Mix Example - ATOMS AND MOLECULES Questions in English

(A) The molar mass of water $(H_2O)$ is $(2 \times 1) + 16 = 18 \,g/mol$.
Number of moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{360 \,g}{18 \,g/mol} = 20 \,moles$.
Thus,statement $(ii)$ is correct.
Number of molecules = $\text{Number of moles} \times \text{Avogadro's number} (N_A)$.
Number of molecules = $20 \times 6.022 \times 10^{23} = 1.2044 \times 10^{25}$ molecules.
Thus,statement $(iv)$ is correct.
Therefore,the correct options are $(ii)$ and $(iv)$.

(B) The statement '$Atoms$ are not able to exist independently' is incorrect. While most atoms are highly reactive and do not exist in a free state, the atoms of noble gases (inert gases) like $Helium$, $Neon$, $Argon$, etc., are chemically inert and exist independently as single atoms.

(C) Nitrogen gas exists as a diatomic molecule in nature. Therefore,its chemical formula is represented as $N_2$.

(D) The chemical symbol for sodium is derived from its Latin name,$Natrium$. Therefore,the correct symbol is $Na$.

(A) To determine which weighs the most,we calculate the mass of each substance using the formula: $\text{Mass} = \text{Number of moles} \times \text{Molar mass}$.
$(a)$ Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \, g/mol$. Mass of $2$ moles $= 2 \times 100 = 200 \, g$.
$(b)$ Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \, g/mol$. Mass of $2$ moles $= 2 \times 44 = 88 \, g$.
$(c)$ Molar mass of sucrose $(C_{12}H_{22}O_{11}) = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \, g/mol$. Mass of $0.2$ moles $= 0.2 \times 342 = 68.4 \, g$.
$(d)$ Molar mass of $H_2O = (2 \times 1) + 16 = 18 \, g/mol$. Mass of $10$ moles $= 10 \times 18 = 180 \, g$.
Comparing the values: $200 \, g > 180 \, g > 88 \, g > 68.4 \, g$. Therefore,$2$ moles of $CaCO_3$ weigh the highest.

(B) The number of atoms is calculated using the formula: $\text{Number of atoms} = \frac{\text{mass of substance}}{\text{molar mass}} \times \text{number of atoms in a molecule} \times N_A$.
$(a)$ For $18 \, g$ of $H_2O$: $\text{Molar mass} = 18 \, g/mol$. Atoms per molecule $= 3$. $\text{Atoms} = \frac{18}{18} \times 3 \times N_A = 3 N_A$.
$(b)$ For $18 \, g$ of $CH_4$: $\text{Molar mass} = 16 \, g/mol$. Atoms per molecule $= 5$. $\text{Atoms} = \frac{18}{16} \times 5 \times N_A = 5.625 N_A$.
$(c)$ For $18 \, g$ of $CO_2$: $\text{Molar mass} = 44 \, g/mol$. Atoms per molecule $= 3$. $\text{Atoms} = \frac{18}{44} \times 3 \times N_A \approx 1.23 N_A$.
$(d)$ For $18 \, g$ of $O_2$: $\text{Molar mass} = 32 \, g/mol$. Atoms per molecule $= 2$. $\text{Atoms} = \frac{18}{32} \times 2 \times N_A = 1.125 N_A$.
Comparing the values,$5.625 N_A$ is the maximum. Thus,$18 \, g$ of $CH_4$ has the maximum number of atoms.

(C) The number of molecules is calculated using the formula: $\text{Number of molecules} = \frac{\text{mass of substance}}{\text{molar mass}} \times N_A$.
For $1\,g$ of $H_2$: $\text{Molar mass} = 2\,g/mol$. $\text{Number of molecules} = \frac{1}{2} \times N_A = 0.5 N_A$.
For $1\,g$ of $CO_2$: $\text{Molar mass} = 44\,g/mol$. $\text{Number of molecules} = \frac{1}{44} \times N_A \approx 0.0227 N_A$.
For $1\,g$ of $N_2$: $\text{Molar mass} = 28\,g/mol$. $\text{Number of molecules} = \frac{1}{28} \times N_A \approx 0.0357 N_A$.
For $1\,g$ of $CH_4$: $\text{Molar mass} = 16\,g/mol$. $\text{Number of molecules} = \frac{1}{16} \times N_A = 0.0625 N_A$.
Comparing the values,$0.5 N_A$ is the largest. Therefore,$1\,g$ of $H_2$ contains the maximum number of molecules.

