The mole concept Questions in English

(C) The density of pure water is approximately $1000 \ g/L$ at $298 \ K$.
The molar mass of water $(H_2O)$ is $18.02 \ g/mol$.
Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$M = \frac{\text{mass}}{\text{molar mass} \times \text{volume in liters}} = \frac{1000 \ g}{18.02 \ g/mol \times 1 \ L} \approx 55.55 \ mol/L$.
Therefore,the molarity of pure water is approximately $55.6 \ M$.

(A) Active mass is defined as molar concentration,which is $\frac{\text{moles}}{\text{volume in liters}}$.
$(a)$ For $N_2$: $\text{Moles} = \frac{7.0 \, g}{28 \, g \, mol^{-1}} = 0.25 \, mol$. Active mass $= \frac{0.25 \, mol}{1 \, L} = 0.25 \, mol \, L^{-1}$.
$(b)$ For $O_2$: $\text{Moles} = \frac{8.0 \, g}{32 \, g \, mol^{-1}} = 0.25 \, mol$. Active mass $= \frac{0.25 \, mol}{1 \, L} = 0.25 \, mol \, L^{-1}$.
$(c)$ For $NH_3$: $\text{Moles} = \frac{34.0 \, g}{17 \, g \, mol^{-1}} = 2.0 \, mol$. Active mass $= \frac{2.0 \, mol}{1 \, L} = 2.0 \, mol \, L^{-1}$.
Thus,the active masses are $0.25, 0.25, 2.0$.

(B) The molar mass of $N_2 = 28 \ g/mol$ and $O_2 = 32 \ g/mol$.
Number of moles of $N_2 (n_{N_2}) = \frac{7}{28} = 0.25 \ mol$.
Number of moles of $O_2 (n_{O_2}) = \frac{8}{32} = 0.25 \ mol$.
Mole fraction of $O_2 (x_{O_2}) = \frac{n_{O_2}}{n_{N_2} + n_{O_2}} = \frac{0.25}{0.25 + 0.25} = \frac{0.25}{0.50} = 0.5$.

(B) The ratio $\frac{N}{n}$ represents the number of molecules per mole of a substance.
This is defined as the Avogadro constant $(N_A)$.
The value of the Avogadro constant is $6.022 \times 10^{23} \ \text{mol}^{-1}$.
Therefore,the correct option is $B$.

(D) According to Avogadro's Law,at constant temperature and pressure,equal volumes of gases contain an equal number of moles $(n)$.
Since $n = \frac{m}{M}$,the mass $(m)$ is directly proportional to the molar mass $(M)$ of the gas $(m = n \times M)$.
The molar masses are: $M(O_2) = 32 \ g/mol$,$M(Ne) = 20 \ g/mol$,and $M(CH_4) = 16 \ g/mol$.
Therefore,the order of increasing mass is $M(CH_4) < M(Ne) < M(O_2)$,which corresponds to $C < B < A$.

(C) The molar mass of magnesium $(Mg)$ is $24 \ g/mol$.
Given mass of $Mg = 0.004 \ g$.
Number of moles of $Mg = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{0.004}{24} \ mol$.
Number of atoms = $\text{Number of moles} \times N_A$.
Number of atoms = $\frac{0.004}{24} \times 6.022 \times 10^{23} \approx 10^{20}$ atoms.

(A) The mass of $1 \ mol$ of a substance is defined as the mass of $N_A$ particles of that substance.
Currently,the mass of $1 \ mol$ ($6.022 \times 10^{23}$ atoms) of carbon is $12 \ g$.
If the Avogadro number $(N_A)$ is changed to $6.022 \times 10^{20} \ mol^{-1}$,the mass of $1 \ mol$ of carbon would become the mass of $6.022 \times 10^{20}$ atoms,which is $\frac{12 \times 6.022 \times 10^{20}}{6.022 \times 10^{23}} = 12 \times 10^{-3} \ g$.
Since the number of particles defining a mole has changed,the mass of one mole of carbon changes.
The ratios of chemical species in balanced equations and elements in compounds are based on the law of definite proportions and conservation of mass,which are independent of the numerical value of $N_A$.

