The mole concept Questions in English

(C) The Avogadro constant is defined as the number of constituent particles (atoms,molecules,or ions) per mole of a given substance.
By definition,one mole of any substance contains $6.022 \times 10^{23}$ particles.
Since one gram molecular mass of a substance is equivalent to one mole,it contains exactly $6.022 \times 10^{23}$ molecules.
Therefore,the correct definition is the number of molecules present in one gram molecular mass of a substance.

(B) $1 \, \text{mol}$ of $CH_4$ contains $4 \, \text{mol}$ of hydrogen atoms,which is equivalent to $4 \, \text{g}$ atoms of hydrogen.

(B) The chemical formula is $CuSO_4 \cdot 5H_2O$.
Breaking down the atoms:
$Cu = 1$
$S = 1$
$O = 4$
$5H_2O = 5 \times (2 \text{ atoms of } H + 1 \text{ atom of } O) = 5 \times 3 = 15$
Total number of atoms $= 1 + 1 + 4 + 15 = 21$.

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

(A) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
According to the mole concept,$1 \ mol$ of water contains $6.022 \times 10^{23}$ molecules and has a mass of $18 \ g$.
Mass of $1$ molecule of water $= \frac{18 \ g}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23} \ g$.
Converting this to kilograms: $2.99 \times 10^{-23} \ g \times \frac{1 \ kg}{1000 \ g} \approx 3 \times 10^{-26} \ kg$.

(C) Let $N_A$ be the Avogadro number.
Number of moles of $A = x/40$.
Number of atoms of $A = (x/40) \times N_A = y$.
Thus,$x = (40 \times y) / N_A$.
Number of moles of $B = 2x/80 = x/40$.
Number of atoms of $B = (x/40) \times N_A$.
Substituting the value of $x$,we get: $((40 \times y) / (40 \times N_A)) \times N_A = y$.
Therefore,the number of atoms in $2x \ g$ of $B$ is $y$.

(A) The molar mass of nitrogen $(N)$ is $14 \ g/mol$. Since the mass of electrons is negligible,the molar mass of nitride ions $(N^{3 -})$ is also $14 \ g/mol$.
One mole of $N^{3 -}$ ions contains $1 \ mol$ of $N$ atoms. The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$,so it has $5$ valence electrons. The $N^{3 -}$ ion gains $3$ electrons,resulting in $5 + 3 = 8$ valence electrons per ion.
Therefore,$1 \ mol$ $(14 \ g)$ of $N^{3 -}$ ions contains $8 \ N_A$ valence electrons.
For $4.2 \ g$ of $N^{3 -}$ ions,the number of valence electrons is:
$\frac{8 \ N_A \times 4.2 \ g}{14 \ g} = 2.4 \ N_A$.

(D) The molar mass of $CuSO_4 \cdot 5H_2O$ is calculated as: $63.5 + 32 + (4 \times 16) + 5 \times (2 + 16) = 63.5 + 32 + 64 + 90 = 249.5 \ g/mol$.
$6.022 \times 10^{23}$ molecules of $CuSO_4 \cdot 5H_2O$ weigh $249.5 \ g$.
Therefore,the weight of $1 \times 10^{22}$ molecules is:
$\text{Weight} = \frac{249.5 \times 1 \times 10^{22}}{6.022 \times 10^{23}} \ g$.
$\text{Weight} = \frac{249.5}{60.22} \ g \approx 4.143 \ g$.
Since $4.143 \ g$ is not exactly $4.159 \ g$,the correct option is $D$.

(B) At $NTP$,$1 \ mole$ of an ideal gas occupies $22400 \ mL$ and contains $6.023 \times 10^{23}$ molecules.
Therefore,the number of molecules in $1 \ mL$ at $NTP$ is calculated as:
$\text{Number of molecules} = \frac{6.023 \times 10^{23}}{22400}$
$= 0.0002688 \times 10^{23} = 2.69 \times 10^{19}$ molecules.

(D) The molar mass of water $(H_2O)$ is $18 \, g \, mol^{-1}$.
Given density $(d)$ = $1 \, g \, cm^{-3}$.
Since $d = \frac{mass}{volume}$,the volume of $1 \, mole$ $(18 \, g)$ of water is $V = \frac{18 \, g}{1 \, g \, cm^{-3}} = 18 \, cm^3$.
$1 \, mole$ of water contains $6.022 \times 10^{23}$ molecules.
Therefore,the volume occupied by $1$ molecule is $\frac{18 \, cm^3}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23} \, cm^3 \approx 3.0 \times 10^{-23} \, cm^3$.

