbonate is an ionic compound made of $\text{Ca}^{2+}$Ca^2+ and $\text{CO}_3^{2-}$CO₃^2- ions, so we call this its formula unit mass rather than its molecular mass.

Answer: (b) $\text{XCl}_2$XCl₂.

With $12$12 electrons, X has the configuration $2,8,2$2,8,2. It loses its $2$2 outer electrons to form the ion $\text{X}^{2+}$X^2+, so its valency is $2$2. Each chloride ion is $\text{Cl}^-$Cl^-, so two are needed: $\text{XCl}_2$XCl₂ (X is magnesium, giving $\text{MgCl}_2$MgCl₂).

Answer: (a) Both A and R are true and R is the correct explanation of A. Because atoms are only rearranged and none are created or destroyed, the total mass stays constant — this is exactly the law of conservation of mass.

One atomic mass unit (u) is defined as one-twelfth ($\frac{1}{12}$1/12) of the mass of one atom of carbon-$12$12.

The carbon-$12$12 atom is taken as the standard, and its mass is fixed at exactly $12$12 u; all other atomic masses are expressed relative to it.

Molecular mass of $\text{H}_2\text{SO}_4$H₂SO₄:

$=(2\times1)+(1\times32)+(4\times16)$=(2×1)+(1×32)+(4×16)

$=2+32+64=98\ \text{u}.$=2+32+64=98 u.

We compare the mass ratio of copper to oxygen in both samples.

Experiment 1: copper $=1.28$=1.28 g, so oxygen $=1.60-1.28=0.32$=1.60-1.28=0.32 g.
$\text{Cu}:\text{O}=1.28:0.32=4:1.$Cu:O=1.28:0.32=4:1.

Experiment 2: copper $=2.76$=2.76 g, so oxygen $=3.45-2.76=0.69$=3.45-2.76=0.69 g.
$\text{Cu}:\text{O}=2.76:0.69=4:1.$Cu:O=2.76:0.69=4:1.

In both samples copper and oxygen combine in the same ratio $4:1$4:1 by mass, regardless of the source. This agrees with the law of constant proportions.

Water is a covalent compound made of separate $\text{H}_2\text{O}$H₂O molecules, so the sum of the atomic masses in one molecule is its molecular mass. Sodium chloride is an ionic compound: it has no separate molecules, only a large arrangement of $\text{Na}^+$Na^+ and $\text{Cl}^-$Cl^- ions in a fixed $1:1$1:1 ratio. Its simplest unit, the formula unit $\text{NaCl}$NaCl, is used instead.

$\text{NaCl}=23+35.5=58.5$NaCl=23+35.5=58.5 u.

$\text{K}_2\text{CO}_3=(2\times39)+12+(3\times16)=78+12+48=138$K₂CO₃=(2×39)+12+(3×16)=78+12+48=138 u.

1. Law of conservation of mass: In a chemical reaction, mass can neither be created nor destroyed — the total mass of the products equals the total mass of the reactants. Example: when $12$12 g of carbon burns in $32$32 g of oxygen, exactly $44$44 g of carbon dioxide is formed ($12+32=44$12+32=44).

2. Law of constant (definite) proportions: A pure chemical compound always contains the same elements combined together in the same fixed proportion by mass, whatever its source. Example: pure water from any source always has hydrogen and oxygen in the mass ratio $1:8$1:8.

Dalton's atomic theory (relevant postulates):

1. All matter is made of tiny indivisible particles called atoms.
2. Atoms are neither created nor destroyed in a chemical reaction — this explains the law of conservation of mass.
3. Atoms of a given element are identical in mass and chemical properties, while atoms of different elements differ.
4. Atoms combine in small whole-number ratios to form compounds, and the relative number and kind of atoms in a given compound are fixed — this explains the law of constant proportions.

Criss-cross method: Write the symbols (or ions) of the two combining species side by side, with the valency of each written above it. Then cross over — the valency of the first becomes the subscript of the second and vice-versa — and reduce the subscripts to the simplest whole-number ratio. Polyatomic ions taken more than once are enclosed in brackets.

(i) Aluminium oxide: Al$^{3}$³ and O$^{2}$² → cross over → $\text{Al}_2\text{O}_3$Al₂O₃.

(ii) Calcium chloride: Ca$^{2}$² and Cl$^{1}$¹ → $\text{CaCl}_2$CaCl₂.

(iii) Sodium carbonate: Na$^{1}$¹ and (CO$_3$₃)$^{2}$² → $\text{Na}_2\text{CO}_3$Na₂CO₃.

(iv) Ammonium sulphate: (NH$_4$₄)$^{1}$¹ and (SO$_4$₄)$^{2}$² → $(\text{NH}_4)_2\text{SO}_4$(NH₄)₂SO₄.

(v) Calcium hydroxide: Ca$^{2}$² and (OH)$^{1}$¹ → $\text{Ca(OH)}_2$Ca(OH)₂.

Formula unit mass of aluminium oxide, $\text{Al}_2\text{O}_3$Al₂O₃:
$=(2\times27)+(3\times16)=54+48=102\ \text{u}.$=(2×27)+(3×16)=54+48=102 u.

(i) The law of conservation of mass.

(ii) Mass of reactants $=5.3+6.0=11.3$=5.3+6.0=11.3 g. Mass of products $=2.2+0.9+8.2=11.3$=2.2+0.9+8.2=11.3 g. The two are equal, so mass is neither created nor destroyed.

(iii) One product, carbon dioxide, is a gas. In an open flask it would escape into the air, the products would appear lighter, and the law would seem to fail.

(iv) Atoms can neither be created nor destroyed in a chemical reaction; they are only rearranged, so the total mass stays the same.