(C) To find the number of water molecules,we calculate the number of moles in each option:
$A$. $1.8 \ g$ of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$
$B$. $18 \ g$ of $H_2O = \frac{18 \ g}{18 \ g/mol} = 1 \ mol$
$C$. $18 \ moles$ of $H_2O = 18 \ mol$
$D$. $18$ molecules of $H_2O = \frac{18}{6.022 \times 10^{23}} \ mol$
Since the number of molecules is directly proportional to the number of moles $(N = n \times N_A)$,the option with the highest number of moles will have the maximum number of molecules.
Therefore,$18 \ moles$ of water contains the maximum number of molecules.

(D) Number of moles $= \frac{\text{number of molecules}}{N_A} = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3} \ mol$.
Molar concentration $= \frac{n \times 1000}{V_{\text{solution } (mL)}} = \frac{10^{-3} \times 1000}{100}$.
Molar concentration $= 0.01 \ M$.

(A) The molar mass of water $(H_2O)$ is $18.02 \ g \ mol^{-1}$.
Given density of water is $1 \ g \ cm^{-3}$.
Volume of $1 \ mol$ of water $= \text{Mass} / \text{Density} = 18.02 \ g / 1 \ g \ cm^{-3} = 18.02 \ cm^3$.
Since $1 \ mol$ contains $6.022 \times 10^{23}$ molecules,the volume occupied by one molecule is:
$V = 18.02 \ cm^3 / 6.022 \times 10^{23} \text{ molecules} \approx 2.99 \times 10^{-23} \ cm^3 \approx 3.0 \times 10^{-23} \ cm^3$.

(D) In one mole of $Mg_{3}(PO_{4})_{2}$,there are $8 \ mol$ of oxygen atoms.
Let $M$ be the number of moles of $Mg_{3}(PO_{4})_{2}$.
The number of moles of oxygen atoms is given by $M \times 8$.
According to the question,$M \times 8 = 0.25 \ mol$.
Therefore,$M = \frac{0.25}{8} = 0.03125 \ mol$.
This can be expressed as $3.125 \times 10^{-2} \ mol$.

(B) Let the mass of $O_2$ be $w$ and the mass of $N_2$ be $4w$.
The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $N_2$ is $28 \ g/mol$.
The number of molecules is proportional to the number of moles $(n = \frac{\text{mass}}{\text{molar mass}})$.
Number of molecules of $O_2$ $(N_{O_2})$ = $\frac{w}{32} \times N_A$
Number of molecules of $N_2$ $(N_{N_2})$ = $\frac{4w}{28} \times N_A$
Ratio $\frac{N_{O_2}}{N_{N_2}} = \frac{w/32}{4w/28} = \frac{w}{32} \times \frac{28}{4w} = \frac{28}{128} = \frac{7}{32}$.

(D) The number of atoms is calculated as: $\text{Number of atoms} = \text{moles} \times N_A \times \text{atomicity}$.
$1$. For $CO_2$: $\text{moles} = \frac{10}{44} \approx 0.227$. $\text{Atomicity} = 3$. $\text{Atoms} = 0.227 \times 3 \times N_A = 0.681 N_A$.
$2$. For $NH_3$: $\text{moles} = \frac{10}{17} \approx 0.588$. $\text{Atomicity} = 4$. $\text{Atoms} = 0.588 \times 4 \times N_A = 2.352 N_A$.
$3$. For $O_2$: $\text{moles} = \frac{10}{32} = 0.3125$. $\text{Atomicity} = 2$. $\text{Atoms} = 0.3125 \times 2 \times N_A = 0.625 N_A$.
Comparing the values: $2.352 N_A (NH_3) > 0.681 N_A (CO_2) > 0.625 N_A (O_2)$.
Thus,the decreasing order is $NH_3 > CO_2 > O_2$.

(A) Given: Volume of $SO_2 = 5.6 \ L$ at $NTP$.
Number of moles = $\frac{5.6}{22.4} = 0.25 \ mol$.
Since $1 \ molecule$ of $SO_2$ contains $3 \ atoms$ $(1 \ S + 2 \ O)$,
Total atoms = $0.25 \times 3 \times N_A = 0.75 \ N_A = \frac{3}{4} N_A$.