(C) At $STP$ ($0^o C$ and $1 \ atm$ pressure),the molar volume of any ideal gas is $22.4 \ L \ mol^{-1}$.
Since the given volume is $22.4 \ L$,the number of moles is calculated as:
$n = \frac{V}{V_m} = \frac{22.4 \ L}{22.4 \ L \ mol^{-1}} = 1 \ mol$.

(B) At $STP$,$22400 \ cc$ of any gas contains $6.022 \times 10^{23}$ molecules (Avogadro's number).
Number of molecules = $\frac{\text{Given volume}}{\text{Molar volume at } STP} \times N_A$
Number of molecules = $\frac{1.12 \times 10^{-7} \ cc}{22400 \ cc/mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$
Number of molecules = $\frac{1.12 \times 6.022}{22400} \times 10^{16} \approx 3.01 \times 10^{12}$ molecules.

(D) Volume of air = $1 \ L = 1000 \ mL$.
Volume of $O_2$ in $1 \ L$ of air = $21 \%$ of $1000 \ mL = 210 \ mL$.
At $STP$,$1 \ mole$ of any gas occupies $22400 \ mL$.
Number of moles of $O_2 = \frac{\text{Volume of } O_2 \text{ in } mL}{22400 \ mL/mol} = \frac{210}{22400} \approx 0.009375 \ mol$.
Rounding to the given options,the correct answer is $0.0093 \ mol$.

(D) At $STP$ ($0^{\circ}C$ and $1 \ atm$),$22.4 \ L$ of any gas contains $6.022 \times 10^{23}$ molecules (Avogadro's number).
Number of moles in $8.96 \ L = \frac{8.96 \ L}{22.4 \ L/mol} = 0.4 \ mol$.
Number of molecules $= \text{moles} \times N_A = 0.4 \times 6.022 \times 10^{23}$.
$= 2.4088 \times 10^{23} = 24.088 \times 10^{22} \approx 24.08 \times 10^{22}$.

(B) The molecular weight of $NH_3$ is $17 \, g/mol$.
According to the mole concept,$17 \, g$ of $NH_3$ contains $6.022 \times 10^{23}$ molecules.
Therefore,the number of moles in $4.25 \, g$ of $NH_3$ is $\frac{4.25 \, g}{17 \, g/mol} = 0.25 \, mol$.
The number of molecules $= \text{moles} \times N_A = 0.25 \times 6.022 \times 10^{23} \approx 1.505 \times 10^{23}$ molecules.
Thus,the correct option is $(B)$.

(A) The number of molecules is directly proportional to the number of moles $(n = \frac{\text{mass}}{\text{molar mass}})$.
For option $(A)$:
$n(O_2) = \frac{16 \ g}{32 \ g/mol} = 0.5 \ mol$
$n(N_2) = \frac{14 \ g}{28 \ g/mol} = 0.5 \ mol$
Since the number of moles is the same,the number of molecules is also the same.

(A) The molar mass of $CO_2$ is $44 \ g/mol$.
$4.4 \ g$ of $CO_2$ corresponds to $\frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$ of $CO_2$.
Each molecule of $CO_2$ contains $2$ oxygen atoms.
Therefore,$0.1 \ mol$ of $CO_2$ contains $0.1 \times 2 = 0.2 \ mol$ of oxygen atoms.
The number of oxygen atoms is $0.2 \times 6.022 \times 10^{23} \approx 1.2 \times 10^{23}$ atoms.

(A) Density of water at room temperature is $1 \ g/mL$.
Mass of the drop $= \text{Density} \times \text{Volume} = 1 \ g/mL \times 0.0018 \ mL = 0.0018 \ g$.
Molar mass of water $(H_2O)$ $= 18 \ g/mol$.
Number of moles $= \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.0018 \ g}{18 \ g/mol} = 1 \times 10^{-4} \ mol$.
Number of water molecules $= \text{Number of moles} \times N_A = 1 \times 10^{-4} \times 6.023 \times 10^{23} = 6.023 \times 10^{19}$.

