The mole concept Questions in English

(C) $1$ mole of $Mg_3(PO_4)_2$ contains $8$ moles of oxygen atoms.
Since $8$ moles of oxygen atoms are present in $1$ mole of $Mg_3(PO_4)_2$,
Therefore,$0.50$ moles of oxygen atoms will be present in $\frac{0.50}{8} = 0.0625$ moles of $Mg_3(PO_4)_2$.
Thus,the correct answer is $0.0625$ moles.

(C) Step $1$: Calculate the number of moles of $Al$ atoms.
$n(Al) = \frac{\text{mass of } Al}{\text{molar mass of } Al} = \frac{54.0 \ g}{27.0 \ g/mol} = 2.0 \ mol$.
Step $2$: Since the number of atoms is the same,the number of moles of $Mg$ atoms must also be $2.0 \ mol$.
Step $3$: Calculate the mass of $Mg$ atoms.
$\text{mass of } Mg = n(Mg) \times \text{molar mass of } Mg = 2.0 \ mol \times 24.0 \ g/mol = 48.0 \ g$.

(B) $1$. For option $A$: Molar mass of $Al = 27 \ g/mol$. Number of moles $= \frac{27 \ g}{27 \ g/mol} = 1 \ mol$. Number of atoms $= 1 \times N_A$ (Avogadro's number). This statement is correct.
$2$. For option $B$: The term '$27 \ g-atom$' refers to $27 \ moles$ of $Al$ atoms. Therefore,$27 \ moles$ of atoms are present,not $1 \ mole$. This statement is incorrect.
$3$. For option $C$: Mass of $1 \ atom$ of $Al = \frac{\text{Molar mass}}{N_A} = \frac{27}{6.022 \times 10^{23}} \approx 4.48 \times 10^{-23} \ g = 44.8 \times 10^{-24} \ g$. The value $45.09 \times 10^{-24} \ g$ is often used as an approximation or is considered incorrect based on standard $N_A$ values. However,option $B$ is fundamentally incorrect by definition of '$g-atom$'.
$4$. Since the question asks for the incorrect statement,and $B$ is clearly incorrect,$B$ is the answer.

(D) At $NTP$,the molar volume of an ideal gas is $22.4 \, L \, mol^{-1}$.
Given volume $V = 1 \, m^3 = 1000 \, L$.
Number of moles $n = \frac{V}{22.4 \, L \, mol^{-1}} = \frac{1000}{22.4} \approx 44.6 \, mol$.

(A) The molar mass of $CaCl_2$ is $40 + 2 \times 35.5 = 111 \ g/mol$.
Number of moles of $CaCl_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{333 \ g}{111 \ g/mol} = 3 \ mol$.
Each molecule of $CaCl_2$ dissociates to give $2 \ Cl^{-}$ ions.
Therefore,$3 \ mol$ of $CaCl_2$ contains $3 \times 2 = 6 \ mol$ of $Cl^{-}$ ions.
The number of $Cl^{-}$ ions is $6 \ N_A$.

(D) The number of moles of $He$ is calculated as: $n = \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{24 \ g}{4 \ g/mol} = 6 \ mol$.
The number of atoms is given by: $\text{Number of atoms} = n \times N_A = 6 \times N_A = 6 \, N_A$.

(A) The number of molecules is the same if the number of moles is the same.
$n = \frac{\text{Given mass}}{\text{Molar mass}}$.
For $A$: $n(O_2) = \frac{16}{32} = 0.5 \ mol$; $n(N_2) = \frac{14}{28} = 0.5 \ mol$.
Since the number of moles is equal,the number of molecules is the same.

(C) The molar mass of $N_3^-$ is $(3 \times 14) = 42 \ g/mol$.
Number of moles of $N_3^- = \frac{4.2 \ g}{42 \ g/mol} = 0.1 \ mol$.
Each $N_3^-$ ion has $(3 \times 5) + 1 = 16$ valence electrons.
Total number of valence electrons $= 0.1 \ mol \times 16 \times N_A = 1.6 \ N_A$.