(B) The molar mass of hydrogen atom $(H)$ is approximately $1.008 \ g/mol$.
Using Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$,the mass of one atom is calculated as:
$\text{Mass of one atom} = \frac{\text{Molar mass}}{N_A} = \frac{1.008 \ g/mol}{6.022 \times 10^{23} \ mol^{-1}} \approx 1.67 \times 10^{-24} \ g$.
Thus,the order of magnitude is $10^{-24} \ g$.

(C) The term $g$-ion refers to the number of moles of the ion.
Given,$2.1 \ g$-ion of $Cl^{-}$ means $2.1 \ \text{moles}$ of $Cl^{-}$ ions.
One $Cl$ atom has $17$ electrons. $A$ $Cl^{-}$ ion has $17 + 1 = 18$ electrons.
Total number of electrons = $(\text{Number of moles}) \times (\text{Number of electrons per ion}) \times (N_A)$.
Total number of electrons = $2.1 \times 18 \times 6.023 \times 10^{23}$.

(B) The molar mass of $D_2O$ is $(2 \times 2) + 16 = 20 \ g/mol$.
Number of moles of $D_2O = \frac{5 \ g}{20 \ g/mol} = 0.25 \ mol$.
In one molecule of $D_2O$ ($D = _1^2H$,$O = _8^{16}O$):
Neutrons in $2D = 2 \times (2 - 1) = 2$.
Neutrons in $O = 16 - 8 = 8$.
Total neutrons per molecule = $2 + 8 = 10$.
Total neutrons in $0.25 \ mol = 0.25 \times 10 \times N_A = 2.5 \ N_A$.

(A) The chemical formula for sodium carbonate decahydrate is $Na_{2}CO_{3} \cdot 10H_{2}O$.
Each molecule of $Na_{2}CO_{3} \cdot 10H_{2}O$ contains $3$ oxygen atoms in the carbonate part and $10$ oxygen atoms in the water molecules,totaling $13$ oxygen atoms per formula unit.
Therefore,$1 \ mol$ of $Na_{2}CO_{3} \cdot 10H_{2}O$ contains $13 \ mol$ of oxygen atoms.
For $0.2 \ mol$ of $Na_{2}CO_{3} \cdot 10H_{2}O$,the number of moles of oxygen atoms is $0.2 \times 13 = 2.6 \ mol$.
The number of oxygen atoms is calculated as $2.6 \times N_{A} = 2.6 \times 6.022 \times 10^{23} \approx 1.5657 \times 10^{24}$.
Rounding to the nearest provided option,the result is $1.56 \times 10^{24}$.

(A) The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Moles of $CO_2 = \frac{\text{given mass}}{\text{molar mass}} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.
Number of molecules of $CO_2 = \text{moles} \times N_A = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules.
Since $1$ molecule of $CO_2$ contains $2$ oxygen atoms,the number of oxygen atoms $= 2 \times (6.022 \times 10^{22}) = 12.044 \times 10^{22} = 1.2044 \times 10^{23}$ atoms.
Thus,the number of oxygen atoms is approximately $1.2 \times 10^{23}$.

(D) The molar mass of water $(H_2O)$ is $18 \, g/mol$.
Given density of water is $1 \, g/mL$,so $18 \, g$ of water occupies $18 \, mL$.
Since $1 \, mol$ of water contains $6.022 \times 10^{23}$ molecules,the volume of $6.022 \times 10^{23}$ molecules is $18 \, mL$.
Therefore,the volume of $1$ molecule of water is $\frac{18 \, mL}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23} \, mL$.