(D) The mass of one atom of an element is calculated by dividing its atomic mass by the Avogadro constant $(N_A)$.
The atomic mass of oxygen is $16 \, u$.
The Avogadro constant $(N_A)$ is approximately $6.023 \times 10^{23} \, \text{mol}^{-1}$.
Therefore,the mass of one atom of oxygen $= \frac{\text{Atomic mass}}{N_A} = \frac{16}{6.023 \times 10^{23}} \, g$.

(A) The formula for the number of moles is: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$.
The molar mass of sucrose $(C_{12}H_{22}O_{11})$ is $342 \, g/mol$.
The number of moles of sucrose $= \frac{3.42 \, g}{342 \, g/mol} = 0.01 \, mol$.
Each molecule of sucrose contains $11$ oxygen atoms.
Therefore,the number of oxygen atoms from sucrose $= 0.01 \times 11 \times N_A = 0.11 \, N_A$.
The molar mass of water $(H_2O)$ is $18 \, g/mol$.
The number of moles of water $= \frac{18 \, g}{18 \, g/mol} = 1 \, mol$.
Each molecule of water contains $1$ oxygen atom.
Therefore,the number of oxygen atoms from water $= 1 \times 1 \times N_A = 1 \, N_A$.
Total number of oxygen atoms $= 0.11 \, N_A + 1 \, N_A = 1.11 \, N_A$.
Using $N_A = 6.022 \times 10^{23}$,the total number of oxygen atoms $= 1.11 \times 6.022 \times 10^{23} = 6.68442 \times 10^{23} \approx 6.68 \times 10^{23}$.

(B) change in the physical state of matter involves the absorption or release of energy.
When a substance changes from a solid to a liquid (melting) or from a liquid to a gas (vaporization),energy is absorbed by the system.
Conversely,when a substance changes from a gas to a liquid (condensation) or from a liquid to a solid (freezing),energy is released by the system.
Therefore,a change in physical state occurs when energy is either given to or taken out from the system.

(C) The correct chemical formula is $BiPO_4$ (Bismuth phosphate).
In $BiPO_4$,Bismuth $(Bi^{3+})$ and Phosphate $(PO_4^{3-})$ are both trivalent,so they combine in a $1:1$ ratio.
In option $(A)$,Calcium $(Ca^{2+})$ is bivalent and Chlorine $(Cl^-)$ is monovalent,so the correct formula should be $CaCl_2$.
In option $(B)$,Sodium $(Na^+)$ is monovalent and Sulphate $(SO_4^{2-})$ is bivalent,so the correct formula should be $Na_2SO_4$.
In option $(D)$,Sodium $(Na^+)$ is monovalent and Sulphur $(S^{2-})$ is bivalent,so the correct formula should be $Na_2S$.

(N/A) Copper $(II)$ bromide: The valency of $Cu$ is $2$ and $Br$ is $1$. Thus,the formula is $CuBr_2$.
$(b)$ Aluminium $(III)$ nitrate: The valency of $Al$ is $3$ and $NO_3$ is $1$. Thus,the formula is $Al(NO_3)_3$.
$(c)$ Calcium $(II)$ phosphate: The valency of $Ca$ is $2$ and $PO_4$ is $3$. Thus,the formula is $Ca_3(PO_4)_2$.
$(d)$ Iron $(III)$ sulphide: The valency of $Fe$ is $3$ and $S$ is $2$. Thus,the formula is $Fe_2S_3$.
$(e)$ Mercury $(II)$ chloride: The valency of $Hg$ is $2$ and $Cl$ is $1$. Thus,the formula is $HgCl_2$.
$(f)$ Magnesium $(II)$ acetate: The valency of $Mg$ is $2$ and $CH_3COO$ is $1$. Thus,the formula is $Mg(CH_3COO)_2$.

(N/A) To form neutral compounds,the total positive charge must equal the total negative charge.
$1$. For $Cu^{2+}$:
- With $Cl^{-}$: $CuCl_{2}$
- With $SO_{4}^{2-}$: $CuSO_{4}$
- With $PO_{4}^{3-}$: $Cu_{3}(PO_{4})_{2}$
$2$. For $Na^{+}$:
- With $Cl^{-}$: $NaCl$
- With $SO_{4}^{2-}$: $Na_{2}SO_{4}$
- With $PO_{4}^{3-}$: $Na_{3}PO_{4}$
$3$. For $Fe^{3+}$:
- With $Cl^{-}$: $FeCl_{3}$
- With $SO_{4}^{2-}$: $Fe_{2}(SO_{4})_{3}$
- With $PO_{4}^{3-}$: $FePO_{4}$

(N/A) To identify the ions, we look at the dissociation of ionic compounds in an aqueous solution. Covalent compounds do not dissociate into ions.

