The mole concept Questions in English

(C) The molarity $(M)$ of a solution is defined as the number of moles of solute $(n)$ per liter of solution ($V$ in $L$):
$M = \frac{n}{V(L)}$
Given:
$M = 0.5 \ M$
$V = 2 \ L$
Substituting the values into the formula:
$0.5 = \frac{n}{2}$
$n = 0.5 \times 2 = 1 \ mol$
Therefore,the number of moles is $1$.

(B) Given mass of methane $(CH_4) = 16 \ g$.
Molar mass of $CH_4 = 12 + (4 \times 1) = 16 \ g/mol$.
Moles of $CH_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{16 \ g}{16 \ g/mol} = 1 \ mol$.
Since $1 \ mol$ of any substance contains $6.022 \times 10^{23}$ molecules (Avogadro's number),
Therefore,$1 \ mol$ of $CH_4$ contains $6.02 \times 10^{23}$ molecules.

(A) The number of moles $(n)$ is given by the formula: $n = \frac{w}{M}$,where $w$ is the mass in grams and $M$ is the molar mass.
Given: $n = 0.25 \, \text{mol}$,Molar mass of $H_2SO_4$ $(M)$ = $(2 \times 1) + 32 + (4 \times 16) = 98 \, g/mol$.
Rearranging the formula to find mass $(w)$: $w = n \times M$.
$w = 0.25 \times 98 = 24.5 \, g$.
Therefore,the correct option is $(A)$.

(B) The number of moles of urea is calculated as: $n = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3} \ mol$.
The volume of the solution is $100 \ mL = 0.1 \ L$.
The molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V(L)} = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M$.

(A) The molar mass of $SO_2Cl_2$ is calculated as: $32 + (2 \times 16) + (2 \times 35.5) = 32 + 32 + 71 = 135 \ g/mol$.
The number of moles $(n)$ is calculated using the formula: $n = \frac{\text{mass}}{\text{molar mass}}$.
$n = \frac{13.5 \ g}{135 \ g/mol} = 0.1 \ mol$.

(A) Step $1$: Calculate the number of moles of $H_2$ $(n_{H_2} = \frac{1 \ g}{2 \ g/mol} = 0.5 \ mol)$.
Step $2$: Calculate the number of moles of $O_2$ $(n_{O_2} = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol)$.
Step $3$: Calculate the mole fraction of $H_2$ $(x_{H_2} = \frac{n_{H_2}}{n_{H_2} + n_{O_2}} = \frac{0.5}{0.5 + 0.25} = \frac{0.5}{0.75} = 0.667)$.

(C) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
The number of moles is calculated using the formula: $\text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}}$.
$\text{Moles of water} = \frac{180 \ g}{18 \ g/mol} = 10 \ mol$.
Therefore,the correct option is $(C)$.

(C) Step $1$: Calculate the number of moles of $CO_2$ $(n_{CO_2} = \frac{44 \ g}{44 \ g/mol} = 1 \ mol)$.
Step $2$: Calculate the number of moles of $N_2$ $(n_{N_2} = \frac{14 \ g}{28 \ g/mol} = 0.5 \ mol)$.
Step $3$: Calculate the mole fraction of $CO_2$ $(X_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{N_2}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{1}{3/2} = 2/3)$.

(B) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Number of moles of water = $\frac{90 \ g}{18 \ g/mol} = 5 \ mol$.
The molar mass of acetic acid $(CH_3COOH)$ is $60 \ g/mol$.
Number of moles of acetic acid = $\frac{300 \ g}{60 \ g/mol} = 5 \ mol$.
Total number of moles = $5 \ mol + 5 \ mol = 10 \ mol$.

(D) The chemical formula is $Na_4[Fe(CN)_6]$.
One mole of $Na_4[Fe(CN)_6]$ contains $4 \ mol$ of $Na$ atoms.
Therefore,$2 \ mol$ of $Na_4[Fe(CN)_6]$ contains $2 \times 4 = 8 \ mol$ of $Na$ atoms.
The number of $Na$ atoms $= 8 \times N_A = 8 \times 6.022 \times 10^{23} = 48.176 \times 10^{23}$.
Rounding to the nearest integer,the value is $48 \times 10^{23}$.

