The mole concept Questions in English

(C) At $STP$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.7 \ L \ mol^{-1}$ (using the current $IUPAC$ standard of $1 \ bar$ pressure).
Given volume $V = 1 \ m^3 = 1000 \ L$.
The number of moles $n$ is calculated as:
$n = \frac{V}{V_m} = \frac{1000 \ L}{22.7 \ L \ mol^{-1}} \approx 44.05 \ mol$.
If using the older $STP$ definition ($1 \ atm$ pressure),the molar volume is $22.4 \ L \ mol^{-1}$.
$n = \frac{1000 \ L}{22.4 \ L \ mol^{-1}} \approx 44.64 \ mol$.
Rounding to the nearest value provided in the options,the correct answer is $44.6$.

(C) The molar mass of sodium hydroxide $(NaOH)$ is calculated as:
$M = 23.0 + 16.0 + 1.0 = 40.0 \ g/mol$.
Number of moles $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{0.160 \ g}{40.0 \ g/mol} = 0.004 \ mol$.
Since $1 \ mol = 1000 \ mmol$,the number of millimoles is:
$0.004 \ mol \times 1000 \ mmol/mol = 4.0 \ mmol$.

(A) $1$. Molar mass of methane $(CH_4)$ = $12 + (4 \times 1) = 16 \ g/mol$.
$2$. Number of moles of $CH_4$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{1.6 \ g}{16 \ g/mol} = 0.1 \ mol$.
$3$. Number of molecules of $CH_4$ = $0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules.
$4$. Each molecule of $CH_4$ contains $6$ (from $C$) + $4 \times 1$ (from $H$) = $10$ electrons.
$5$. Total number of electrons = $6.022 \times 10^{22} \times 10 = 6.022 \times 10^{23}$ electrons.

(B) $1$. Calculate the molar mass of methane $(CH_4)$: $12 + 4 \times 1 = 16 \ g/mol$.
$2$. Calculate the number of moles in $3.2 \ g$ of $CH_4$: $n = \frac{3.2 \ g}{16 \ g/mol} = 0.2 \ mol$.
$3$. Calculate the number of molecules: $0.2 \ mol \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.2044 \times 10^{23} \text{ molecules}$.
$4$. Determine the number of electrons in one molecule of $CH_4$: Carbon has $6$ electrons and each Hydrogen has $1$ electron,so $6 + 4(1) = 10$ electrons per molecule.
$5$. Calculate the total number of electrons: $1.2044 \times 10^{23} \text{ molecules} \times 10 \text{ electrons/molecule} = 1.2044 \times 10^{24} \text{ electrons}$.

(A) Step $1$: Calculate the number of moles of urea.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.4 \ g}{60 \ g \ mol^{-1}} = 0.09 \ mol$.
Step $2$: Calculate the number of molecules.
$\text{Number of molecules} = n \times N_A = 0.09 \ mol \times 6.022 \times 10^{23} \ mol^{-1} = 5.4198 \times 10^{22} \approx 5.42 \times 10^{22}$ molecules.

(B) The formula to calculate mass is: $\text{Mass} = \text{Number of moles} \times \text{Molar mass}$.
Given: $\text{Number of moles} = 0.25 \ mol$,$\text{Molar mass} = 56 \ g \ mol^{-1}$.
$\text{Mass} = 0.25 \ mol \times 56 \ g \ mol^{-1} = 14 \ g$.
To convert the mass into kilograms: $14 \ g = 14 \times 10^{-3} \ kg = 1.4 \times 10^{-2} \ kg$.
Therefore,the correct option is $B$.

(C) The chemical formula of urea is $NH_2CONH_2$ or $CH_4N_2O$.
The molar mass of urea is $(12 + 4 \times 1 + 2 \times 14 + 16) = 60 \ g/mol$.
The number of moles of urea in $5.4 \ g$ is $n = \frac{5.4}{60} = 0.09 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,the number of hydrogen atoms = $n \times 4 \times N_A$,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Number of hydrogen atoms = $0.09 \times 4 \times 6.022 \times 10^{23} = 0.36 \times 6.022 \times 10^{23} = 2.16792 \times 10^{23} \approx 2.168 \times 10^{23}$.