(A) To find the number of atoms,use the formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{atomicity}$.
$A) \ 1 \ g$ of $C_4H_{10}$: $\text{molar mass} = 58 \ g/mol$,$\text{atoms} = \frac{1}{58} \times 14 \times N_A \approx 0.241 \ N_A$.
$B) \ 1 \ g$ of $N_2$: $\text{molar mass} = 28 \ g/mol$,$\text{atoms} = \frac{1}{28} \times 2 \times N_A \approx 0.071 \ N_A$.
$C) \ 1 \ g$ of $Ag$: $\text{molar mass} = 108 \ g/mol$,$\text{atoms} = \frac{1}{108} \times 1 \times N_A \approx 0.009 \ N_A$.
$D) \ 1 \ g$ of $H_2O$: $\text{molar mass} = 18 \ g/mol$,$\text{atoms} = \frac{1}{18} \times 3 \times N_A \approx 0.166 \ N_A$.
Comparing the values,$1 \ g$ of $C_4H_{10}$ has the greatest number of atoms.

(C) The chemical formula $C_4H_6O_6$ indicates that $1 \ mol$ of the compound contains $6 \ mol$ of oxygen atoms.
To find the number of gram atoms (moles) of oxygen in $0.8 \ mol$ of $C_4H_6O_6$,we multiply the moles of the compound by the number of oxygen atoms per molecule:
$\text{Moles of O} = 0.8 \ mol \times 6 = 4.8 \ mol$.
Thus,there are $4.8$ gram atoms of oxygen.

(D) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
The number of moles $(n)$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}} = \frac{180 \ g}{18 \ g/mol} = 10 \ moles$.
The number of molecules is calculated as $n \times N_A = 10 \times 6.023 \times 10^{23} = 6.023 \times 10^{24}$ molecules.
Therefore,the correct option is $D$.

(D) First,calculate the atomic mass of the element:
Atomic mass $= \text{mass of one atom} \times N_A = 3.32 \times 10^{-23} \ g \times 6.022 \times 10^{23} \ mol^{-1} \approx 20 \ g/mol$.
Next,convert the total mass to grams:
$20 \ kg = 20,000 \ g$.
Gram atoms (moles) $= \frac{\text{Total mass}}{\text{Atomic mass}} = \frac{20,000 \ g}{20 \ g/mol} = 1000 \ \text{mol}$.

(C) The molar mass of $NH_3$ is $14 + (3 \times 1) = 17 \ g/mol$.
Moles of $NH_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{34 \ g}{17 \ g/mol} = 2 \ mol$.
Number of $NH_3$ molecules $= \text{Moles} \times N_A = 2 \ N_A$.
Each $NH_3$ molecule contains $7$ (from $N$) $+ 3 \times 1$ (from $H$) $= 10$ electrons.
Total number of electrons $= 2 \ N_A \times 10 = 20 \ N_A$.

(C) Number of moles of urea $= \frac{\text{Number of molecules}}{N_A} = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3} \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{10^{-3} \ mol}{200 \ mL \times 10^{-3} \ L/mL} = \frac{10^{-3}}{0.2} = 0.005 \ M$.

(A) The molar mass of $H_2SO_4$ is $98 \, g/mol$.
Total moles of $H_2SO_4$ initially $= \frac{392 \times 10^{-3} \, g}{98 \, g/mol} = 4 \times 10^{-3} \, mol$.
Moles of $H_2SO_4$ removed $= \frac{1.204 \times 10^{21}}{6.02 \times 10^{23} \, mol^{-1}} = 0.2 \times 10^{-2} \, mol = 2 \times 10^{-3} \, mol$.
Moles of $H_2SO_4$ left $= (4 \times 10^{-3}) - (2 \times 10^{-3}) = 2.0 \times 10^{-3} \, mol$.

(D) The chemical formula of phosphorus trichloride is $PCl_{3}$.
Each molecule of $PCl_{3}$ contains $1$ atom of phosphorus and $3$ atoms of chlorine,totaling $4$ atoms per molecule.
Therefore,$1 \, mol$ of $PCl_{3}$ contains $4 \, mol$ of atoms.
For $1.4 \, mol$ of $PCl_{3}$,the total number of moles of atoms is $1.4 \times 4 = 5.6 \, mol$.
Using Avogadro's number $(N_{A} = 6.022 \times 10^{23} \, mol^{-1})$,the total number of atoms is $5.6 \times 6.022 \times 10^{23} = 3.37232 \times 10^{24}$ atoms.
Rounding to the nearest provided option,the correct answer is $3.3712 \times 10^{24}$.