(C) The molar mass of $N^{3-}$ is approximately $14 \, g/mol$ (as electrons have negligible mass).
Number of moles of $N^{3-} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.2 \, g}{14 \, g/mol} = 0.3 \, mol$.
One nitrogen atom $(N)$ has $5$ valence electrons. An $N^{3-}$ ion has $5 + 3 = 8$ valence electrons.
Total number of valence electrons = (Number of moles) $\times$ (Avogadro's number) $\times$ (Valence electrons per ion).
Total valence electrons = $0.3 \times 6.022 \times 10^{23} \times 8$.
Total valence electrons = $1.8066 \times 10^{23} \times 8 = 1.44528 \times 10^{24} \approx 1.44 \times 10^{24}$.

(A) The molar mass of $NaCl = 23 + 35.5 = 58.5 \, g/mol$.
Given that $1000 \, g$ of $NaCl$ costs $Rs. 4$.
Therefore,the cost of $1$ mole $(58.5 \, g)$ of $NaCl = \frac{58.5 \times 4}{1000} = 0.234 \, Rs$.
The molar mass of table sugar $(C_{12}H_{22}O_{11}) = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \, g/mol$.
Given that $1000 \, g$ of sugar costs $Rs. 30$.
Therefore,the cost of $1$ mole $(342 \, g)$ of sugar $= \frac{342 \times 30}{1000} = 10.26 \, Rs$.
Thus,the prices are $0.234 \, Rs$ and $10.26 \, Rs$ respectively.

(C) The number of molecules in a given mass is calculated as: $\text{Number of molecules} = \frac{\text{Mass}}{\text{Molar Mass}} \times N_A$,where $N_A$ is Avogadro's number $(6.022 \times 10^{23} \text{ mol}^{-1})$.
For $1 \text{ g}$ of each substance to have the same number of molecules,their molar masses must be equal.
Let's calculate the molar masses:
$(I) \, N_2O = (2 \times 14) + 16 = 44 \text{ g/mol}$; $CO = 12 + 16 = 28 \text{ g/mol}$. (Not equal)
$(II) \, N_2 = 28 \text{ g/mol}$; $CO_2 = 12 + (2 \times 16) = 44 \text{ g/mol}$. (Not equal)
$(III) \, N_2 = 28 \text{ g/mol}$; $CO = 28 \text{ g/mol}$. (Equal)
$(IV) \, N_2O = 44 \text{ g/mol}$; $CO_2 = 44 \text{ g/mol}$. (Equal)
Thus,pairs $(III)$ and $(IV)$ contain the same number of molecules for $1 \text{ g}$ of each substance.

(C) Given that moles of $SO_2$ = moles of $CH_4$.
The molar mass of $SO_2$ is $64 \, g/mol$ and the molar mass of $CH_4$ is $16 \, g/mol$.
Using the formula $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$,we have:
$\frac{1.24}{64} = \frac{W}{16}$
$W = \frac{1.24 \times 16}{64}$
$W = \frac{1.24}{4} = 0.31 \, g$.

(A) To find the number of atoms,we calculate the number of moles of molecules and multiply by the number of atoms per molecule.
$A)$ $1 \ g$ of $C_4H_{10}$: Molar mass = $58 \ g/mol$. Moles of molecules = $1/58$. Atoms per molecule = $14$. Total atoms $\propto 14/58 \approx 0.241$.
$B)$ $1 \ g$ of $N_2$: Molar mass = $28 \ g/mol$. Moles of molecules = $1/28$. Atoms per molecule = $2$. Total atoms $\propto 2/28 \approx 0.071$.
$C)$ $1 \ g$ of $Ag$: Molar mass = $108 \ g/mol$. Moles of atoms = $1/108 \approx 0.009$.
$D)$ $1 \ g$ of $H_2O$: Molar mass = $18 \ g/mol$. Moles of molecules = $1/18$. Atoms per molecule = $3$. Total atoms $\propto 3/18 \approx 0.167$.
Comparing the values,$1 \ g$ of butane has the highest number of atoms.