Compound	Cation	Anion
$(a)$ $CH_3COONa$	$Na^+$	$CH_3COO^-$
$(b)$ $NaCl$	$Na^+$	$Cl^-$
$(c)$ $H_2$	None (Covalent)	None (Covalent)
$(d)$ $NH_4NO_3$	$NH_4^+$	$NO_3^-$

(N/A) To determine the chemical formula,we use the valency of each element:
$(a)$ Calcium $(Ca^{2+})$ and Fluorine $(F^{-})$: The formula is $CaF_{2}$.
$(b)$ Hydrogen $(H^{+})$ and Sulphur $(S^{2-})$: The formula is $H_{2}S$.
$(c)$ Nitrogen $(N^{3-})$ and Hydrogen $(H^{+})$: The formula is $NH_{3}$.
$(d)$ Carbon $(C^{4+})$ and Chlorine $(Cl^{-})$: The formula is $CCl_{4}$.
$(e)$ Sodium $(Na^{+})$ and Oxygen $(O^{2-})$: The formula is $Na_{2}O$.
$(f)$ Carbon $(C^{4+})$ and Oxygen $(O^{2-})$: Carbon can form $CO$ (Carbon monoxide) and $CO_{2}$ (Carbon dioxide) depending on the conditions.

(A, B, C, E) Incorrect,the correct symbol of cobalt is $Co$. The first letter must be capitalized and the second letter must be lowercase.
$(b)$ Incorrect,the correct symbol of carbon is $C$. Symbols consisting of a single letter must be capitalized.
$(c)$ Incorrect,the correct symbol of aluminium is $Al$. The first letter must be capitalized and the second letter must be lowercase.
$(d)$ Correct,the symbol for helium is $He$.
$(e)$ Incorrect,the correct symbol of sodium is $Na$. The symbol is derived from its Latin name,$Natrium$.

(N/A) To find the ratio by mass,we divide the total mass of each element in the formula by the atomic mass of the elements.
$(a)$ Ammonia $(NH_3)$: Atomic masses are $N=14$,$H=1$. Ratio $N:H = 14 : (3 \times 1) = 14:3$.
$(b)$ Carbon monoxide $(CO)$: Atomic masses are $C=12$,$O=16$. Ratio $C:O = 12:16 = 3:4$.
$(c)$ Hydrogen chloride $(HCl)$: Atomic masses are $H=1$,$Cl=35.5$. Ratio $H:Cl = 1:35.5 = 2:71$.
$(d)$ Aluminium fluoride $(AlF_3)$: Atomic masses are $Al=27$,$F=19$. Ratio $Al:F = 27 : (3 \times 19) = 27:57 = 9:19$.
$(e)$ Magnesium sulphide $(MgS)$: Atomic masses are $Mg=24$,$S=32$. Ratio $Mg:S = 24:32 = 3:4$.

(A) $CO_{3}^{2-}$ consists of $1$ carbon atom and $3$ oxygen atoms,totaling $1 + 3 = 4$ atoms.
$(b)$ $PO_{4}^{3-}$ consists of $1$ phosphorus atom and $4$ oxygen atoms,totaling $1 + 4 = 5$ atoms.
$(c)$ $P_{2}O_{5}$ consists of $2$ phosphorus atoms and $5$ oxygen atoms,totaling $2 + 5 = 7$ atoms.
$(d)$ $CO$ consists of $1$ carbon atom and $1$ oxygen atom,totaling $1 + 1 = 2$ atoms.

(C) The molar mass of water $(H_2O)$ is $18 \, g/mol$.
The mass of one molecule of water is $\frac{18}{N_A} \, g$,where $N_A$ is the Avogadro number.
In one molecule of water $(H_2O)$,there are $2$ hydrogen atoms and $1$ oxygen atom.
Hydrogen $(^1H)$ has $0$ neutrons,while Oxygen $(^{16}O)$ has $8$ neutrons.
Therefore,the total number of neutrons in one molecule of water is $8$.
The mass of $8$ neutrons is $\frac{8}{N_A} \, g$.
The fraction of the mass of water due to neutrons is calculated as: $\frac{\text{Mass of neutrons}}{\text{Total mass of water}} = \frac{8/N_A}{18/N_A} = \frac{8}{18}$.