(D) The number of molecules is directly proportional to the number of moles $(n = \text{mass} / \text{molar mass})$.
For $A$: $n(SO_2) = 64 \ g / 64 \ g \ mol^{-1} = 1 \ mol$.
For $B$: $n(CO_2) = 44 \ g / 44 \ g \ mol^{-1} = 1 \ mol$.
For $C$: $n(O_3) = 48 \ g / 48 \ g \ mol^{-1} = 1 \ mol$.
For $D$: $n(H_2) = 8 \ g / 2 \ g \ mol^{-1} = 4 \ mol$.
Since $H_2$ has the highest number of moles $(4 \ mol)$,it contains the maximum number of molecules.

(D) At $STP$,$1 \, mole$ of any gas occupies $22.4 \, L$ (or $22.4 \, dm^3$).
Number of moles of $O_2 = \frac{224 \, dm^3}{22.4 \, dm^3/mol} = 10 \, mol$.
Since $O_2$ is a diatomic molecule,$1 \, mole$ of $O_2$ contains $2 \, moles$ of oxygen atoms.
Total moles of oxygen atoms $= 10 \times 2 = 20 \, mol$.
Number of atoms $= 20 \times 6.022 \times 10^{23} \approx 1.2 \times 10^{25}$ atoms.

(C) The number of atoms is calculated as: $\text{Number of atoms} = \text{atomicity} \times \text{moles} \times N_A$.
$A$. For $36 \ g \ H_2O$: $\text{moles} = \frac{36}{18} = 2 \ mol$. $\text{Atomicity} = 3$. $\text{Total atoms} = 2 \times 3 \times N_A = 6 N_A$.
$B$. For $28 \ g \ CO$: $\text{moles} = \frac{28}{28} = 1 \ mol$. $\text{Atomicity} = 2$. $\text{Total atoms} = 1 \times 2 \times N_A = 2 N_A$.
$C$. For $46 \ g \ C_2H_5OH$: $\text{moles} = \frac{46}{46} = 1 \ mol$. $\text{Atomicity} = 9$. $\text{Total atoms} = 1 \times 9 \times N_A = 9 N_A$.
$D$. For $54 \ g \ N_2O_5$: $\text{moles} = \frac{54}{108} = 0.5 \ mol$. $\text{Atomicity} = 7$. $\text{Total atoms} = 0.5 \times 7 \times N_A = 3.5 N_A$.
Comparing the values,$46 \ g \ C_2H_5OH$ contains the largest number of atoms.

(C) Number of moles of glucose = $\frac{0.9 \ g}{180 \ g/mol} = 0.005 \ mol$.
Number of hydrogen atoms in glucose = $0.005 \times 12 \times N_A = 0.06 \ N_A$.
Now,checking option $C$:
Number of moles of ethane $(C_2H_6)$ = $\frac{0.3 \ g}{30 \ g/mol} = 0.01 \ mol$.
Number of hydrogen atoms in ethane = $0.01 \times 6 \times N_A = 0.06 \ N_A$.
Since the number of hydrogen atoms is the same,the correct option is $C$.

(B) The chemical formula $[Cu(NH_3)_4]SO_4$ indicates that $1 \text{ mole}$ of the compound contains $4 \text{ moles}$ of ammonia $(NH_3)$ molecules.
Number of moles of $NH_3$ molecules = $\frac{\text{Total number of molecules}}{\text{Avogadro's number}} = \frac{2.4 \times 10^{24}}{6.022 \times 10^{23}} \approx 4 \text{ moles}$.
Since $4 \text{ moles}$ of $NH_3$ are present in $1 \text{ mole}$ of $[Cu(NH_3)_4]SO_4$,the number of moles of $[Cu(NH_3)_4]SO_4$ is $\frac{4}{4} = 1 \text{ mole}$.