(D) Given mass of gold $= 19.7 \ kg = 19700 \ g$.
Atomic mass of gold $(Au)$ $= 197 \ g/mol$.
Number of moles of gold $= \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{19700 \ g}{197 \ g/mol} = 100 \ mol$.
Number of atoms $= \text{Number of moles} \times \text{Avogadro's number} (N_A)$.
Number of atoms $= 100 \times 6.022 \times 10^{23} = 6.022 \times 10^{25} \text{ atoms}$.
Thus,the correct option is $(D)$.

(C) The molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g/mol$.
Number of moles in $10 \ g$ of $CaCO_3 = \frac{10 \ g}{100 \ g/mol} = 0.1 \ mol$.
Number of molecules $= 0.1 \times 6.023 \times 10^{23} = 6.023 \times 10^{22}$ molecules.
Number of protons in one molecule of $CaCO_3$ $(Ca=20, C=6, O=8)$: $20 + 6 + (3 \times 8) = 50$ protons.
Total number of protons $= 50 \times 6.023 \times 10^{22} = 3.0115 \times 10^{24}$.

(C) According to Avogadro's hypothesis,equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Since the volume $(100 \ mL)$,temperature,and pressure $(STP)$ are identical for all three gases ($O_2$,$NH_3$,and $CO_2$),the number of molecules in each will be the same.

(D) Density of water $(d)$ is $1 \ g/mL$.
Mass of $1 \ L$ $(1000 \ mL)$ of water is $1000 \ g$.
Molar mass of water $(H_2O)$ is $18 \ g/mol$.
Number of moles of water = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{1000 \ g}{18 \ g/mol} = 55.55 \, mol$.
Number of molecules = $\text{Moles} \times N_A = 55.55 \, N_A$.

(A) $1 \ mole$ of $H_2$ contains $6.023 \times 10^{23}$ molecules.
Each molecule of $H_2$ contains $2$ electrons.
Therefore,the total number of electrons in $1 \ mole$ of $H_2$ is $2 \times 6.023 \times 10^{23} = 12.046 \times 10^{23}$.

(A) The chemical formula of barium carbonate is $BaCO_3$.
In $1$ mole of $BaCO_3$,there are $3$ moles of oxygen atoms.
Therefore,to find the number of moles of $BaCO_3$ containing $1.5$ moles of oxygen atoms,we use the unitary method:
$\text{Moles of } BaCO_3 = \frac{1 \text{ mole of } BaCO_3}{3 \text{ moles of } O} \times 1.5 \text{ moles of } O = 0.5 \text{ moles}$.

(B) The number of molecules present in $1 \ mL$ of an ideal gas at $STP$ is known as the Loschmidt number.
$22400 \ mL$ of gas at $STP$ contains $6.022 \times 10^{23}$ molecules.
Therefore,$1 \ mL$ of gas contains $\frac{6.022 \times 10^{23}}{22400} \approx 2.69 \times 10^{19}$ molecules.

(B) The molar mass of hydrogen gas $(H_2)$ is $2 \ g/mol$.
$2 \ g$ of $H_2$ contains $6.022 \times 10^{23}$ molecules.
Therefore,$1 \ g$ of $H_2$ contains $\frac{6.022 \times 10^{23}}{2} = 3.011 \times 10^{23}$ molecules.
The value is $3.01 \times 10^{23}$ molecules.

(A) The molecular weight of $SO_2Cl_2$ is calculated as:
$32 + 32 + (2 \times 35.5) = 135 \ g/mol$.
Since $135 \ g$ of $SO_2Cl_2$ corresponds to $1 \ g$-molecule,
the number of $g$-molecules in $13.5 \ g$ is:
$\frac{13.5 \ g}{135 \ g/mol} = 0.1 \ mol$.

(A) To find the number of molecules,we use the formula: $\text{Number of molecules} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_A$.
$(A)$ $34 \ g$ of $H_2O$ (Molar mass = $18 \ g/mol$): $\frac{34}{18} \times N_A \approx 1.89 \times N_A$ molecules.
$(B)$ $28 \ g$ of $CO_2$ (Molar mass = $44 \ g/mol$): $\frac{28}{44} \times N_A \approx 0.64 \times N_A$ molecules.
$(C)$ $46 \ g$ of $CH_3OH$ (Molar mass = $32 \ g/mol$): $\frac{46}{32} \times N_A \approx 1.44 \times N_A$ molecules.
$(D)$ $54 \ g$ of $N_2O_5$ (Molar mass = $108 \ g/mol$): $\frac{54}{108} \times N_A = 0.5 \times N_A$ molecules.
Comparing the values,$34 \ g$ of water contains the largest number of molecules.