(D) The number of molecules is directly proportional to the number of moles $(n = \frac{\text{mass}}{\text{molar mass}})$.
$A) \, n = \frac{18 \, g}{18 \, g/mol} = 1 \, mol$
$B) \, n = \frac{22 \, g}{44 \, g/mol} = 0.5 \, mol$
$C) \, n = \frac{16 \, g}{16 \, g/mol} = 1 \, mol$
$D) \, n = \frac{20 \, g}{46 \, g/mol} \approx 0.435 \, mol$
Since $0.435 < 0.5 < 1$,the sample with $20 \, g \, C_2H_5OH$ has the least number of molecules.

(C) The number of moles of $Al$ is equal to the number of moles of $Mg$ because the number of atoms is the same.
Using the formula: $\frac{W_{Al}}{M_{Al}} = \frac{W_{Mg}}{M_{Mg}}$
Given: $W_{Al} = 54 \ g$,$M_{Al} = 27 \ g/mol$,$M_{Mg} = 24 \ g/mol$
Substituting the values: $\frac{54}{27} = \frac{W_{Mg}}{24}$
$2 = \frac{W_{Mg}}{24}$
$W_{Mg} = 2 \times 24 = 48 \ g$

(A) The molar mass of $SO_2Cl_2 = 32 + (2 \times 16) + (2 \times 35.5) = 32 + 32 + 71 = 135 \ g/mol$.
Number of gram-molecules (moles) $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{13.5 \ g}{135 \ g/mol} = 0.1 \ mol$.

(B) The molar mass of $H_2O$ is $18 \ g/mol$.
Moles of $H_2O = \frac{27 \ g}{18 \ g/mol} = 1.5 \ mol$.
Each molecule of $H_2O$ contains $2$ atoms of $H$.
Therefore,moles of $H$ atoms $= 1.5 \ mol \times 2 = 3 \ mol$.
Number of $H$ atoms $= 3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{24}$.

(D) The number of moles of $C$ atoms is given by $n = \frac{1.3699 \times 10^{23}}{6.022 \times 10^{23}} = 0.2275 \ mol$.
Since one molecule of $CO_2$ contains one $C$ atom,the moles of $CO_2$ are equal to the moles of $C$ atoms,which is $0.2275 \ mol$.
The molar mass of $CO_2$ is $44 \ g/mol$.
Mass of $CO_2 = \text{moles} \times \text{molar mass} = 0.2275 \ mol \times 44 \ g/mol = 10 \ g$.

(A) The molar mass of ethanol $[C_2H_5OH]$ is $(2 \times 12) + (6 \times 1) + 16 = 46 \, g/mol$.
Number of moles of ethanol $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{0.46 \, g}{46 \, g/mol} = 0.01 \, mol = 10^{-2} \, mol$.
Each molecule of $C_2H_5OH$ contains $6$ hydrogen atoms.
Therefore,the number of moles of $H$ atoms $= 6 \times 10^{-2} \, mol$.
Number of $H$ atoms $= \text{moles} \times N_A = 6 \times 10^{-2} \times 6.022 \times 10^{23} \approx 3.61 \times 10^{22}$ atoms.
Thus,the correct option is $A$.

(D) Moles of $Fe = \frac{280}{56} = 5 \ mol$.
Number of atoms in $Fe = 5 \times N_A$.
Moles of $N$ in $35 \ g = \frac{35}{14} = 2.5 \ mol$.
Number of atoms in $N = 2.5 \times N_A$.
Moles of $H$ in $10 \ g = \frac{10}{1} = 10 \ mol$.
Number of atoms in $H = 10 \times N_A$.
Since $5 \times N_A$ is not equal to $2.5 \times N_A$ and is not half of $10 \times N_A$ (which is $5 \times N_A$),the correct answer is $D$.

(B) The chemical formula $Mg_3(PO_4)_2$ contains $8$ oxygen atoms per formula unit.
Therefore,$1$ mole of $Mg_3(PO_4)_2$ contains $8$ moles of oxygen atoms.
To find the moles of $Mg_3(PO_4)_2$ containing $0.25$ moles of oxygen atoms,we use the ratio:
$\text{Moles of } Mg_3(PO_4)_2 = \frac{\text{Moles of oxygen atoms}}{8}$
$\text{Moles of } Mg_3(PO_4)_2 = \frac{0.25}{8} = 0.03125 = 3.125 \times 10^{-2}$.