(C) The atomic number of oxygen is $8$,so a neutral oxygen atom has $8$ electrons.
For the oxide ion ${}_{8}^{18}O^{2-}$,the number of electrons is $8 + 2 = 10$ electrons per ion.
The number of ions in $2 \times 10^{-3} \ mol$ is $2 \times 10^{-3} \times N_A$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Total electrons = $(\text{Number of ions}) \times (\text{Electrons per ion})$
Total electrons = $(2 \times 10^{-3} \times 6.022 \times 10^{23}) \times 10$
Total electrons = $12.044 \times 10^{21} \approx 1.2 \times 10^{22}$.

(D) The molar mass of $NH_3 = 14 + (3 \times 1) = 17 \ g/mol$.
Number of moles of $NH_3 = \frac{4.25 \ g}{17 \ g/mol} = 0.25 \ mol$.
Number of molecules of $NH_3 = 0.25 \times 6.022 \times 10^{23} \approx 1.5055 \times 10^{23}$.
Each molecule of $NH_3$ contains $4$ atoms ($1$ nitrogen and $3$ hydrogen).
Total number of atoms $= 4 \times 1.5055 \times 10^{23} \approx 6.022 \times 10^{23} \approx 6 \times 10^{23}$.

(A) The number of moles of $CO_2$ is calculated as $n = \frac{11.2 \, L}{22.4 \, L/mol} = 0.5 \, mol$.
The number of molecules of $CO_2$ is $0.5 \times N_A$.
Since each molecule of $CO_2$ contains $3$ atoms ($1$ carbon and $2$ oxygen atoms),the total number of atoms is $3 \times 0.5 \times N_A = 1.5 \, N_A$.

(C) The number of protons in one $NH_{4}^{+}$ ion is calculated as: $7 (\text{for } N) + 4 (\text{for } H) = 11$ protons.
The molar mass of $NH_{4}^{+}$ is $14 + 4 = 18 \ g/mol$.
The number of moles of $NH_{4}^{+}$ in $1.8 \ g$ is $\frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$.
The number of $NH_{4}^{+}$ ions is $0.1 \ N_{A}$.
Therefore,the total number of protons is $11 \times 0.1 \ N_{A} = 1.1 \ N_{A}$.

(B) Step $1$: Calculate the number of atoms in $8 \ g$ of $O_2$.
Molar mass of $O_2 = 32 \ g/mol$.
Moles of $O_2 = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol$.
Number of $O_2$ molecules $= 0.25 \times N_A$.
Number of $O$ atoms $= 2 \times (0.25 \times N_A) = 0.5 \ N_A$.
Step $2$: Calculate the number of atoms in the options.
$(A)$ $14 \ g$ of $CO$: Molar mass $= 28 \ g/mol$. Moles $= \frac{14}{28} = 0.5 \ mol$. Total atoms $= 0.5 \times 2 \times N_A = 1.0 \ N_A$.
$(B)$ $7 \ g$ of $CO$: Moles $= \frac{7}{28} = 0.25 \ mol$. Total atoms $= 0.25 \times 2 \times N_A = 0.5 \ N_A$.
$(C)$ $11 \ g$ of $CO_2$: Molar mass $= 44 \ g/mol$. Moles $= \frac{11}{44} = 0.25 \ mol$. Total atoms $= 0.25 \times 3 \times N_A = 0.75 \ N_A$.
$(D)$ $22 \ g$ of $CO_2$: Moles $= \frac{22}{44} = 0.5 \ mol$. Total atoms $= 0.5 \times 3 \times N_A = 1.5 \ N_A$.
Conclusion: $7 \ g$ of $CO$ contains $0.5 \ N_A$ atoms,which is the same as $8 \ g$ of $O_2$.

(D) To find the number of molecules,we calculate the number of moles for each option:
$A$: $22.4 \times 10^3 \ mL = 22.4 \ L$. At $STP$,$1 \ mole$ of gas occupies $22.4 \ L$. So,$22.4 \ L = 1 \ mole$.
$B$: Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \ g/mol$. Moles = $22 \ g / 44 \ g/mol = 0.5 \ mole$.
$C$: At $STP$,$11.2 \ L / 22.4 \ L/mol = 0.5 \ mole$.
$D$: Given as $0.1 \ mole$.
Comparing the values: $1 \ mole > 0.5 \ mole > 0.1 \ mole$. Therefore,$0.1 \ mole$ of $CO_2$ gas has the smallest number of molecules.