(D) Total number of sugar molecules required $= (\text{Population}) \times (\text{Molecules per person})$
$= 3 \times 10^{10} \times 100 = 3 \times 10^{12} \text{ molecules}$.
Number of moles of sugar $= \frac{\text{Total molecules}}{N_A} = \frac{3 \times 10^{12}}{6.022 \times 10^{23}} \approx 0.5 \times 10^{-11} \text{ mol}$.
Mass of sugar $= \text{Moles} \times \text{Molar mass} = 0.5 \times 10^{-11} \times 342 \ g/mol$.
$= 171 \times 10^{-11} \ g$.

(B) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
According to the mole concept,$1 \ mol$ of water contains $N_A$ (Avogadro's number,$6.022 \times 10^{23}$) molecules.
Therefore,the mass of $N_A$ molecules of water is $18 \ g$.
The mass of $1$ molecule of water is calculated as $\frac{18}{N_A} \ g$.
Substituting $N_A = 6.022 \times 10^{23}$,we get $\frac{18}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23} \ g$.

(B) Mass of $K-40$ in $370 \ mg$ of $K = \frac{370 \times 0.012}{100} \ mg = 0.0444 \ mg = 0.0444 \times 10^{-3} \ g$.
Number of moles of $K-40 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.0444 \times 10^{-3} \ g}{40 \ g/mol} = 1.11 \times 10^{-6} \ mol$.
Number of atoms of $K-40 = \text{moles} \times N_A = 1.11 \times 10^{-6} \times 6.022 \times 10^{23} \approx 6.69 \times 10^{17} \ \text{atoms}$.

(B) Number of moles of $Fe = \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{558.5}{55.85} = 10 \ mol$.
Number of atoms in $10 \ mol$ of $Fe = 10 \times N_A$.
Now,check option $B$: $60 \ g$ of $C$ contains $\frac{60}{12} = 5 \ mol$ of $C$.
Twice the number of atoms in $60 \ g$ of $C = 2 \times (5 \times N_A) = 10 \times N_A$.
Since both values are equal,option $B$ is correct.

(C) The molar mass of $N_2$ is $28 \ g/mol$.
Calculation of moles:
$\text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{56 \ g}{28 \ g/mol} = 2 \ mol$.
Calculation of volume at $STP$:
$1 \ mol$ of any gas occupies $22.4 \ L$ at $STP$.
Therefore,$2 \ mol$ of $N_2$ occupies $2 \times 22.4 \ L = 44.8 \ L$.

(C) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.
The number of moles of glucose is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.23 \ g}{180 \ g/mol} \approx 0.02905 \ mol$.
The number of molecules is calculated by multiplying the number of moles by Avogadro's constant $(N_A = 6.022 \times 10^{23} \ mol^{-1})$:
Number of molecules $= 0.02905 \ mol \times 6.022 \times 10^{23} \ mol^{-1} \approx 1.75 \times 10^{22}$ molecules.

(C) The molar mass of $CO_2$ is $44 \ g/mol$.
Number of moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.
Number of molecules of $CO_2 = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \ \text{molecules}$.
Since each molecule of $CO_2$ contains $2$ oxygen atoms,
Number of oxygen atoms = $2 \times 6.022 \times 10^{22} = 1.2044 \times 10^{23} \ \text{atoms}$.

(C) Molar mass of common salt $(NaCl)$ = $58.5 \ g/mol$. Molar mass of sugar $(C_{12}H_{22}O_{11})$ = $342 \ g/mol$.
Number of moles in $1 \ kg$ $(1000 \ g)$ of salt = $\frac{1000}{58.5} \approx 17.09 \ mol$.
Number of moles in $1 \ kg$ $(1000 \ g)$ of sugar = $\frac{1000}{342} \approx 2.92 \ mol$.
Cost of $1 \ mole$ of salt = $\frac{7 \times 58.5}{1000} = 0.4095 \ rupees$.
Cost of $1 \ mole$ of sugar = $\frac{14 \times 342}{1000} = 4.788 \ rupees$.
Comparing the two,the cost of $1 \ mole$ of sugar is significantly higher than the cost of $1 \ mole$ of salt.