(N/A) Yes,the solubility of a substance is a temperature-dependent property.
Generally,the solubility of solids in liquids increases with an increase in temperature.
For example,you can dissolve a larger amount of sugar in hot water compared to the same volume of cold water because the kinetic energy of the solvent molecules increases,facilitating the breakdown of the solute.

(N/A) Atomicity is the total number of atoms present in one molecule of an element or compound.
$(a) F_2$: Diatomic ($2$ atoms)
$(b) NO_2$: Triatomic ($3$ atoms)
$(c) N_2O$: Triatomic ($3$ atoms)
$(d) C_2H_6$: Octatomic ($8$ atoms)
$(e) P_4$: Tetratomic ($4$ atoms)
$(f) H_2O_2$: Tetratomic ($4$ atoms)
$(g) P_4O_{10}$: Polyatomic ($14$ atoms)
$(h) O_3$: Triatomic ($3$ atoms)
$(i) HCl$: Diatomic ($2$ atoms)
$(j) CH_4$: Pentatomic ($5$ atoms)
$(k) He$: Monoatomic ($1$ atom)
$(l) Ag$: Polyatomic (Metals exist as a giant lattice structure containing a large number of atoms).

(C) $1$. Heating: Sugar is an organic compound $(C_{12}H_{22}O_{11})$ that chars (turns black) upon heating due to dehydration. Salt $(NaCl)$ is an inorganic ionic compound that does not char upon heating.
$2$. Electrical Conductivity: Salt is an ionic compound that dissociates into ions ($Na^+$ and $Cl^-$) in water,making the solution a good conductor of electricity. Sugar is a covalent compound that does not dissociate into ions in water,making the solution a non-conductor of electricity.

(C) The number of moles is calculated using the formula: $\text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}}$.
Given mass of magnesium = $12 \, g$.
Molar mass of magnesium = $24 \, g \, mol^{-1}$.
Therefore,$\text{Number of moles} = \frac{12}{24} = 0.5 \, mol$.

(D) Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \, g \, mol^{-1}$.
Mass of $5 \, mol$ of $CO_2 = 5 \times 44 = 220 \, g$.
Molar mass of $H_2O = (2 \times 1) + 16 = 18 \, g \, mol^{-1}$.
Mass of $5 \, mol$ of $H_2O = 5 \times 18 = 90 \, g$.
Since $220 \, g \neq 90 \, g$,they do not have the same mass.
$(b)$ Molar mass of $Ca = 40 \, g \, mol^{-1}$.
Moles of $Ca = \frac{240 \, g}{40 \, g \, mol^{-1}} = 6 \, mol$.
Molar mass of $Mg = 24 \, g \, mol^{-1}$.
Moles of $Mg = \frac{240 \, g}{24 \, g \, mol^{-1}} = 10 \, mol$.
Ratio of moles of $Ca$ to $Mg = 6:10 = 3:5$.

(N/A) $Ca : C : O_{3} = 40 : 12 : (16 \times 3) = 40 : 12 : 48 = 10 : 3 : 12$
$(b)$ $Mg : Cl_{2} = 24 : (35.5 \times 2) = 24 : 71$
$(c)$ $H_{2} : S : O_{4} = (1 \times 2) : 32 : (16 \times 4) = 2 : 32 : 64 = 1 : 16 : 32$
$(d)$ $C_{2} : H_{6} : O = (12 \times 2) : (1 \times 6) : 16 = 24 : 6 : 16 = 12 : 3 : 8$
$(e)$ $N : H_{3} = 14 : (1 \times 3) = 14 : 3$
$(f)$ $Ca : O_{2} : H_{2} = 40 : (16 \times 2) : (1 \times 2) = 40 : 32 : 2 = 20 : 16 : 1$

(B) The molar mass of $CaCl_{2} = 40 + (2 \times 35.5) = 111 \, g/mol$.
Given mass of $CaCl_{2} = 222 \, g$.
Number of moles of $CaCl_{2} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{222}{111} = 2 \, mol$.
According to the dissociation equation $CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}$,one formula unit of $CaCl_{2}$ produces $3$ ions ($1$ calcium ion and $2$ chloride ions).
Therefore,$1 \, mol$ of $CaCl_{2}$ produces $3 \, mol$ of ions.
$2 \, mol$ of $CaCl_{2}$ will produce $2 \times 3 = 6 \, mol$ of ions.
Total number of ions = $\text{Number of moles} \times \text{Avogadro number} (N_{A})$.
Total number of ions = $6 \times 6.022 \times 10^{23} = 36.132 \times 10^{23} = 3.6132 \times 10^{24}$ ions.