(A) $\text{Number of moles} = 5 = \frac{\text{Given mass (g)}}{\text{Molar mass}}$
$\text{For acetic acid, } M = 60 \ g \ mol^{-1}$
$\text{Given mass (g)} = 5 \ mol \times 60 \ g \ mol^{-1} = 300 \ g$
$\text{Mass in } kg = \frac{300 \ g}{1000} = 0.3 \ kg$

(A) Given mass of carbon $= 3.6 \ kg = 3600 \ g$.
Molar mass of carbon $= 12 \ g/mol$.
The number of moles is calculated as:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{3600 \ g}{12 \ g/mol} = 300 \ mol$.
This can be expressed as $3.0 \times 10^2 \ mol$.

(C) The chemical formula of cytosine is $C_4H_5N_3O$.
From the chemical formula,it is evident that one molecule of cytosine contains $3$ nitrogen atoms.
Therefore,one mole of cytosine contains $3$ moles of nitrogen atoms.

(D) The molar mass of water $(H_2O)$ is $18 \times 10^{-3} \ kg \ mol^{-1}$.
Number of moles $(n) = \frac{\text{Given mass}}{\text{Molar mass}}$.
$n = \frac{9.10 \times 10^{-2} \ kg}{18 \times 10^{-3} \ kg \ mol^{-1}}$.
$n = \frac{91 \times 10^{-3}}{18 \times 10^{-3}} = 5.055 \approx 5.0 \ mol$.

(B) The molar mass of water $(H_2O)$ is $18 \,g/mol$.
$1 \,mol$ of water contains $6.022 \times 10^{23}$ molecules.
Mass of $6.022 \times 10^{23}$ molecules of water $= 18 \,g$.
Therefore, the mass of $1$ molecule of water $= \frac{18 \,g}{6.022 \times 10^{23}} = 2.988 \times 10^{-23} \,g$.
Given that $\text{density} = \frac{\text{mass}}{\text{volume}}$, the volume occupied by $1$ molecule is $\text{Volume} = \frac{\text{mass}}{\text{density}}$.
$\text{Volume} = \frac{2.988 \times 10^{-23} \,g}{1 \,g \,cm^{-3}} = 2.988 \times 10^{-23} \,cm^3 \approx 2.98 \times 10^{-23} \,cm^3$.

(A) The chemical formula of carnallite is $KCl \cdot MgCl_2 \cdot 6 H_2O$.
From the formula,it is evident that $1$ mole of carnallite contains $6$ moles of water molecules.

(C) The number of moles is calculated using the formula: $\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass}}$
Given mass $= 6.9 \times 10^{-2} \ kg = 6.9 \times 10^{-2} \times 10^3 \ g = 69 \ g$
Molar mass of sodium $= 23 \ g \ mol^{-1}$
Number of moles $= \frac{69 \ g}{23 \ g \ mol^{-1}} = 3 \ mol$

(B) The chemical formula of ammonia is $NH_3$.
The molar mass of $NH_3 = 14 + (3 \times 1) = 17 \ g \ mol^{-1}$.
Mass in grams $= \text{moles} \times \text{molar mass} = 2.5 \ mol \times 17 \ g \ mol^{-1} = 42.5 \ g$.
To convert the mass into $kg$,divide by $1000$:
Mass in $kg = \frac{42.5 \ g}{1000} = 0.0425 \ kg = 4.25 \times 10^{-2} \ kg$.