(A) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g/mol$.
Mass = Number of moles $\times$ Molar mass.
Mass = $0.1 \ mol \times 16 \ g/mol = 1.6 \ g$.

(C) According to Avogadro's Law,at constant temperature and pressure,the number of molecules is directly proportional to the volume of the gas $(N \propto V)$.
Therefore,$\frac{N_1}{N_2} = \frac{V_1}{V_2}$.
Given: $V_1 = 500 \ mL$,$N_1 = 6.0 \times 10^{21}$ molecules,$V_2 = 100 \ mL$.
$\frac{6.0 \times 10^{21}}{N_2} = \frac{500 \ mL}{100 \ mL}$.
$N_2 = \frac{6.0 \times 10^{21} \times 100}{500} = \frac{6.0 \times 10^{21}}{5} = 1.2 \times 10^{21}$ molecules.

(B) Moles of $O_2 = \frac{5.6 \, L}{22.4 \, L/mol} = 0.25 \, \text{mol}$.
Since each molecule of $O_2$ contains $2$ atoms of oxygen,the moles of oxygen atoms $= 0.25 \times 2 = 0.5 \, \text{mol}$.
Number of oxygen atoms $= \text{moles} \times N_A = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \, \text{atoms}$.

(B) The number of moles of $N_2$ is $n_1 = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$.
The number of moles of $O_2$ is $n_2 = \frac{8 \ g}{32 \ g/mol} = 0.25 \ mol$.
The mole fraction of $N_2$ is $X_{N_2} = \frac{n_1}{n_1 + n_2} = \frac{0.25}{0.25 + 0.25} = 0.5$.
The mole fraction of $O_2$ is $X_{O_2} = \frac{n_2}{n_1 + n_2} = \frac{0.25}{0.25 + 0.25} = 0.5$.
Therefore,$X_{N_2} = X_{O_2}$.

(A) Step $1$: Calculate the number of moles of $NaOH$.
Number of moles = $\frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{3.01 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.05 \, mol$.
Step $2$: Convert volume to liters.
Volume = $500 \, mL = 0.5 \, L$.
Step $3$: Calculate Molarity $(M)$.
$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.05 \, mol}{0.5 \, L} = 0.1 \, M$.

(B) The chemical formula of ammonium phosphate is $(NH_4)_3PO_4$.
In one mole of $(NH_4)_3PO_4$,there are $12$ moles of hydrogen atoms $(3 \times 4 = 12)$ and $4$ moles of oxygen atoms.
Let $n_H$ be the moles of hydrogen atoms and $n_O$ be the moles of oxygen atoms.
The ratio of moles of hydrogen atoms to oxygen atoms is $\frac{n_H}{n_O} = \frac{12}{4} = 3$.
Given that $n_H = 6$ moles,we have $\frac{6}{n_O} = 3$.
Therefore,$n_O = \frac{6}{3} = 2$ moles.

(D) To find the number of atoms,we calculate the moles of molecules and multiply by the atomicity of the substance.
$A) \ 12 \ g \ He: \text{moles} = \frac{12}{4} = 3 \ mol \text{ atoms}$.
$B) \ 1.8 \ g \ H_2O: \text{moles} = \frac{1.8}{18} = 0.1 \ mol \text{ molecules}$. $\text{Atoms} = 0.1 \times 3 = 0.3 \ mol \text{ atoms}$.
$C) \ 22 \ g \ CO_2: \text{moles} = \frac{22}{44} = 0.5 \ mol \text{ molecules}$. $\text{Atoms} = 0.5 \times 3 = 1.5 \ mol \text{ atoms}$.
$D) \ 2.45 \ g \ H_2SO_4: \text{moles} = \frac{2.45}{98} = 0.025 \ mol \text{ molecules}$. $\text{Atoms} = 0.025 \times 7 = 0.175 \ mol \text{ atoms}$.
Comparing the values: $3 > 1.5 > 0.3 > 0.175$. Thus,$2.45 \ g \ H_2SO_4$ has the minimum number of atoms.