(B) The chemical formula of sodium oxide is $Na_2O$.
The molar mass of $Na_2O = (2 \times 23) + 16 = 46 + 16 = 62 \ g/mol$.
The number of moles is calculated using the formula: $n = \frac{\text{mass}}{\text{molar mass}}$.
$n = \frac{620 \ g}{62 \ g/mol} = 10 \ moles$.
Therefore,the correct option is $B$.

(B) The number of moles of atoms in $2 \ g$ of oxygen $(O)$ is calculated as: $\text{Moles of } O = \frac{\text{mass}}{\text{atomic mass}} = \frac{2 \ g}{16 \ g/mol} = 0.125 \ mol = \frac{1}{8} \ mol$.
Now,let us calculate the moles of atoms for each option:
$A) \ 0.5 \ g$ of hydrogen $(H)$: $\text{Moles} = \frac{0.5}{1} = 0.5 \ mol$.
$B) \ 4 \ g$ of sulphur $(S)$: $\text{Moles} = \frac{4}{32} = 0.125 \ mol = \frac{1}{8} \ mol$.
$C) \ 7 \ g$ of nitrogen $(N)$: $\text{Moles} = \frac{7}{14} = 0.5 \ mol$.
$D) \ 2.3 \ g$ of sodium $(Na)$: $\text{Moles} = \frac{2.3}{23} = 0.1 \ mol$.
Since $4 \ g$ of sulphur contains the same number of moles of atoms as $2 \ g$ of oxygen,the correct option is $B$.

(A) The chemical formula of sucrose is $C_{12}H_{22}O_{11}$.
One molecule of sucrose contains $12$ carbon atoms,$22$ hydrogen atoms,and $11$ oxygen atoms.
Total atoms per molecule = $12 + 22 + 11 = 45$ atoms.
One mole of any substance contains $6.022 \times 10^{23}$ molecules (Avogadro's number).
Therefore,one mole of sucrose contains $45 \times 6.022 \times 10^{23}$ atoms.

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

(C) $1$ molecule of $PCl_3$ contains $4$ atoms ($1$ atom of $P$ and $3$ atoms of $Cl$).
Number of atoms $= ( \text{moles of } PCl_3) \times ( \text{atoms per molecule}) \times N_A$
Number of atoms $= 1.4 \times 4 \times 6.022 \times 10^{23}$
Number of atoms $= 5.6 \times 6.022 \times 10^{23}$
Number of atoms $= 3.37232 \times 10^{24}$ atoms.

(D) The chemical formula of sodium ferrocyanide is $Na_4[Fe(CN)_6]$.
Each molecule of sodium ferrocyanide contains $4$ sodium $(Na)$ atoms.
Therefore,the number of $Na$ atoms in $2$ moles of sodium ferrocyanide is calculated as:
Number of $Na$ atoms = (Number of moles) $\times$ (Number of $Na$ atoms per molecule) $\times$ (Avogadro's number)
Number of $Na$ atoms = $2 \times 4 \times 6.022 \times 10^{23}$
Number of $Na$ atoms = $8 \times 6.022 \times 10^{23} \approx 48.176 \times 10^{23} \approx 48 \times 10^{23}$ atoms.
Thus,the correct option is $(D)$.

(A) According to Avogadro's Law,equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
Since both gases occupy the same volume $(10 \ dm^3)$ at the same temperature and pressure,they must contain the same number of moles.
For the number of molecules to be identical,the molar mass of gas $X$ must be equal to the molar mass of $N_2$ if the masses were equal,or simply,the gases must have the same molecular weight if they are diatomic with the same number of atoms.
The molar mass of $N_2 = 2 \times 14 = 28 \ g/mol$.
Checking the options:
$A$: $CO = 12 + 16 = 28 \ g/mol$.
$B$: $CO_2 = 12 + 32 = 44 \ g/mol$.
$C$: $H_2 = 2 \times 1 = 2 \ g/mol$.
$D$: $NO = 14 + 16 = 30 \ g/mol$.
Therefore,the gas $X$ is $CO$.