(B) The chemical formula of ammonium nitrate is $NH_4NO_3$.
The molar mass of $NH_4NO_3 = (14 + 4 \times 1 + 14 + 3 \times 16) \ g/mol = 80 \ g/mol$.
Number of moles of $NH_4NO_3 = \frac{8 \ g}{80 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of $NH_4NO_3$ contains $2 \ mol$ of nitrogen atoms,
$0.1 \ mol$ of $NH_4NO_3$ contains $0.1 \times 2 = 0.2 \ mol$ of nitrogen atoms.

(A) For $4.25 \ g$ of $NH_3$: Moles of $NH_3 = \frac{4.25 \ g}{17 \ g/mol} = 0.25 \ mol$. Total atoms $= 0.25 \ mol \times (1 + 3) \ atoms/molecule = 1 \ mol$ of atoms.
For $1.8 \ g$ of $H_2O$: Moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times (2 + 1) = 0.3 \ mol$ of atoms.
For $2.8 \ g$ of $N_2$: Moles of $N_2 = \frac{2.8 \ g}{28 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times 2 = 0.2 \ mol$ of atoms.
For $4.4 \ g$ of $CO_2$: Moles of $CO_2 = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$. Total atoms $= 0.1 \ mol \times (1 + 2) = 0.3 \ mol$ of atoms.
Thus,$4.25 \ g$ of $NH_3$ contains $1 \ mol$ of atoms.

(D) At $STP$,$1 \text{ mole}$ of any gas occupies $22.4 \ L$ volume.
Number of moles $= \frac{\text{Given volume}}{\text{Molar volume at } STP} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \text{ mole} = \frac{1}{4} \text{ mole}$.

(D) Step $1$: Calculate the number of moles of $CO_2$ using the given molar volume at $\text{STP}$.
Number of moles $= \frac{\text{Given Volume}}{\text{Molar Volume}} = \frac{45.42 \ L}{22.71 \ L \ mol^{-1}} = 2 \ mol$.
Step $2$: Calculate the number of molecules using Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Number of molecules $= \text{moles} \times N_A = 2 \ mol \times 6.022 \times 10^{23} \ mol^{-1} = 1.2044 \times 10^{24}$ molecules.

(A) The molar mass of urea $(NH_2CONH_2)$ is $14 + 2(1) + 12 + 16 + 14 + 2(1) = 60 \ g/mol$.
Moles of urea $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6.0 \ g}{60 \ g/mol} = 0.1 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,moles of $H$-atoms $= 0.1 \ mol \times 4 = 0.4 \ mol$.
Number of hydrogen atoms $= \text{moles} \times N_A = 0.4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23} \approx 2.4 \times 10^{23}$.

(B) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g/mol$.
Number of moles of methane $(CH_4) = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{16 \ g}{16 \ g/mol} = 1 \ mol$.
At $STP$,the volume occupied by $1 \ mol$ of any ideal gas is $22.4 \ L$.
Since $1 \ L = 1000 \ cm^3$,the volume is $22.4 \times 1000 \ cm^3 = 22400 \ cm^3$.

(B) The number of moles is calculated using the formula:
$n = \frac{\text{Given mass}}{\text{Molar mass}}$
Given mass $= 5.4 \ g$
Molar mass $= 60 \ g/mol$
$n = \frac{5.4}{60} = 0.09 \ mol$

(B) At $STP$,the molar volume of an ideal gas is $22.4 \ L$ or $22400 \ cm^3$.
Number of moles of gas $= \frac{\text{Volume at STP}}{22400 \ cm^3/mol} = \frac{22400 \ cm^3}{22400 \ cm^3/mol} = 1 \ mol$.
Number of molecules $= \text{moles} \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ molecules.