(A) To find the number of molecules,we calculate the number of moles for each substance using the formula: $n = \frac{\text{mass}}{\text{molar mass}}$.
$A) \ 36 \ g \ H_2O: n = \frac{36}{18} = 2 \ \text{moles}$.
$B) \ 28 \ g \ CO: n = \frac{28}{28} = 1 \ \text{mole}$.
$C) \ 46 \ g \ C_2H_5OH: n = \frac{46}{46} = 1 \ \text{mole}$.
$D) \ 54 \ g \ N_2O_5: n = \frac{54}{108} = 0.5 \ \text{moles}$.
Since the number of molecules is directly proportional to the number of moles $(N = n \times N_A)$,the substance with the highest number of moles will have the largest number of molecules. Thus,$36 \ g \ H_2O$ contains the largest number of molecules.

(D) At $NTP$,$1 \ mole$ of any gas occupies $22.4 \ L$ and contains $6.022 \times 10^{23}$ molecules.
For $O_2$ gas,the molar mass is $32 \ g/mol$.
$A$: $32 \ g$ of $O_2 = 32/32 = 1 \ mole$.
$B$: $22.4 \ L$ of $O_2$ at $NTP = 1 \ mole$.
$C$: $6.022 \times 10^{23}$ molecules of $O_2 = 1 \ mole$.
$D$: $16 \ g$ of $O_2 = 16/32 = 0.5 \ mole$.
Therefore,$16 \ g$ of $O_2$ does not represent $1 \ mole$.

(C) The number of moles of $CO_2$ is calculated as: $\text{Moles of } CO_2 = \frac{\text{Volume at } STP}{22.4 \ L/mol} = \frac{2.8 \ L}{22.4 \ L/mol} = 0.125 \ mol$.
Each molecule of $CO_2$ contains $2$ atoms of oxygen.
Therefore,the moles of oxygen $= 0.125 \ mol \times 2 = 0.25 \ mol$.

(D) The chemical formula $[Cu(NH_3)_4]SO_4$ indicates that $1 \ mol$ of the compound contains $4 \ mol$ of $NH_3$ molecules.
First,calculate the total moles of $NH_3$ molecules in the sample:
$\text{Moles of } NH_3 = \frac{2.4 \times 10^{24}}{6.022 \times 10^{23}} \approx 4 \ mol$.
Since $4 \ mol$ of $NH_3$ are present in $1 \ mol$ of $[Cu(NH_3)_4]SO_4$,the number of moles of the compound is $\frac{4 \ mol}{4} = 1 \ mol$.

(B) The chemical formula of the compound is $(COOH)_2 \cdot 2H_2O$.
One molecule of $(COOH)_2 \cdot 2H_2O$ contains $6$ oxygen atoms ($4$ from the oxalic acid part and $2$ from the water of crystallization).
Therefore,$1 \ mol$ of $(COOH)_2 \cdot 2H_2O$ contains $6 \ mol$ of oxygen atoms.
For $0.3 \ mol$ of $(COOH)_2 \cdot 2H_2O$,the number of moles of oxygen atoms is $0.3 \times 6 = 1.8 \ mol$.
Since $1 \ mol$ of atoms is equivalent to $1 \ gram \ atom$,the number of gram atoms of oxygen is $1.8$.

(C) One mole of any substance contains $6.022 \times 10^{23}$ particles (Avogadro's number). Therefore,equal moles of different substances contain the same number of constituent particles. The Assertion is correct.
The number of particles in a given weight depends on the molar mass of the substance $(n = \frac{mass}{molar \ mass})$. Since different substances have different molar masses,equal weights of different substances will contain different numbers of moles and thus different numbers of constituent particles. The Reason is incorrect.