(A) Initial mass of $CO_2 = 200 \ mg = 0.2 \ g$.
Molar mass of $CO_2 = 44 \ g/mol$.
Initial number of moles of $CO_2 = \frac{0.2 \ g}{44 \ g/mol} = 0.004545 \ mol$.
Initial number of molecules $= 0.004545 \times 6.022 \times 10^{23} \approx 2.737 \times 10^{21}$ molecules.
Number of molecules removed $= 10^{21}$.
Remaining molecules $= 2.737 \times 10^{21} - 1 \times 10^{21} = 1.737 \times 10^{21}$ molecules.
Remaining moles $= \frac{1.737 \times 10^{21}}{6.022 \times 10^{23}} \approx 0.288 \times 10^{-2} = 2.88 \times 10^{-3} \ mol$.
Note: Using $6 \times 10^{23}$ as Avogadro's number gives $2.72 \times 10^{21}$ molecules,leading to $1.72 \times 10^{21}$ remaining,which results in $\frac{1.72}{6} \times 10^{-2} \approx 0.286 \times 10^{-2} = 2.86 \times 10^{-3} \ mol$. The closest option is $2.85$.

(C) The molar mass of $CH_4$ is $16 \ g/mol$.
At $STP$,$1 \ mole$ of any gas occupies $22400 \ cm^3$.
The number of moles $n$ is given by $n = \frac{V}{22400} = \frac{112}{22400} = 0.005 \ mol$.
The mass $W$ is calculated as $W = n \times M = 0.005 \times 16 = 0.08 \ g$.
Therefore,the correct option is $C$.

(B) The mass of one proton is approximately $1.673 \times 10^{-24} \ g$.
The mass of one mole of protons is $1.673 \times 10^{-24} \ g \times 6.022 \times 10^{23} \ mol^{-1} \approx 1.008 \ g$.
The mass of one electron is approximately $9.109 \times 10^{-28} \ g$.
The mass of one mole of electrons is $9.109 \times 10^{-28} \ g \times 6.022 \times 10^{23} \ mol^{-1} \approx 5.48 \times 10^{-4} \ g = 0.548 \ mg \approx 0.55 \ mg$.
Thus,the correct option is $(B)$.

(C) One mole of electrons contains $6.022 \times 10^{23}$ electrons.
The mass of one electron is $9.1 \times 10^{-28} \ g$.
Therefore,the mass of one mole of electrons is calculated as:
$\text{Mass} = (6.022 \times 10^{23}) \times (9.1 \times 10^{-28} \ g) \approx 5.48 \times 10^{-4} \ g$.
Converting this to milligrams $(mg)$:
$5.48 \times 10^{-4} \ g \times 1000 \ mg/g = 0.548 \ mg \approx 0.55 \ mg$.

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

(B) Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
As a consequence of this hypothesis,it is established that $1 \, \text{mole}$ (or one gram molecule) of any ideal gas at $N.T.P.$ (Normal Temperature and Pressure,$273.15 \, K$ and $1 \, \text{atm}$) occupies a volume of $22.4 \, L$.

(C) To find the maximum number of molecules,we calculate the number of moles $(n)$ for each option:
$A$: $n = \frac{0.5 \ g}{2 \ g/mol} = 0.25 \ mol$. Molecules $= 0.25 \times N_A = 0.25 N_A$.
$B$: $n = \frac{10 \ g}{32 \ g/mol} = 0.3125 \ mol$. Molecules $= 0.3125 \times N_A = 0.3125 N_A$.
$C$: $n = \frac{15 \ L}{22.4 \ L/mol} \approx 0.67 \ mol$. Molecules $= 0.67 \times N_A = 0.67 N_A$.
$D$: $n = \frac{5 \ L}{22.4 \ L/mol} \approx 0.22 \ mol$. Molecules $= 0.22 \times N_A = 0.22 N_A$.
Comparing the values,$15 \ L$ of $H_2$ gas at $STP$ contains the maximum number of molecules.

(C) At $NTP$,the molar volume of any gas is $22400 \, mL$.
The molar mass of oxygen $(O_2)$ is $32 \, g/mol$.
Thus,$22400 \, mL$ of $O_2$ at $NTP$ weighs $32 \, g$.
The weight of $112 \, mL$ of $O_2$ at $NTP$ is calculated as:
$\text{Weight} = \frac{32 \, g}{22400 \, mL} \times 112 \, mL = 0.16 \, g$.