(B) The chemical formula for sodium oxide is $Na_2O$.
The molar mass of $Na_2O = (2 \times 23) + 16 = 46 + 16 = 62 \, g/mol$.
Number of moles = $\frac{\text{Given mass}}{\text{Molar mass}}$.
Number of moles = $\frac{620 \, g}{62 \, g/mol} = 10 \, mol$.

(B) The dissociation reaction is $CaCO_3 \to Ca^{+2} + CO_3^{-2}$.
Thus,$1$ molecule of $CaCO_3$ consists of one $Ca^{+2}$ ion and one $CO_3^{-2}$ ion.
Given $3.01 \times 10^{23}$ ions of $Ca^{+2}$ and $CO_3^{-2}$ each,the number of $CaCO_3$ molecules is $3.01 \times 10^{23}$.
Since $6.022 \times 10^{23}$ molecules = $1$ mole of $CaCO_3$,
$3.01 \times 10^{23}$ molecules = $0.5$ moles of $CaCO_3$.
The molar mass of $CaCO_3$ is $40 + 12 + (3 \times 16) = 100 \ g/mol$.
Mass = $\text{moles} \times \text{molar mass} = 0.5 \ mol \times 100 \ g/mol = 50 \ g$.

(C) The mass of $1$ electron is $9.1 \times 10^{-28} \, g$.
The mass of $1$ mole of electrons is the mass of $N_A$ electrons.
Mass $= 6.022 \times 10^{23} \times 9.1 \times 10^{-28} \, g$.
Mass $\approx 5.48 \times 10^{-4} \, g = 0.548 \, mg \approx 0.55 \, mg$.

(A) At $NTP$,$22400 \, cc$ of a gas contains $6.022 \times 10^{23}$ molecules.
Therefore,$1 \, cc$ of gas contains $\frac{6.022 \times 10^{23}}{22400}$ molecules.
For $1.12 \times 10^{-7} \, cc$ of gas,the number of molecules is $\frac{6.022 \times 10^{23}}{22400} \times 1.12 \times 10^{-7}$.
$= \frac{6.022 \times 10^{23} \times 1.12 \times 10^{-7}}{2.24 \times 10^4} = 3.011 \times 10^{12}$ molecules.

(A) Moles of sugar $= \frac{1.71 \ g}{342 \ g/mol} = 0.005 \ mol = 5 \times 10^{-3} \ mol$.
Number of sugar molecules $= 5 \times 10^{-3} \times 6.022 \times 10^{23} = 3.011 \times 10^{21}$.
Each molecule of sugar $(C_{12}H_{22}O_{11})$ contains $12$ carbon atoms.
Total number of carbon atoms $= 12 \times 3.011 \times 10^{21} = 3.6132 \times 10^{22} \approx 3.6 \times 10^{22}$.

(B) The molar mass of ozone $(O_3)$ is $3 \times 16 = 48 \ g/mol$.
Number of moles of $O_3 = \frac{8 \ g}{48 \ g/mol} = \frac{1}{6} \ mol$.
Each molecule of $O_3$ contains $3$ oxygen atoms.
Total number of oxygen atoms = $(\text{Number of moles of } O_3) \times 3 \times N_A$.
Total number of oxygen atoms = $\frac{1}{6} \times 3 \times N_A = \frac{1}{2} N_A = \frac{6.02 \times 10^{23}}{2}$.

(A) According to Avogadro's hypothesis,equal volumes of gases at the same temperature and pressure contain an equal number of molecules.
Since the volumes of $N_2$,$Ne$,and $O_3$ are equal ($1 \ L$ each),the number of molecules in each flask is the same (let it be $x$).
Number of atoms in $N_2$ = $2 \times x$
Number of atoms in $Ne$ = $1 \times x$
Number of atoms in $O_3$ = $3 \times x$
The ratio of the total number of atoms is $2x : 1x : 3x$,which simplifies to $2:1:3$.

(A) The molar mass of $CuSO_4 \cdot 5H_2O$ is calculated as:
$M = 63.5 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16) = 63.5 + 32 + 64 + 90 = 249.5 \, g/mol$.
Number of moles $n = \frac{\text{Number of molecules}}{N_A} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0166 \, mol$.
Mass = $n \times M = 0.0166 \times 249.5 \approx 4.14 \, g$.
Given the options,the closest value is $4.159 \, g$.