(C) $\text{Number of moles of gas} = \frac{\text{Given volume}}{22.4 \ dm^3/mol}$
$\text{Number of moles} = \frac{67.2}{22.4} = 3 \ mol$
$\text{Number of molecules} = \text{moles} \times N_A$
$\text{Number of molecules} = 3 \times 6.022 \times 10^{23} \approx 3 \times 6 \times 10^{23}$
$\text{Number of molecules} = 18 \times 10^{23} = 1.8 \times 10^{24}$

(C) The formula for calculating mass from moles is: $\text{Mass} = \text{moles} \times \text{Molar mass}$.
The molar mass of $H_2O$ is $(2 \times 1.008) + 16.00 = 18.016 \ g/mol \approx 18 \ g/mol$.
Given moles $= 0.25 \ mol$.
Therefore,$\text{Mass} = 0.25 \ mol \times 18 \ g/mol = 4.5 \ g$.

(B) The number of molecules is directly proportional to the number of moles $(n = \frac{\text{mass}}{\text{molar mass}})$.
$I. \ n_{Ar} = \frac{13.3 \ g}{39.9 \ g \ mol^{-1}} = 0.33 \ mol$
$II. \ n_{O_{2}} = \frac{24.0 \ g}{32 \ g \ mol^{-1}} = 0.75 \ mol$
$III. \ n_{CO_{2}} = \frac{11 \ g}{44 \ g \ mol^{-1}} = 0.25 \ mol$
$IV. \ n_{CH_{4}} = \frac{16.0 \ g}{16 \ g \ mol^{-1}} = 1.0 \ mol$
Since $CH_{4}$ has the maximum number of moles $(1.0 \ mol)$,it contains the maximum number of molecules.

(D) Volume of $NH_{3}$ gas at $STP = 5.6 \ cm^{3} = 5.6 \times 10^{-3} \ L$ (or $dm^{3}$).
Since $22.4 \ L$ of any gas at $STP$ contains $1 \ mole$.
Number of moles of $NH_{3} = \frac{5.6 \times 10^{-3} \ L}{22.4 \ L/mol} = 2.5 \times 10^{-4} \ mol$.
One molecule of $NH_{3}$ contains $4$ atoms ($1$ Nitrogen and $3$ Hydrogen).
Total number of atoms = (Number of moles) $\times$ (Avogadro's number) $\times$ (Atomicity).
Total number of atoms = $2.5 \times 10^{-4} \times 6.022 \times 10^{23} \times 4 = 6.022 \times 10^{20}$ atoms.

(B) The number of atoms is calculated by the formula: $\text{Number of atoms} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_A$.
Since the mass is $1 \ g$ for all,the number of atoms is inversely proportional to the molar mass.
$I$. $1 \ g \ Fe = \frac{1}{56} \times N_A \approx 0.0178 \times N_A$ atoms.
$II$. $1 \ g \ Au = \frac{1}{197} \times N_A \approx 0.0051 \times N_A$ atoms.
$III$. $1 \ g \ Na = \frac{1}{23} \times N_A \approx 0.0435 \times N_A$ atoms.
$IV$. $1 \ g \ Cu = \frac{1}{63.5} \times N_A \approx 0.0157 \times N_A$ atoms.
Comparing the values,$Na$ has the smallest molar mass,therefore it has the highest number of atoms.

(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$.
Initial amount of $H_2SO_4 = 98 \ mg = 98 \times 10^{-3} \ g = 0.1 \times 10^{-2} \ mol = 1 \times 10^{-3} \ mol$.
Number of molecules in $1 \times 10^{-3} \ mol = 1 \times 10^{-3} \times 6.022 \times 10^{23} = 6.022 \times 10^{20}$ molecules.
Number of molecules removed $= 3.01 \times 10^{20}$.
Number of molecules left $= 6.022 \times 10^{20} - 3.01 \times 10^{20} = 3.012 \times 10^{20}$ molecules.
Number of moles left $= \frac{3.012 \times 10^{20}}{6.022 \times 10^{23}} \approx 0.5 \times 10^{-3} \ mol$.