(A) To find the number of molecules,we calculate the number of moles $(n)$ in each case,as $Number \; of \; molecules = n \times N_A$.
$A$) $18 \; mL$ of water: Since density $d = 1 \; g/mL$,mass $W = 18 \; g$. Moles $n = \frac{18}{18} = 1 \; mol$. Molecules $= 1 \times N_A$.
$B$) $0.18 \; g$ of water: Moles $n = \frac{0.18}{18} = 0.01 \; mol$. Molecules $= 0.01 \times N_A$.
$C$) $0.00224 \; L$ of water vapour at $STP$: Moles $n = \frac{0.00224}{22.4} = 0.0001 \; mol$. Molecules $= 0.0001 \times N_A$.
$D$) $10^{-3} \; mol$ of water: Moles $n = 0.001 \; mol$. Molecules $= 0.001 \times N_A$.
Comparing the values,$1 \times N_A$ is the maximum. Thus,the correct option is $A$.

$i.$ $1$ mole of $C_{2}H_{6}$ contains $2$ moles of carbon atoms.
$\therefore$ Number of moles of carbon atoms in $3$ moles of $C_{2}H_{6} = 2 \times 3 = 6 \text{ moles}$.
$ii.$ $1$ mole of $C_{2}H_{6}$ contains $6$ moles of hydrogen atoms.
$\therefore$ Number of moles of hydrogen atoms in $3$ moles of $C_{2}H_{6} = 3 \times 6 = 18 \text{ moles}$.
$iii.$ $1$ mole of $C_{2}H_{6}$ contains $6.022 \times 10^{23}$ molecules of ethane.
$\therefore$ Number of molecules in $3$ moles of $C_{2}H_{6} = 3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{24} \text{ molecules}$.

(C) $(i).$ $1 \,g$ of $Au_{(s)} = \frac{1}{197} \,mol$ of $Au_{(s)} = \frac{6.022 \times 10^{23}}{197} \approx 3.06 \times 10^{21}$ atoms of $Au_{(s)}$.
$(ii).$ $1 \,g$ of $Na_{(s)} = \frac{1}{23} \,mol$ of $Na_{(s)} = \frac{6.022 \times 10^{23}}{23} \approx 26.2 \times 10^{21}$ atoms of $Na_{(s)}$.
$(iii).$ $1 \,g$ of $Li_{(s)} = \frac{1}{7} \,mol$ of $Li_{(s)} = \frac{6.022 \times 10^{23}}{7} \approx 86.0 \times 10^{21}$ atoms of $Li_{(s)}$.
$(iv).$ $1 \,g$ of $Cl_{2(g)} = \frac{1}{71} \,mol$ of $Cl_{2(g)}$. Since $1 \,mol$ of $Cl_2$ contains $2 \,mol$ of $Cl$ atoms,number of atoms $= \frac{2 \times 6.022 \times 10^{23}}{71} \approx 16.96 \times 10^{21}$ atoms of $Cl$.
Comparing the values,$1 \,g$ of $Li_{(s)}$ has the largest number of atoms.

(N/A) $(i)$ $1$ mole of $Ar = 6.022 \times 10^{23}$ atoms of $Ar$
$\therefore 52$ mol of $Ar = 52 \times 6.022 \times 10^{23}$ atoms of $Ar = 3.13144 \times 10^{25}$ atoms of $Ar$
$(ii)$ $1$ atom of $He = 4$ $u$ of $He$
$4$ $u$ of $He = 1$ atom of $He$
$1$ $u$ of $He = \frac{1}{4}$ atom of $He$
$52$ $u$ of $He = \frac{52}{4}$ atoms of $He = 13$ atoms of $He$
$(iii)$ $4$ $g$ of $He = 6.022 \times 10^{23}$ atoms of $He$
$\therefore 52$ $g$ of $He = \frac{6.022 \times 10^{23} \times 52}{4}$ atoms of $He = 7.8286 \times 10^{24}$ atoms of $He$

(A) Molar mass of dinitrogen $(N_2) = 28 \,g \,mol^{-1}$.
Number of moles of $N_2 = \frac{1.4 \,g}{28 \,g \,mol^{-1}} = 0.05 \,mol$.
Number of molecules of $N_2 = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22}$ molecules.
Each $N$ atom has $7$ electrons,so $1$ molecule of $N_2$ has $14$ electrons.
Total electrons $= 14 \times 3.011 \times 10^{22} = 4.2154 \times 10^{23}$ electrons.