(A) The number of molecules is calculated using the formula: $\text{Number of molecules} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_A$.
Given $N_A = 6.02 \times 10^{23} \ \text{molecules/mol}$.
We need $6.02 \times 10^{22}$ molecules,which is $0.1 \ \text{mol}$.
For $A$: $2.8 \ g \ N_2$ $(M = 28 \ g/mol)$ = $0.1 \ \text{mol}$. Since $N_2$ is a molecule,it contains $0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}$ molecules.
For $B$: $1.6 \ g \ O_2$ $(M = 32 \ g/mol)$ = $0.05 \ \text{mol}$.
For $C$: $Na$ is an atom,not a molecule.
For $D$: $1.4 \ g \ O_2$ $(M = 32 \ g/mol)$ = $0.04375 \ \text{mol}$.
Thus,$A$ is the correct answer.

(A) The molar mass of $NH_3$ is $14 + (3 \times 1) = 17 \ g/mol$.
Number of moles of $NH_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.25}{17} = 0.25 \ mol$.
Number of molecules = $\text{Number of moles} \times N_A$.
Number of molecules = $0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}$ molecules.

(D) The molar mass of $Na_2SO_4 \cdot 10H_2O$ is calculated as: $2(23) + 32 + 4(16) + 10(18) = 46 + 32 + 64 + 180 = 322 \ g/mol$.
One mole of $Na_2SO_4 \cdot 10H_2O$ contains $4$ oxygen atoms from $SO_4^{2-}$ and $10$ oxygen atoms from $10H_2O$,totaling $14$ moles of oxygen atoms.
The number of moles of $Na_2SO_4 \cdot 10H_2O$ in $64.4 \ g$ is: $n = \frac{64.4 \ g}{322 \ g/mol} = 0.2 \ mol$.
Therefore,the moles of oxygen atoms = $0.2 \ mol \times 14 = 2.8 \ mol$.

(C) $22.4 \ L$ of oxygen gas corresponds to $1 \ mol$ of molecules at $STP$.
Number of mole molecules in $56 \ L = \frac{56}{22.4} = 2.5 \ mol$ of molecules.
We know that,$\text{mole atoms} = \text{atomicity} \times \text{mole molecules}$.
For oxygen $(O_2)$,atomicity is $2$.
Therefore,$\text{mole atoms} = 2 \times 2.5 = 5 \ mol$ of atoms.

(D) The molar mass of water $(H_2O) = (2 \times 1) + 16 = 18 \ g/mol$.
The number of moles in $1.8 \ g$ of water $= \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol = \frac{1}{10} \ mol$.
The number of molecules of water $= \text{moles} \times N_A = \frac{N_A}{10}$.
Each molecule of water $(H_2O)$ contains $3$ atoms ($2$ hydrogen atoms and $1$ oxygen atom).
Therefore,the total number of atoms $= 3 \times \text{number of molecules} = 3 \times \frac{N_A}{10}$.

(C) The molar mass of water $(H_2O)$ is $18 \text{ g/mol}$.
The mass of one molecule of water is calculated as $\frac{18 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 2.989 \times 10^{-23} \text{ g}$.
Given the density of water is $1 \text{ g/cm}^3$,the volume of one molecule is calculated using the formula $\text{Volume} = \frac{\text{Mass}}{\text{Density}}$.
$\text{Volume} = \frac{2.989 \times 10^{-23} \text{ g}}{1 \text{ g/cm}^3} \approx 3 \times 10^{-23} \text{ cm}^3$.

(C) $1$ mole of $Mg_{3}(PO_{4})_{2}$ contains $8$ moles of $O$ atoms.
$8$ moles of $O$ atoms $\equiv 1$ mole of $Mg_{3}(PO_{4})_{2}$.
Therefore,$0.25$ moles of $O$ atoms $\equiv \frac{0.25}{8}$ moles of $Mg_{3}(PO_{4})_{2}$.
Calculation: $\frac{0.25}{8} = 0.03125 = 3.125 \times 10^{-2}$ moles.