(D) Since the mixture contains $50\%$ helium and $50\%$ methane by volume,we can assume a total of $100 \text{ moles}$ of the gas mixture.
This means there are $50 \text{ moles}$ of $He$ and $50 \text{ moles}$ of $CH_4$.
The molar mass of $He$ is $4 \text{ g/mol}$ and the molar mass of $CH_4$ is $16 \text{ g/mol}$.
Mass of $He = 50 \text{ moles} \times 4 \text{ g/mol} = 200 \text{ g}$.
Mass of $CH_4 = 50 \text{ moles} \times 16 \text{ g/mol} = 800 \text{ g}$.
Total mass of the mixture $= 200 \text{ g} + 800 \text{ g} = 1000 \text{ g}$.
Percent by weight of $CH_4 = \frac{\text{Mass of } CH_4}{\text{Total mass}} \times 100 = \frac{800}{1000} \times 100 = 80\%$.

(D) The molar mass of $NaCl$ is $58.5 \ g/mol$. The number of moles of $NaCl$ is $n_{NaCl} = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$.
The molar mass of $H_2O$ is $18 \ g/mol$. The number of moles of $H_2O$ is $n_{H_2O} = \frac{90 \ g}{18 \ g/mol} = 5 \ mol$.
The mole fraction of $NaCl$ is given by $\chi_{NaCl} = \frac{n_{NaCl}}{n_{NaCl} + n_{H_2O}} = \frac{0.1}{0.1 + 5} = \frac{0.1}{5.1} \approx 0.0196$.

(D) To find the number of molecules,we calculate the number of moles $(n = \text{mass} / \text{molar mass})$ and multiply by Avogadro's number $(N_A)$:
$A) \ 16 \ g \ O_2: n = 16 / 32 = 0.5 \ mol$
$B) \ 16 \ g \ NO_2: n = 16 / 46 \approx 0.348 \ mol$
$C) \ 7 \ g \ N_2: n = 7 / 28 = 0.25 \ mol$
$D) \ 2 \ g \ H_2: n = 2 / 2 = 1.0 \ mol$
Since $H_2$ has the highest number of moles $(1.0 \ mol)$,it contains the maximum number of molecules.

(A) $1$. Calculate the number of moles of ethanol $(C_2H_5OH)$:
$n_{\text{ethanol}} = \frac{\text{mass}}{\text{molar mass}} = \frac{414 \ g}{46 \ g/mol} = 9 \ mol$.
$2$. Calculate the number of moles of water $(H_2O)$:
$n_{\text{water}} = \frac{18 \ g}{18 \ g/mol} = 1 \ mol$.
$3$. Calculate the mole fraction of water $(x_{\text{water}})$:
$x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_{\text{ethanol}}} = \frac{1}{1 + 9} = \frac{1}{10} = 0.1$.

(B) The molar mass of ammonia $(NH_3)$ is $14 + (3 \times 1) = 17 \ g/mol$.
Number of moles of $NH_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.25 \ g}{17 \ g/mol} = 0.25 \ mol$.
Number of molecules = $\text{Number of moles} \times N_A = 0.25 \times 6.022 \times 10^{23} \approx 1.505 \times 10^{23}$.
Thus,the number of molecules is approximately $1.5 \times 10^{23}$.

(C) The number of molecules in a sample is calculated as: $\text{Number of molecules} = \frac{\text{Given mass } (W)}{\text{Molar mass } (M)} \times N_A$,where $N_A$ is Avogadro's number.
$A$: For $25 \ g$ of $CO_2$ $(M = 44 \ g/mol)$: $\frac{25}{44} \times N_A \approx 0.568 \ N_A$
$B$: For $46 \ g$ of $C_2H_5OH$ $(M = 46 \ g/mol)$: $\frac{46}{46} \times N_A = 1 \ N_A$
$C$: For $36 \ g$ of $H_2O$ $(M = 18 \ g/mol)$: $\frac{36}{18} \times N_A = 2 \ N_A$
$D$: For $54 \ g$ of $N_2O_5$ $(M = 108 \ g/mol)$: $\frac{54}{108} \times N_A = 0.5 \ N_A$
Comparing the values,$36 \ g$ of $H_2O$ contains the largest number of molecules.