(N/A) The Avogadro number is $6.02 \times 10^{23}$.
The time required in seconds is calculated as:
$\text{Time} = \frac{6.02 \times 10^{23}}{10^{10}} \ s = 6.02 \times 10^{13} \ s$.
To convert this into years:
$\text{Time in years} = \frac{6.02 \times 10^{13}}{60 \times 60 \times 24 \times 365} \ \text{years} \approx 1.909 \times 10^{6} \ \text{years}$.
Hence,the time taken would be approximately $1.909 \times 10^{6} \ \text{years}$.

Atoms and molecules are extremely small in size,and their numbers in even a small amount of any substance are very large.
In the $SI$ system,the mole (symbol: $mol$) was introduced as the seventh base quantity for the amount of a substance.
One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly $12 \ g$ (or $0.012 \ kg$) of the ${}^{12}C$ isotope.
To determine this number precisely,the mass of a carbon-$12$ atom was determined by a mass spectrometer and found to be $1.992648 \times 10^{-23} \ g$.
Knowing that one mole of carbon weighs $12 \ g$,the number of atoms in it is calculated as:
$\frac{12 \ g/mol \ {}^{12}C}{1.992648 \times 10^{-23} \ g/{}^{12}C \text{ atom}} = 6.0221367 \times 10^{23} \text{ atoms}/mol$.
This number of entities in $1 \ mol$ is known as the 'Avogadro constant',denoted by $N_{A}$,in honor of Amedeo Avogadro. Therefore,$1 \ mol$ of hydrogen atoms $= 6.022 \times 10^{23}$ atoms.

(C) The number of atoms is calculated as: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{atomicity}$.
$(i)$ $1 \ g$ $Au$: $\frac{1}{197} \times N_A \approx 0.00507 \times N_A$ atoms.
$(ii)$ $1 \ g$ $Na$: $\frac{1}{23} \times N_A \approx 0.04347 \times N_A$ atoms.
$(iii)$ $1 \ g$ $Li$: $\frac{1}{7} \times N_A \approx 0.14285 \times N_A$ atoms.
$(iv)$ $1 \ g$ $Cl_2$: $\frac{1}{71} \times N_A \times 2 \approx 0.02816 \times N_A$ atoms.
Comparing the values,$1 \ g$ $Li$ has the maximum number of atoms.

The mass of a $^{12}C$ (carbon-$12$) atom is calculated using the molar mass and Avogadro's constant $(N_{A})$.
$1$ mole of $^{12}C$ atoms has a mass of $12 \ g$ and contains $6.022 \times 10^{23}$ atoms.
Therefore,the mass of $1$ atom of $^{12}C$ is given by:
$\text{Mass of one atom} = \frac{\text{Molar mass}}{N_{A}}$
$= \frac{12 \ g}{6.022 \times 10^{23}}$
$= 1.992648 \times 10^{-23} \ g \approx 1.99 \times 10^{-23} \ g$

(N/A) $(i)$ $1 \ mol$ of $Ar = 6.022 \times 10^{23}$ atoms of $Ar$.
$52 \ mol$ of $Ar = 52 \times 6.022 \times 10^{23} = 3.131 \times 10^{25}$ atoms of $Ar$.
$(ii)$ Atomic mass of $He = 4 \ u$.
$4 \ u$ is the mass of $1$ atom of $He$.
Therefore,$52 \ u$ of $He$ contains $= \frac{52}{4} = 13$ atoms of $He$.
$(iii)$ Gram atomic mass of $He = 4 \ g$.
$4 \ g$ of $He$ contains $6.022 \times 10^{23}$ atoms.
Therefore,$52 \ g$ of $He$ contains $= \frac{6.022 \times 10^{23} \times 52}{4} = 7.8286 \times 10^{24}$ atoms.