(D) The molar mass of oxygen $(O_2)$ is $32 \ g/mol$.
Calculation of mass:
Mass $= \text{moles} \times \text{molar mass} = \frac{1}{4} \times 32 \ g = 8 \ g$.
Calculation of volume:
At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$.
Volume $= \text{moles} \times 22.4 \ L = \frac{1}{4} \times 22.4 \ L = 5.6 \ L$.

(B) Density = Mass / Volume.
Mass of water = $1 \ g/mL \times 18 \ mL = 18 \ g$.
Moles of water = $18 \ g / 18 \ g/mol = 1 \ mol$.
$1 \ mol$ of $H_2O$ contains $6.022 \times 10^{23}$ molecules of $H_2O$.
Each $H_2O$ molecule contains $2$ electrons from $2H$ atoms and $8$ electrons from $1O$ atom,so total electrons per molecule = $2 + 8 = 10 \ e^-$.
Total electrons = $1 \ mol \times 6.022 \times 10^{23} \text{ molecules/mol} \times 10 \text{ electrons/molecule} = 6.022 \times 10^{24} \ e^-$.

(A) Given: Number of molecules $= 6.022 \times 10^{20}$.
Volume of solution $= 100 \ mL = 0.1 \ L$.
Number of moles of urea $= \frac{\text{Number of molecules}}{N_A} = \frac{6.022 \times 10^{20}}{6.022 \times 10^{23}} = 10^{-3} \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$.
$M = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M$.

(D) At $NTP$,the molar volume of an ideal gas is $22.4 \ L \ mol^{-1}$.
We know that $1 \ m^3 = 1000 \ L$.
Number of moles $n = \frac{\text{Volume in Liters}}{\text{Molar Volume at NTP}}$.
$n = \frac{1000 \ L}{22.4 \ L \ mol^{-1}} \approx 44.6 \ mol$.

(C) At $STP$,$22400 \ mL$ of gas contains $6.022 \times 10^{23}$ molecules (Avogadro's number,$N_A$).
Number of molecules $= \frac{\text{Given volume}}{\text{Molar volume at } STP} \times N_A$
Number of molecules $= \frac{1.12 \times 10^{-7} \ mL}{22400 \ mL/mol} \times 6.022 \times 10^{23} \ mol^{-1}$
Number of molecules $= \frac{1.12 \times 10^{-7}}{2.24 \times 10^4} \times 6.022 \times 10^{23}$
Number of molecules $= 0.5 \times 10^{-11} \times 6.022 \times 10^{23}$
Number of molecules $= 3.011 \times 10^{12}$

(B) The Avogadro number is $N_A = 6.023 \times 10^{23}$.
Time taken in seconds $= \frac{6.023 \times 10^{23}}{10^{20}} = 6023 \text{ seconds}$.
To convert seconds into hours,divide by $3600$:
Time in hours $= \frac{6023}{3600} \approx 1.673 \text{ hours}$.

(C) To find the number of atoms,we calculate the number of moles and multiply by the atomicity of the molecule.
$1$. For $16 \, g \, O_2$: Moles = $16 / 32 = 0.5 \, mol$. Atoms = $0.5 \times 2 \times N_A = 1.0 \, N_A$.
$2$. For $14 \, g \, N_2$: Moles = $14 / 28 = 0.5 \, mol$. Atoms = $0.5 \times 2 \times N_A = 1.0 \, N_A$.
$3$. For $2 \, g \, H_2$: Moles = $2 / 2 = 1.0 \, mol$. Atoms = $1.0 \times 2 \times N_A = 2.0 \, N_A$.
$4$. For $6 \, g \, I_2$: Moles = $6 / 254 \approx 0.023 \, mol$. Atoms = $0.023 \times 2 \times N_A \approx 0.046 \, N_A$.
Comparing the values,$2 \, g \, H_2$ contains the maximum number of atoms.

(A) The molar mass of $SO_2Cl_2$ is calculated as: $32 + (2 \times 16) + (2 \times 35.5) = 32 + 32 + 71 = 135 \ g/mol$.
Number of moles of $SO_2Cl_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{13.5 \ g}{135 \ g/mol} = 0.1 \ mol$.
Total number of molecules = $\text{Number of moles} \times N_A = 0.1 \times N_A = 0.1 \ N_A$.