(D) The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g/mol$.
Number of moles of $CO_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{176 \ g}{44 \ g/mol} = 4 \ mol$.
Since each molecule of $CO_2$ contains $2$ atoms of oxygen,the number of moles of oxygen atoms $= 4 \times 2 = 8 \ mol$.
Number of oxygen atoms $= \text{moles} \times N_A = 8 \times 6.022 \times 10^{23} = 48.176 \times 10^{23} = 4.8176 \times 10^{24}$ atoms.
Rounding to the nearest option,the value is approximately $4.816 \times 10^{24}$.

(B) Density of water $= 1 \ g/mL = 1 \ g/cm^{3}$.
Mass of $1 \ cm^{3}$ of water $=$ density $\times$ volume $= 1 \ g/cm^{3} \times 1 \ cm^{3} = 1 \ g$.
Molar mass of water $(H_{2}O)$ $= (2 \times 1) + 16 = 18 \ g/mol$.
Number of moles of water $= \frac{\text{mass}}{\text{molar mass}} = \frac{1}{18} \ mol$.
Number of molecules $= \text{moles} \times N_{A} = \frac{1}{18} \times 6.022 \times 10^{23} \approx 3.34 \times 10^{22}$ molecules.

(B) $1 \, mol$ of $N^{3-}$ ions contains $10 \, mol$ of electrons $(7 + 3 = 10)$.
$1 \, mol$ of $O^{2-}$ ions contains $10 \, mol$ of electrons $(8 + 2 = 10)$,which equals $10 \times 6.023 \times 10^{23} = 6.023 \times 10^{24}$ electrons.
Option $B$ states $6.023 \times 10^{22}$ electrons,which is incorrect.
$1 \, mol$ of $CH_4$ contains $(6 + 4 \times 1) = 10 \, mol$ of protons.
$1 \, mol$ of $H_2O$ contains $(2 \times 1 + 8) = 10 \, mol$ of protons.

(B) The molar mass of $H_2O$ is $18 \ g/mol$.
Given volume $= 18 \ mL$ and density $= 1 \ g/mL$,so the mass of water is $18 \ g$.
Number of moles $= \frac{\text{mass}}{\text{molar mass}} = \frac{18 \ g}{18 \ g/mol} = 1 \ mol$.
Number of molecules $= 1 \ mol \times 6.023 \times 10^{23} \text{ molecules/mol} = 6.023 \times 10^{23} \text{ molecules}$.
Each $H_2O$ molecule contains $2$ electrons from $2$ hydrogen atoms and $8$ electrons from $1$ oxygen atom,totaling $10$ electrons per molecule.
Total electrons $= 6.023 \times 10^{23} \text{ molecules} \times 10 \text{ electrons/molecule} = 6.023 \times 10^{24} \text{ electrons}$.

(A) Given: Volume of one drop of water $(V)$ = $0.0018 \ mL$.
Density of water $( ho)$ = $1 \ g/mL$.
Mass of one drop $(m)$ = $V \times \rho = 0.0018 \ mL \times 1 \ g/mL = 0.0018 \ g$.
Moles of water in one drop = $\frac{\text{mass}}{\text{molar mass}} = \frac{0.0018 \ g}{18 \ g/mol} = 10^{-4} \ mol$.
Number of water molecules in one drop = $\text{moles} \times N_A = 10^{-4} \times 6.023 \times 10^{23} = 6.023 \times 10^{19}$.
Number of water molecules in two drops = $2 \times 6.023 \times 10^{19} = 12.046 \times 10^{19}$.

(B) $1 \, \text{mole}$ of $H_2O$ contains $1 \, \text{mole}$ of oxygen atoms,which is $16 \, g$ of oxygen.
$\therefore$ $3.6 \, \text{moles}$ of $H_2O$ contains $3.6 \times 16 \, g$ of oxygen.
$= 57.6 \, g$.