$(i)$ $1 \ mol$ of ethane $(C_2H_6)$ contains $2 \ mol$ of carbon atoms. Therefore,$3 \ mol$ of ethane contains $3 \times 2 = 6 \ mol$ of carbon atoms.
$(ii)$ $1 \ mol$ of ethane $(C_2H_6)$ contains $6 \ mol$ of hydrogen atoms. Therefore,$3 \ mol$ of ethane contains $3 \times 6 = 18 \ mol$ of hydrogen atoms.
$(iii)$ $1 \ mol$ of ethane contains $6.022 \times 10^{23}$ molecules. Therefore,$3 \ mol$ of ethane contains $3 \times 6.022 \times 10^{23} = 18.066 \times 10^{23}$ molecules of ethane.

(B) $0.50 \ mol$ $Na_2CO_3$ represents the absolute amount of substance (number of moles) of $Na_2CO_3$ present.
$0.50 \ M$ $Na_2CO_3$ represents the molar concentration (molarity),which is defined as the number of moles of solute present in $1 \ L$ of solution.
Therefore,$0.50 \ M$ $Na_2CO_3$ means $0.50 \ mol$ of $Na_2CO_3$ is dissolved in $1 \ L$ of solution.

(N/A) The symbol for the $SI$ unit of mole is $mol$.
One mole is defined as the amount of a substance that contains as many elementary entities as there are atoms in exactly $12 \ g$ $(0.012 \ kg)$ of the $^{12}C$ isotope.
$12 \ g$ of $^{12}C$ isotope $= 1 \ mole$.

(B) The atomic number of nitrogen $(N)$ is $7$,which means a neutral nitrogen atom has $7$ electrons.
For the $N^{3-}$ ion,it has gained $3$ electrons,so the total number of electrons in one $N^{3-}$ ion is $7 + 3 = 10$ electrons.
One mole of $N^{3-}$ ions contains $6.022 \times 10^{23}$ ions.
Therefore,the total number of electrons in one mole of $N^{3-}$ ions is $10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}$ electrons.

(B) $1$. Calculate the number of moles of $_{16}^{32}S$: $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.192 \ g}{32 \ g/mol} = 0.006 \ mol$.
$2$. Calculate the number of atoms: $\text{Atoms} = 0.006 \times 6.022 \times 10^{23} = 3.6132 \times 10^{21} \text{ atoms}$.
$3$. For $_{16}^{32}S$,the number of protons = $16$,electrons = $16$,and neutrons = $32 - 16 = 16$.
$4$. Total subatomic particles per atom = $16 + 16 + 16 = 48$.
$5$. Total number of particles = $3.6132 \times 10^{21} \times 48 = 1.734 \times 10^{23}$.

(N/A) In $1811$,Italian scientist Amedeo Avogadro combined the conclusions of Dalton's atomic theory and Gay-Lussac's law of combining volumes,which is now known as Avogadro's Law.
Avogadro's Law: It states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Mathematical Formula: According to Avogadro's Law,as temperature and pressure remain constant,the volume depends upon the number of molecules of the gas,or in other words,the amount of the gas.
$V \propto n$ (constant $T$ and $p$) .....(Eq.-$i$)
where,$n =$ number of moles of the gas.
$V = k_4 n$ (Eq.-$ii$)
Molar volume at $STP = 22.7 \ L$.
Atoms in one mole $= 6.022 \times 10^{23}$.
The relation between volume and density of a gas can be obtained as $M = k_4 d$.
Calculation of moles of gas:
Gaseous mole $(n)$ $= \frac{\text{Weight of gas (} m \text{)}}{\text{Molecular mass of gas (} M \text{)}}$ ....(Eq.-$i$)
$\therefore n = \frac{m}{M}$
where,$m =$ weight of gas,$M =$ molecular mass of gas.
According to Avogadro's law formula:
$V = k_4 n$ ....(Eq.-$ii$)
$\therefore V = k_4 \frac{m}{M}$ .....(Eq.-$iii$)
Thus,$M = k_4 \left( \frac{m}{V} \right)$ .....(Eq.-$iv$)
Since,$\frac{m}{V} =$ density of gas $= d$
$\therefore M = k_4 d$ (Eq.-$v$)
Conclusion: The density of a gas is directly proportional to its molecular mass.