(D) To find the number of oxygen atoms,we calculate the number of moles of atoms for each case:
$1$. For $1 \ g$ of $O$: Moles of atoms = $\frac{1}{16} = 0.0625 \ mol$.
$2$. For $1 \ g$ of $O_2$: Moles of molecules = $\frac{1}{32} \ mol$. Moles of atoms = $2 \times \frac{1}{32} = \frac{1}{16} = 0.0625 \ mol$.
$3$. For $1 \ g$ of $O_3$: Moles of molecules = $\frac{1}{48} \ mol$. Moles of atoms = $3 \times \frac{1}{48} = \frac{1}{16} = 0.0625 \ mol$.
Since the number of moles of oxygen atoms is the same $(0.0625 \ mol)$ in all three cases,the number of atoms is identical.

(D) Density of water is $1 \, g/mL$. Therefore,$1 \, mL$ of water has a mass of $1 \, g$.
Molar mass of water $(H_2O)$ is $18 \, g/mol$.
Number of moles in $1 \, mL$ of water = $\frac{1 \, g}{18 \, g/mol} = 0.0556 \, mol$.
Number of molecules in $1 \, mL$ of water = $0.0556 \times 6.022 \times 10^{23} \approx 3.346 \times 10^{22}$ molecules.
Since $1 \, mL$ contains $20$ drops,the number of molecules in $1$ drop = $\frac{3.346 \times 10^{22}}{20} = 1.673 \times 10^{21}$ molecules.

(A) The molar mass of methane $(CH_4)$ is calculated as:
$M = 12 + (4 \times 1) = 16 \ g/mol$.
Given number of moles $(n)$ = $0.1 \ mol$.
Mass $(m)$ = $n \times M$.
$m = 0.1 \ mol \times 16 \ g/mol = 1.6 \ g$.

(A) According to Avogadro's Law,equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
Since the volume of $N_2$ and gas $x$ are equal ($10 \ dm^3$ each) at the same temperature and pressure,they must contain the same number of molecules.
However,the question implies identifying a gas that is chemically equivalent or has the same molar mass as $N_2$ $(28 \ g/mol)$.
$N_2$ molar mass $= 2 \times 14 = 28 \ g/mol$.
$CO$ molar mass $= 12 + 16 = 28 \ g/mol$.
Thus,$CO$ is the gas $x$.

(A) The number of atoms is proportional to the number of moles $(n = \text{mass} / \text{atomic mass})$.
$1$. For $24 \ g$ of $C$: $n = 24 / 12 = 2 \ mol$.
$2$. For $56 \ g$ of $Fe$: $n = 56 / 56 = 1 \ mol$.
$3$. For $27 \ g$ of $Al$: $n = 27 / 27 = 1 \ mol$.
$4$. For $108 \ g$ of $Ag$: $n = 108 / 108 = 1 \ mol$.
Since $2 \ mol$ is the highest value,$24 \ g$ of $C$ contains the greatest number of atoms.

(A) The molar mass of $CO_2$ is calculated as: $12 + (2 \times 16) = 44 \ g/mol$.
Given mass of $CO_2 = 44 \ g$.
Number of moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{44 \ g}{44 \ g/mol} = 1 \ mol$.
Number of molecules = $\text{Number of moles} \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ molecules.

(C) The number of moles $n$ is given by $n = \frac{W}{M} = \frac{V}{22400 \ cm^3}$.
Given $V = 112 \ cm^3$ and molar mass of $CH_4$ $(M)$ = $12 + 4(1) = 16 \ g/mol$.
Substituting the values: $\frac{W}{16} = \frac{112}{22400}$.
$W = \frac{112 \times 16}{22400} = \frac{1792}{22400} = 0.08 \ g$.

(A) The molar mass of $SO_2$ is calculated as: $M(SO_2) = 32 + (2 \times 16) = 64 \ g/mol$.
Number of moles $(n)$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{64 \ g}{64 \ g/mol} = 1 \ mol$.
Number of molecules = $n \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ molecules.