(A) Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \ g/mol$.
Number of moles of $CO_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.
$1$ molecule of $CO_2$ contains $2$ oxygen atoms.
Therefore,$1$ mole of $CO_2$ contains $2 \times 6.022 \times 10^{23}$ oxygen atoms.
Number of oxygen atoms in $0.1 \ mol$ of $CO_2 = 0.1 \times 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$ oxygen atoms.
Thus,the number of oxygen atoms is approximately $1.2 \times 10^{23}$.

(B) $1$ mole of $O_2 = 6.022 \times 10^{23}$ molecules $= 32 \ g$ of $O_2$.
$1$ molecule of $O_2 = \frac{32}{6.022 \times 10^{23}} \ g$ of $O_2$.
Therefore,$1.8 \times 10^{22}$ molecules of $O_2 = \frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} \ g$.
$= 0.960 \ g$ of $O_2$.

(D) The molar mass of $H_{2}O$ is $18 \ g/mol$.
$18 \ g$ of $H_{2}O$ contains $6.022 \times 10^{23}$ molecules.
Therefore,$0.018 \ g$ of $H_{2}O$ contains:
$\text{Number of molecules} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_{A}$
$= \frac{0.018 \ g}{18 \ g/mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$
$= 0.001 \times 6.022 \times 10^{23}$
$= 6.022 \times 10^{20} \text{ molecules}$.

(A) $1 \ mol$ of $NH_3 = 17 \ g = 22400 \ cm^3$ at $STP$.
Since $22400 \ cm^3$ of $NH_3$ at $STP$ has a mass of $17 \ g$.
Therefore,the mass of $112 \ cm^3$ of $NH_3$ at $STP$ is calculated as:
$\text{Mass} = \frac{17 \ g}{22400 \ cm^3} \times 112 \ cm^3 = 0.085 \ g$.

(B) Given volume of water $= 18 \ mL$ and density $= 1 \ g \ mL^{-1}$.
Mass of water $= \text{density} \times \text{volume} = 1 \ g \ mL^{-1} \times 18 \ mL = 18 \ g$.
Molar mass of $H_2O = (2 \times 1) + 16 = 18 \ g \ mol^{-1}$.
Number of moles of $H_2O = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
One molecule of $H_2O$ contains $2 \times 1 (\text{from } H) + 8 (\text{from } O) = 10$ electrons.
Total number of electrons in $1 \ mol$ of $H_2O = 1 \ mol \times 6.022 \times 10^{23} \text{ molecules/mol} \times 10 \text{ electrons/molecule}$.
Total number of electrons $= 6.022 \times 10^{24}$.

(C) The molar mass of methane $(CH_{4})$ is calculated as: $12 + (4 \times 1) = 16 \ g/mol$.
Mass of $0.1 \ mol$ of methane $= \text{Number of moles} \times \text{Molar mass}$.
Mass $= 0.1 \ mol \times 16 \ g/mol = 1.6 \ g$.

(D) The molar mass of oxygen atoms $(O)$ is $16 \ g/mol$. However,oxygen gas exists as $O_2$.
Number of moles of $O_2 = \frac{80 \ g}{32 \ g/mol} = 2.5 \ mol$.
Number of oxygen atoms $= 2.5 \ mol \times 2 \times N_A = 5 \times N_A$.
For hydrogen gas $(H_2)$,the molar mass is $2 \ g/mol$.
Number of atoms in $5 \ g$ of $H_2$:
Moles of $H_2 = \frac{5 \ g}{2 \ g/mol} = 2.5 \ mol$.
Number of hydrogen atoms $= 2.5 \ mol \times 2 \times N_A = 5 \times N_A$.
Thus,$80 \ g$ of oxygen contains the same number of atoms as $5 \ g$ of hydrogen.

(B) The chemical formula for hydrogen chloride is $HCl$.
One molecule of $HCl$ contains one atom of chlorine.
Therefore,$6.02 \times 10^{25}$ molecules of $HCl$ contain $6.02 \times 10^{25}$ atoms of chlorine.
The number of moles (gram molecules) of chlorine atoms is calculated as:
$\text{Number of moles} = \frac{\text{Total number of atoms}}{\text{Avogadro's number}} = \frac{6.02 \times 10^{25}}{6.02 \times 10^{23}} = 100$.