(C) sodium atom $(Na)$ and a sodium ion $(Na^+)$ differ by exactly one electron.
For $100 \, mol$ of sodium atoms and $100 \, mol$ of sodium ions, the difference in mass is due to $100 \, mol$ of electrons.
Given, mass of $100 \, mol$ of electrons $= 5.48002 \, g$.
Mass of $1 \, mol$ of electrons $= \frac{5.48002}{100} \, g = 0.0548002 \, g$.
Since $1 \, mol$ contains $6.022 \times 10^{23}$ electrons (Avogadro's number), the mass of one electron is calculated as:
Mass of one electron $= \frac{0.0548002 \, g}{6.022 \times 10^{23}} \approx 9.108 \times 10^{-26} \, g$.
Wait, recalculating: $0.0548002 / 6.022 \approx 0.0091 \times 10^{-23} = 9.1 \times 10^{-26} \, g$ is incorrect. Let's re-evaluate: $5.48002 \, g / 100 = 0.0548002 \, g$. Dividing by $6.022 \times 10^{23}$ gives $9.1 \times 10^{-26} \, g$. Converting to $kg$: $9.1 \times 10^{-29} \, kg$.
Correction: The standard mass of an electron is $9.1 \times 10^{-31} \, kg$. The provided value $5.48002 \, g$ for $100 \, mol$ is consistent with $9.1 \times 10^{-31} \, kg$ because $100 \times 6.022 \times 10^{23} \times 9.1 \times 10^{-31} \, kg = 5.48002 \times 10^{-2} \, kg = 5.48002 \, g$.

(D) $1$. Calculate the molar mass of $HgS$: $M(HgS) = 200.6 + 32 = 232.6 \,g \,mol^{-1}$.
$2$. Determine the mass fraction of mercury in $HgS$: $\text{Mass fraction of } Hg = \frac{200.6}{232.6}$.
$3$. Calculate the mass of mercury in $225 \,g$ of $HgS$: $\text{Mass of } Hg = \left( \frac{200.6}{232.6} \right) \times 225 \,g = 194.04 \,g$.

(A) The mass of one mole of screws is calculated by multiplying the mass of one screw by Avogadro's number $(N_A = 6.022 \times 10^{23})$.
Mass of $1 \text{ mole of screws} = 4.11 \, g \times 6.022 \times 10^{23} \approx 2.475 \times 10^{24} \, g$.
Converting this to kilograms: $2.475 \times 10^{24} \, g = 2.475 \times 10^{21} \, kg$.
Now,compare this with the mass of the Earth $(5.98 \times 10^{24} \, kg)$:
Ratio $= \frac{\text{Mass of the Earth}}{\text{Mass of 1 mole of screws}} = \frac{5.98 \times 10^{24} \, kg}{2.475 \times 10^{21} \, kg} \approx 2.416 \times 10^3 \approx 2400$.
Thus,the Earth is approximately $2400$ times heavier than one mole of steel screws.

(B) We know that $1 \, mol$ of any substance contains $6.022 \times 10^{23}$ particles (Avogadro's number).
To find the number of moles of oxygen atoms,we use the formula:
$\text{Number of moles} = \frac{\text{Given number of atoms}}{\text{Avogadro's number}}$
$\text{Number of moles} = \frac{2.58 \times 10^{24}}{6.022 \times 10^{23}}$
$\text{Number of moles} = \frac{25.8 \times 10^{23}}{6.022 \times 10^{23}} \approx 4.28 \, mol$
Therefore,the sample contains $4.28 \, mol$ of oxygen atoms.

(A) The mass of sodium atoms taken by Krish $= (5 \times 23) \, g = 115 \, g$.
The mass of carbon atoms taken by Raunak $= (5 \times 12) \, g = 60 \, g$.
Since $115 \, g > 60 \, g$,Krish's container is heavier.
$(b)$ Both containers have the same number of atoms because they contain the same number of moles of atoms ($5 \, mol$ each). The number of atoms is calculated as $n \times N_A$,where $n$ is the number of moles and $N_A$ is Avogadro's constant $(6.022 \times 10^{23} \, mol^{-1})$.

