Atomic Foundations of Matter — CBSE Class 9 Science Important Questions

Answer: (b) $6.022\times10^{23}$6.022×10^23.

One mole of any substance contains Avogadro's number, $6.022\times10^{23}$6.022×10^23, of particles (atoms, molecules or ions).

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 g/mol.

$\text{CaCO}_3=40+12+(3\times16)=40+12+48=100$CaCO_3=40+12+(3×16)=40+12+48=100 g/mol.

Answer: (b) $3.011\times10^{23}$3.011×10^23.

Moles of water $=\dfrac{9}{18}=0.5$=9/18=0.5 mol. Number of molecules $=0.5\times6.022\times10^{23}=3.011\times10^{23}$=0.5×6.022×10^23=3.011×10^23.

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 $\text{H}_2\text{SO}_4$H_2SO_4:

$=(2\times1)+(1\times32)+(4\times16)$=(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.

Number of moles:
$n=\frac{\text{given mass}}{\text{molar mass}}=\frac{22}{44}=0.5\ \text{mol}.$n=given mass/molar mass=22/44=0.5 mol.

Number of molecules:
$N=n\times N_A=0.5\times6.022\times10^{23}=3.011\times10^{23}\ \text{molecules}.$N=n× N_A=0.5×6.022×10^23=3.011×10^23 molecules.

The relationship used can be summarised as:

Step 1 — moles from number of molecules:
$n=\frac{\text{number of molecules}}{N_A}=\frac{3.011\times10^{23}}{6.022\times10^{23}}=0.5\ \text{mol}.$n=number of molecules/N_A=3.011×10^236.022×10^23=0.5 mol.

Step 2 — molar mass of $\text{NH}_3$NH_3:
$14+(3\times1)=17\ \text{g/mol}.$14+(3×1)=17 g/mol.

Step 3 — mass:
$\text{mass}=n\times\text{molar mass}=0.5\times17=8.5\ \text{g}.$mass=n×molar mass=0.5×17=8.5 g.

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 → $\text{Al}_2\text{O}_3$Al_2O_3.

(ii) Calcium chloride: Ca$^{2}$^2 and Cl$^{1}$^1 → $\text{CaCl}_2$CaCl_2.

(iii) Sodium carbonate: Na$^{1}$^1 and (CO$_3$_3)$^{2}$^2 → $\text{Na}_2\text{CO}_3$Na_2CO_3.

(iv) Ammonium sulphate: (NH$_4$_4)$^{1}$^1 and (SO$_4$_4)$^{2}$^2 → $(\text{NH}_4)_2\text{SO}_4$(NH_4)_2SO_4.

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

Molar mass of aluminium oxide, $\text{Al}_2\text{O}_3$Al_2O_3:
$=(2\times27)+(3\times16)=54+48=102\ \text{g/mol}.$=(2×27)+(3×16)=54+48=102 g/mol.

(i) One mole is the amount of a substance that contains $6.022\times10^{23}$6.022×10^23 (Avogadro's number) of particles (atoms, molecules or ions).

(ii) Molar mass of $\text{O}_2=2\times16=\mathbf{32}$O_2=2×16=32 g/mol.

(iii) Molar mass of $\text{CO}_2=12+(2\times16)=44$CO_2=12+(2×16)=44 g/mol.
$n=\frac{88}{44}=\mathbf{2}\ \textbf{moles}.$n=88/44=2 moles.

(iv) Number of molecules $=n\times N_A=2\times6.022\times10^{23}=\mathbf{1.2044\times10^{24}}$=n× N_A=2×6.022×10^23=1.2044×10^24 molecules.