(D) The number of oxygen atoms is calculated as: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{number of oxygen atoms in one molecule}$.
For $2 \ g$ of $CO$ $(M = 28 \ g/mol)$: $\frac{2}{28} \times 1 = 0.071 \ N_A$.
For $2 \ g$ of $CO_2$ $(M = 44 \ g/mol)$: $\frac{2}{44} \times 2 = 0.091 \ N_A$.
For $2 \ g$ of $SO_2$ $(M = 64 \ g/mol)$: $\frac{2}{64} \times 2 = 0.0625 \ N_A$.
For $2 \ g$ of $H_2O$ $(M = 18 \ g/mol)$: $\frac{2}{18} \times 1 = 0.111 \ N_A$.
Comparing the values,$2 \ g$ of $H_2O$ contains the maximum number of oxygen atoms.

(B) Initial weight of $H_2SO_4$ $(W)$ = $98 \ mg = 98 \times 10^{-3} \ g$.
Initial number of moles $(n_1)$ = $\frac{98 \times 10^{-3} \ g}{98 \ g/mol} = 10^{-3} \ mol$.
Number of moles removed $(n_2)$ = $\frac{\text{Number of molecules}}{N_A} = \frac{3.01 \times 10^{20}}{6.02 \times 10^{23}} = 0.5 \times 10^{-3} \ mol$.
Remaining moles = $n_1 - n_2 = 10^{-3} - 0.5 \times 10^{-3} = 0.5 \times 10^{-3} \ mol = 5 \times 10^{-4} \ mol$.

(A) Density of water = $1.0 \text{ g cm}^{-3}$.
Volume of water = $1.0 \text{ L} = 1000 \text{ mL}$.
Mass of $1.0 \text{ L}$ of water = $1000 \text{ mL} \times 1.0 \text{ g mL}^{-1} = 1000 \text{ g}$.
Molar mass of $H_2O = (2 \times 1.008) + 16.00 = 18.016 \text{ g mol}^{-1} \approx 18 \text{ g mol}^{-1}$.
In $18 \text{ g}$ of $H_2O$,the mass of hydrogen is $2 \text{ g}$.
Therefore,the mass of hydrogen in $1000 \text{ g}$ of $H_2O = (2 \text{ g} / 18 \text{ g}) \times 1000 \text{ g} = 111.11 \text{ g}$.
This is equal to $1.11 \times 10^2 \text{ g}$.

(C) Step $1$: Calculate the mass of one mole of element $X$ (molar mass).
Mass of one atom = $6.643 \times 10^{-23} \ g$.
Molar mass = Mass of one atom $\times$ Avogadro's number $(N_A)$.
Molar mass = $(6.643 \times 10^{-23} \ g) \times (6.022 \times 10^{23} \ mol^{-1}) \approx 40 \ g/mol$.
Step $2$: Calculate the number of moles in $50 \ kg$ $(50,000 \ g)$.
Number of moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{50,000 \ g}{40 \ g/mol} = 1250 \ moles$.

(A) $1$ mole of $O_{2(g)}$ at $STP$ contains Avogadro number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$ of $O_2$ molecules.
Since $1$ mole of $O_2$ has a molar mass of $32 \ g/mol$,it corresponds to $32 \ g$ of oxygen gas.
Therefore,the correct statement is that it contains $6.022 \times 10^{23}$ molecules of oxygen.