(N/A) To solve this, we use the relations: $\text{Number of particles} = \text{moles} \times 6.022 \times 10^{23}$ and $\text{Mass} = \text{moles} \times \text{Molar mass}$.
$1$. For $CO_2$: Given particles = $3.011 \times 10^{23}$. Moles = $(3.011 \times 10^{23}) / (6.022 \times 10^{23}) = 0.5 \, \text{mol}$. Mass = $0.5 \times 44 \, \text{g/mol} = 22 \, \text{g}$.
$2$. For $Na$ atom: Given mass = $115 \, \text{g}$. Molar mass of $Na = 23 \, \text{g/mol}$. Moles = $115 / 23 = 5 \, \text{mol}$. Particles = $5 \times 6.022 \times 10^{23} = 3.011 \times 10^{24}$.
$3$. For $MgCl_2$: Given moles = $0.5$. Molar mass = $24 + (2 \times 35.5) = 95 \, \text{g/mol}$. Mass = $0.5 \times 95 = 47.5 \, \text{g}$. Particles = $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$.
$4$. For $H_2O$: Given moles = $2$. Particles = $2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$.

(B) To find the number of moles,we divide the total number of entities by Avogadro's number $(N_A = 6.023 \times 10^{23} \, \text{mol}^{-1})$.
Number of moles of stars $= \frac{10^{22}}{6.023 \times 10^{23}}$
Number of moles of stars $= \frac{1}{6.023 \times 10} \approx 0.0166 \, \text{mol}$.

$(a)$ $10^{3}$ is kilo $(k)$
$(b)$ $10^{-1}$ is deci $(d)$
$(c)$ $10^{-2}$ is centi $(c)$
$(d)$ $10^{-6}$ is micro $(\mu)$
$(e)$ $10^{-9}$ is nano $(n)$
$(f)$ $10^{-12}$ is pico $(p)$

To convert units to kilograms $(kg)$,we use the following conversion factors:
$1 \,g = 10^{-3} \,kg$
$1 \,mg = 10^{-6} \,kg$
$(a)$ $5.84 \times 10^{-3} \,mg = 5.84 \times 10^{-3} \times 10^{-6} \,kg = 5.84 \times 10^{-9} \,kg$
$(b)$ $58.34 \,g = 58.34 \times 10^{-3} \,kg = 5.834 \times 10^{-2} \,kg$
$(c)$ $0.584 \,g = 0.584 \times 10^{-3} \,kg = 5.84 \times 10^{-4} \,kg$
$(d)$ $5.873 \times 10^{-21} \,g = 5.873 \times 10^{-21} \times 10^{-3} \,kg = 5.873 \times 10^{-24} \,kg$

(A) $Mg^{2+}$ ion and a $Mg$ atom differ by two electrons.
$10^{3} \, \text{mol}$ of $Mg^{2+}$ ions and $Mg$ atoms would differ by $10^{3} \times 2 \, \text{mol}$ of electrons.
Number of electrons $= 2 \times 10^{3} \times 6.022 \times 10^{23} = 12.044 \times 10^{26} \, \text{electrons}$.
Mass of $2 \times 10^{3} \, \text{mol}$ of electrons $= 2 \times 10^{3} \times 6.022 \times 10^{23} \times 9.1 \times 10^{-31} \, \text{kg}$.
$= 12.044 \times 10^{26} \times 9.1 \times 10^{-31} \, \text{kg}$.
$= 109.6004 \times 10^{-5} \, \text{kg}$.
$= 1.096 \times 10^{-3} \, \text{kg}$.

(B) $(i)$ For $100 \, g$ of $N_2$:
Number of moles $= \frac{100 \, g}{28 \, g/mol} \approx 3.57 \, mol$.
Number of molecules $= 3.57 \times 6.022 \times 10^{23} \approx 21.505 \times 10^{23}$.
Since each $N_2$ molecule has $2$ atoms,total atoms $= 2 \times 21.505 \times 10^{23} = 43.01 \times 10^{23}$.
$(ii)$ For $100 \, g$ of $NH_3$:
Number of moles $= \frac{100 \, g}{17 \, g/mol} \approx 5.88 \, mol$.
Number of molecules $= 5.88 \times 6.022 \times 10^{23} \approx 35.42 \times 10^{23}$.
Since each $NH_3$ molecule has $4$ atoms ($1$ Nitrogen $+ 3$ Hydrogen),total atoms $= 4 \times 35.42 \times 10^{23} = 141.69 \times 10^{23}$.
Comparing the two,$100 \, g$ of $NH_3$ has more atoms.