(A) To find the number of atoms,we calculate the moles and multiply by the atomicity and Avogadro's number $(N_A = 6.023 \times 10^{23} \ mol^{-1})$.
$A$) $4.0 \ g$ of $H_2$: Moles $= \frac{4}{2} = 2 \ mol$. Atoms $= 2 \times 2 \times N_A = 4 N_A$.
$B$) $71.0 \ g$ of $Cl_2$: Moles $= \frac{71}{71} = 1 \ mol$. Atoms $= 1 \times 2 \times N_A = 2 N_A$.
$C$) $127.0 \ g$ of $I_2$: Moles $= \frac{127}{254} = 0.5 \ mol$. Atoms $= 0.5 \times 2 \times N_A = 1 N_A$.
$D$) $48.0 \ g$ of $Mg$: Moles $= \frac{48}{24} = 2 \ mol$. Atoms $= 2 \times 1 \times N_A = 2 N_A$.
Comparing the values,$4.0 \ g$ of hydrogen has the largest number of atoms $(4 N_A)$.

(D) For $16.0 \ g$ $O_3$: Molar mass $= 48 \ g/mol$. Moles $= 16/48 = 1/3 \ mol$. Number of $O$ atoms $= 3 \times (1/3) \times N_A = N_A$.
For $28.0 \ g$ $CO$: Molar mass $= 28 \ g/mol$. Moles $= 28/28 = 1 \ mol$. Number of $O$ atoms $= 1 \times 1 \times N_A = N_A$.
For $16.0 \ g$ $O_2$: Molar mass $= 32 \ g/mol$. Moles $= 16/32 = 0.5 \ mol$. Number of $O$ atoms $= 2 \times 0.5 \times N_A = N_A$.
Thus,the ratio of oxygen atoms is $N_A : N_A : N_A$,which simplifies to $1 : 1 : 1$.

(C) The number of moles of the gas is $0.1 \ mol$.
Since the gas is triatomic,each molecule contains $3$ atoms.
The number of molecules $= \text{moles} \times N_A = 0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22} \ \text{molecules}$.
The number of atoms $= \text{number of molecules} \times 3 = 6.02 \times 10^{22} \times 3 = 1.806 \times 10^{23} \ \text{atoms}$.

(D) $1$. For $CO_2$: $n = \frac{11.2 \, L}{22.4 \, L/mol} = 0.5 \, mol$. Number of atoms = $0.5 \times 3 \times N_A = 1.5 N_A$.
$2$. For $H_2$: $n = \frac{2 \, g}{2 \, g/mol} = 1 \, mol$. Number of atoms = $1 \times 2 \times N_A = 2 N_A$.
$3$. For $C$: $2 \, g$-atom means $2 \, moles$ of $C$ atoms. Number of atoms = $2 N_A$.
$4$. For $SO_2$: $3 \, g$-molecule means $3 \, moles$ of $SO_2$ molecules. Number of atoms = $3 \times 3 \times N_A = 9 N_A$.
Comparing all,$9 N_A$ is the highest.

(A) According to Avogadro's Law,under the same conditions of pressure,temperature,and volume,the number of moles of different gases must be equal.
$n_A = n_B$
$\frac{w_A}{M_A} = \frac{w_B}{M_B}$
Given $w_A = 0.11 \, g$ $(CO_2)$,$M_A = 44 \, g/mol$,and $w_B = 0.07 \, g$ (unknown gas).
$\frac{0.11}{44} = \frac{0.07}{M_B}$
$M_B = \frac{0.07 \times 44}{0.11} = \frac{7 \times 44}{11} = 7 \times 4 = 28 \, g/mol$.
The molar mass of the unknown gas is $28 \, g/mol$.
Checking the options:
$CO$: $12 + 16 = 28 \, g/mol$.
$CO_2$: $12 + 32 = 44 \, g/mol$.
$C_2H_6$: $2(12) + 6(1) = 30 \, g/mol$.
Since $CO$ has a molar mass of $28 \, g/mol$,the unknown gas is $CO$.