(C) $1$. Calculate the number of atoms in $6 \,g$ of $H_2O$:
$n(H_2O) = \frac{6 \,g}{18 \,g/mol} = \frac{1}{3} \,mol$.
Since each $H_2O$ molecule has $3$ atoms ($2$ $H$ and $1$ $O$),total atoms $= 3 \times \frac{1}{3} \,N_A = N_A$.
$2$. Evaluate options:
$A$: $n(He) = \frac{0.4}{4} = 0.1 \,mol$. Atoms $= 0.1 \,N_A$.
$B$: $n(CO_2) = \frac{22}{44} = 0.5 \,mol$. Atoms $= 3 \times 0.5 \,N_A = 1.5 \,N_A$.
$C$: $n(H_2) = \frac{1}{2} = 0.5 \,mol$. Atoms $= 2 \times 0.5 \,N_A = N_A$.
$D$: $n(CO) = \frac{12}{28} \approx 0.43 \,mol$. Atoms $= 2 \times 0.43 \,N_A = 0.86 \,N_A$.
Thus,$1 \,g$ of $H_2$ has the same number of atoms.

(C) The chemical formula for sodium ferrocyanide is $Na_4[Fe(CN)_6]$.
Each molecule of $Na_4[Fe(CN)_6]$ contains $4$ sodium ions $(Na^+)$.
Therefore,$1 \ mol$ of $Na_4[Fe(CN)_6]$ contains $4 \ mol$ of $Na^+$ ions.
For $0.5 \ mol$ of $Na_4[Fe(CN)_6]$,the number of moles of $Na^+$ ions is $0.5 \times 4 = 2 \ mol$.
The number of $Na^+$ ions is calculated as $2 \times N_A$,where $N_A \approx 6.022 \times 10^{23} \ mol^{-1}$.
Number of ions $= 2 \times 6.022 \times 10^{23} = 12.044 \times 10^{23} \approx 12 \times 10^{23}$.

(C) The chemical formula of ethane is $C_2H_6$ $(CH_3-CH_3)$.
Each carbon atom has $6$ electrons and each hydrogen atom has $1$ electron.
Total electrons in one molecule of ethane $= (2 \times 6) + (6 \times 1) = 12 + 6 = 18$ electrons.
One mole of ethane contains $N_A$ molecules,where $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Total electrons in one mole of ethane $= 18 \times 6.022 \times 10^{23} = 108.396 \times 10^{23} \approx 108.4 \times 10^{23}$.
Therefore,the correct option is $(C)$.

(C) The number of particles in a substance is given by $N = n \times N_A$,where $n$ is the number of moles and $N_A$ is Avogadro's constant $(6.022 \times 10^{23} \ mol^{-1})$.
Since $N_A$ is constant,equal moles $(n)$ of different substances contain the same number of constituent particles. Thus,Assertion $(A)$ is true.
However,the number of moles is given by $n = \frac{w}{M}$,where $w$ is the weight and $M$ is the molar mass.
Since different substances have different molar masses $(M)$,equal weights $(w)$ of different substances will result in different number of moles $(n)$,and consequently,a different number of constituent particles.
Therefore,Reason $(R)$ is false.

(A) Number of moles of ethyl alcohol $(C_2H_5OH)$ = $\frac{138}{46} = 3 \ mol$.
Number of moles of water $(H_2O)$ = $\frac{72}{18} = 4 \ mol$.
Mole fraction of alcohol $(X_{C_2H_5OH})$ = $\frac{3}{3+4} = \frac{3}{7}$.
Mole fraction of water $(X_{H_2O})$ = $\frac{4}{3+4} = \frac{4}{7}$.
The ratio of mole fraction of alcohol to water is $\frac{X_{C_2H_5OH}}{X_{H_2O}} = \frac{3/7}{4/7} = \frac{3}{4}$.

(B) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Number of moles of water $= \frac{0.36 \ g}{18 \ g/mol} = 0.02 \ mol$.
Number of water molecules $= 0.02 \times 6.023 \times 10^{23} = 1.205 \times 10^{22}$ molecules.
Since each molecule of water $(H_2O)$ contains $1$ oxygen atom,the number of oxygen atoms is equal to the number of water molecules.
Therefore,the number of oxygen atoms $= 1.205 \times 10^{22}$.