(A) The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \, g/mol$.
Number of moles of $NaCl = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{5.85 \, g}{58.5 \, g/mol} = 0.1 \, mol$.
Each formula unit of $NaCl$ dissociates into two ions: one $Na^+$ ion and one $Cl^-$ ion.
Therefore,total moles of ions $= 0.1 \, mol \times 2 = 0.2 \, mol$.
Number of ions $= \text{Moles of ions} \times \text{Avogadro constant}$.
Number of ions $= 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$ (approximately $1.2042 \times 10^{23}$ based on standard calculation).
Thus,the correct option is $A$.

(B) The mass of the gold sample is $1 \, g$.
Since the sample contains $90\%$ gold, the mass of gold present is $0.9 \, g$.
The atomic mass of gold $(Au)$ is approximately $197 \, g/mol$.
The number of moles of gold is calculated as: $\text{moles} = \frac{\text{mass}}{\text{atomic mass}} = \frac{0.9 \, g}{197 \, g/mol} \approx 0.004568 \, mol$.
The number of atoms is calculated by multiplying the number of moles by Avogadro's number $(N_A = 6.022 \times 10^{23} \, atoms/mol)$: $\text{Number of atoms} = 0.004568 \times 6.022 \times 10^{23} \approx 2.75 \times 10^{21} \approx 2.77 \times 10^{21}$ atoms.

(N/A) Molecular compounds are formed when atoms of different elements combine in definite proportions to form molecules. Examples include $H_2O$ (water),$NH_3$ (ammonia),and $CO_2$ (carbon dioxide).
Ionic compounds are composed of metals and non-metals containing charged species known as ions. An ion is a charged particle,which can be either positively charged (cation) or negatively charged (anion). Examples include $NaCl$ (sodium chloride) and $CaO$ (calcium oxide).

(A) The molar mass of an aluminium atom is $27 \,g \,mol^{-1}$.
An aluminium atom $(Al)$ loses $3$ electrons to form an aluminium ion $(Al^{3+})$.
For $1 \,mol$ of $Al^{3+}$ ions,$3 \,mol$ of electrons are lost.
The mass of $1 \,mol$ of electrons is calculated by multiplying the mass of one electron by Avogadro's number $(N_A = 6.022 \times 10^{23} \,mol^{-1})$.
Mass of $3 \,mol$ of electrons $= 3 \times (9.1 \times 10^{-28} \,g) \times (6.022 \times 10^{23} \,mol^{-1})$.
$= 27.3 \times 6.022 \times 10^{-5} \,g \,mol^{-1} \approx 164.4 \times 10^{-5} \,g \,mol^{-1} = 0.001644 \,g \,mol^{-1}$.
Since the ion has lost mass (electrons),the aluminium atom is heavier than the aluminium ion by the mass of the lost electrons,which is approximately $0.0016 \,g$.

(B) Mass of silver $= m \, g$
Mass of gold $= \frac{m}{100} \, g$
Number of atoms of silver $= \frac{\text{mass}}{\text{atomic mass}} \times N_A = \frac{m}{108} \times N_A$
Number of atoms of gold $= \frac{\text{mass}}{\text{atomic mass}} \times N_A = \frac{m}{100 \times 197} \times N_A$
Ratio of number of atoms of gold to silver $= \frac{\text{Number of atoms of gold}}{\text{Number of atoms of silver}}$
$= \frac{\frac{m}{100 \times 197} \times N_A}{\frac{m}{108} \times N_A} = \frac{108}{100 \times 197}$
$= \frac{108}{19700} \approx \frac{1}{182.41}$
Thus,the ratio is $1: 182.41$.

(C) The molar mass of methane $(CH_4)$ is $16 \text{ g/mol}$ and ethane $(C_2H_6)$ is $30 \text{ g/mol}$.
The mass of $1.5 \times 10^{20}$ molecules of $CH_4$ is given by: $\text{Mass} = \frac{1.5 \times 10^{20}}{N_A} \times 16 \text{ g}$.
Let $x$ be the number of molecules of $C_2H_6$. The mass of $x$ molecules of $C_2H_6$ is: $\text{Mass} = \frac{x}{N_A} \times 30 \text{ g}$.
Since the masses are equal:
$\frac{1.5 \times 10^{20} \times 16}{N_A} = \frac{x \times 30}{N_A}$.
Solving for $x$:
$x = \frac{1.5 \times 10^{20} \times 16}{30} = \frac{24 \times 10^{20}}{30} = 0.8 \times 10^{20}$ molecules.

(N/A) Law of conservation of mass.
$(b)$ Polyatomic ion.
$(c)$ The formula unit mass of $Ca_3(PO_4)_2$ is calculated as: $(3 \times 40) + 2 \times (31 + 4 \times 16) = 120 + 2 \times (31 + 64) = 120 + 2 \times 95 = 120 + 190 = 310 \text{ u}$.
$(d)$ The formula of sodium carbonate is $Na_2CO_3$ and the formula of ammonium sulphate is $(NH_4)_2SO_4$.

(N/A) The completed crossword puzzle corresponds to the following elements:
$(1)$ Silver
$(2)$ Gold
$(3)$ Iron
$(4)$ Copper
$(5)$ Phosphorus
$(6)$ Mercury
$(7)$ Hydrogen
$(8)$ Lead

(A) The completed crossword puzzle reveals the names of the $11$ elements: Chlorine $(Cl)$,Hydrogen $(H)$,Argon $(Ar)$,Oxygen $(O)$,Xenon $(Xe)$,Nitrogen $(N)$,Helium $(He)$,Fluorine $(F)$,Krypton $(Kr)$,Radon $(Rn)$,and Neon $(Ne)$.
$(b)$ Inert gases (Noble gases) are elements that have a stable electron configuration. From the given crossword,there are $6$ inert gases present: Helium $(He)$,Neon $(Ne)$,Argon $(Ar)$,Krypton $(Kr)$,Xenon $(Xe)$,and Radon $(Rn)$.

(N/A) Caustic potash: $KOH$
Molecular mass: $39 + 16 + 1 = 56 \, g \, mol^{-1}$
$(b)$ Baking soda: $NaHCO_3$
Molecular mass: $23 + 1 + 12 + (3 \times 16) = 84 \, g \, mol^{-1}$
$(c)$ Limestone: $CaCO_3$
Molecular mass: $40 + 12 + (3 \times 16) = 100 \, g \, mol^{-1}$
$(d)$ Caustic soda: $NaOH$
Molecular mass: $23 + 16 + 1 = 40 \, g \, mol^{-1}$
$(e)$ Ethanol: $C_2H_5OH$
Molecular mass: $(2 \times 12) + (6 \times 1) + 16 = 46 \, g \, mol^{-1}$
$(f)$ Common salt: $NaCl$
Molecular mass: $23 + 35.5 = 58.5 \, g \, mol^{-1}$

(D) The balanced chemical equation for photosynthesis is: $6CO_2 + 6H_2O \xrightarrow{\text{Chlorophyll, Sunlight}} C_6H_{12}O_6 + 6O_2$.
From the equation,$1$ mole of glucose $(C_6H_{12}O_6)$ is produced from $6$ moles of water $(H_2O)$.
The molar mass of glucose is $(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180\, g/mol$.
The molar mass of water is $(2 \times 1) + 16 = 18\, g/mol$.
Thus,$180\, g$ of glucose requires $6 \times 18 = 108\, g$ of water.
Therefore,$18\, g$ of glucose requires: $\frac{108}{180} \times 18 = 10.8\, g$ of water.
Given the density of water is $1\, g\, cm^{-3}$,the volume of water consumed is: $\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{10.8\, g}{1\, g\, cm^{-3}} = 10.8\, cm^3$.

(N/A) $(i)$ Cation: $Al^{3+}$ (Aluminium ion)
$(ii)$ Anion: $O^{2-}$ (Oxide ion)

(N/A) Atomicity is defined as the total number of atoms present in one molecule of an element.
$(i)$ The chemical formula of a sulphur molecule is $S_{8}$. Since it contains $8$ atoms of sulphur,its atomicity is $8$ (Polyatomic).
$(ii)$ The chemical formula of a phosphorus molecule is $P_{4}$. Since it contains $4$ atoms of phosphorus,its atomicity is $4$ (Tetraatomic).