Molecular Mass Questions in English

(C) $(i)$ Molecular mass of $H_2$ (hydrogen) $= \text{Atomic mass of hydrogen} \times 2 = 1 \times 2 = 2\,u$.
$(ii)$ Molecular mass of $O_2$ (oxygen) $= \text{Atomic mass of oxygen} \times 2 = 16 \times 2 = 32\,u$.
$(iii)$ Molecular mass of $Cl_2$ (chlorine) $= \text{Atomic mass of chlorine} \times 2 = 35.5 \times 2 = 71\,u$.

(D) $(i)$ Molecular mass of $CO_2$ (carbon dioxide) = (Atomic mass of carbon $\times 1$) + (Atomic mass of oxygen $\times 2$) = $12 + (16 \times 2) = 12 + 32 = 44\,u$.
$(ii)$ Molecular mass of $CH_4$ (methane) = (Atomic mass of carbon $\times 1$) + (Atomic mass of hydrogen $\times 4$) = $12 + (1 \times 4) = 12 + 4 = 16\,u$.
$(iii)$ Molecular mass of $C_2H_6$ (ethane) = (Atomic mass of carbon $\times 2$) + (Atomic mass of hydrogen $\times 6$) = $(12 \times 2) + (1 \times 6) = 24 + 6 = 30\,u$.

(A) $(i)$ Molecular mass of $C_2H_4$ (ethene):
$= (\text{Atomic mass of C} \times 2) + (\text{Atomic mass of H} \times 4)$
$= (12 \times 2) + (1 \times 4) = 24 + 4 = 28\,u$.
$(ii)$ Molecular mass of $NH_3$ (ammonia):
$= (\text{Atomic mass of N} \times 1) + (\text{Atomic mass of H} \times 3)$
$= (14 \times 1) + (1 \times 3) = 14 + 3 = 17\,u$.
$(iii)$ Molecular mass of $CH_3OH$ (methanol):
$= (\text{Atomic mass of C} \times 1) + (\text{Atomic mass of H} \times 4) + (\text{Atomic mass of O} \times 1)$
$= (12 \times 1) + (1 \times 4) + (16 \times 1) = 12 + 4 + 16 = 32\,u$.

(B) $(i)$ Formula unit mass of $ZnO = 65 + 16 = 81\,u$.
$(ii)$ Formula unit mass of $Na_2O = (23 \times 2) + 16 = 46 + 16 = 62\,u$.
$(iii)$ Formula unit mass of $K_2CO_3 = (39 \times 2) + 12 + (16 \times 3) = 78 + 12 + 48 = 138\,u$.

(B) Molar mass of $C_2H_2 = (2 \times \text{Atomic mass of } C) + (2 \times \text{Atomic mass of } H)$
$= (2 \times 12) + (2 \times 1) = 24 + 2 = 26\,u$.
$(b)$ Molar mass of $S_8 = 8 \times \text{Atomic mass of } S$
$= 8 \times 32 = 256\,u$.

(C) Molar mass of $P_4 = 4 \times$ Atomic mass of $P = 4 \times 31 = 124\,u$.
$(b)$ Molar mass of $HCl = $ Atomic mass of $H + $ Atomic mass of $Cl = 1 + 35.5 = 36.5\,u$.
$(c)$ Molar mass of $HNO_3 = $ Atomic mass of $H + $ Atomic mass of $N + (3 \times$ Atomic mass of $O) = 1 + 14 + (3 \times 16) = 15 + 48 = 63\,u$.

(D) Molar mass of $N$ atom $=$ Atomic mass of $N = 14\, g/mol$.
Mass of $1$ mole of $N$ atoms $= 1 \times 14 = 14\, g$.
$(b)$ Molar mass of $Al$ atom $= 27\, g/mol$.
Mass of $4$ moles of $Al$ atoms $= 4 \times 27 = 108\, g$.
$(c)$ Molar mass of $Na_2SO_3 = (23 \times 2) + 32 + (16 \times 3) = 46 + 32 + 48 = 126\, g/mol$.
Mass of $10$ moles of $Na_2SO_3 = 10 \times 126 = 1260\, g$.

(A) For oxygen gas $(O_2)$:
Molar mass of $O_2 = 16 \times 2 = 32\, \text{g/mol}$.
Number of moles $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{12\, \text{g}}{32\, \text{g/mol}} = 0.375\, \text{mol}$.
$(b)$ For water $(H_2O)$:
Molar mass of $H_2O = (2 \times 1) + 16 = 18\, \text{g/mol}$.
Number of moles $= \frac{20\, \text{g}}{18\, \text{g/mol}} \approx 1.11\, \text{mol}$.
$(c)$ For carbon dioxide $(CO_2)$:
Molar mass of $CO_2 = 12 + (2 \times 16) = 44\, \text{g/mol}$.
Number of moles $= \frac{22\, \text{g}}{44\, \text{g/mol}} = 0.5\, \text{mol}$.

(B) The molar mass of oxygen atoms $(O)$ is $16 \, g/mol$.
Mass of $0.2$ mole of oxygen atoms $= \text{Number of moles} \times \text{Molar mass} = 0.2 \times 16 = 3.2 \, g$.
$(b)$ The molar mass of water molecules $(H_2O)$ is $(2 \times 1) + 16 = 18 \, g/mol$.
Mass of $0.5$ mole of water molecules $= \text{Number of moles} \times \text{Molar mass} = 0.5 \times 18 = 9.0 \, g$.

(B) Molar mass of sulphur $(S_8) = 32 \times 8 = 256 \, g/mol$.
Number of moles of $S_8$ in $16 \, g = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{16}{256} = 0.0625 \, mol$.
Number of molecules = $\text{Number of moles} \times \text{Avogadro constant} (N_A)$.
Number of molecules = $0.0625 \times 6.022 \times 10^{23} = 3.76375 \times 10^{22} \approx 3.76 \times 10^{22}$ molecules.
Wait,re-calculating: $0.0625 \times 6.022 \times 10^{23} = 0.376375 \times 10^{23} = 3.76 \times 10^{22}$.
Correction: The provided option $3.76 \times 10^{22}$ is not listed,but $3.76 \times 10^{22}$ is $0.376 \times 10^{23}$. Let us re-verify: $16/256 = 1/16 = 0.0625$. $0.0625 \times 6.022 \times 10^{23} = 0.376375 \times 10^{23}$. Thus,the correct value is $0.376 \times 10^{23}$.

(D) $1$. Calculate the molar mass of aluminium oxide $(Al_{2}O_{3})$:
Molar mass = $(2 \times 27) + (3 \times 16) = 54 + 48 = 102 \, g/mol$.
$2$. Determine the number of moles in $0.051 \, g$ of $Al_{2}O_{3}$:
Moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{0.051}{102} = 0.0005 \, mol = 5 \times 10^{-4} \, mol$.
$3$. Relate moles of $Al_{2}O_{3}$ to moles of $Al^{3+}$ ions:
One mole of $Al_{2}O_{3}$ contains $2$ moles of $Al^{3+}$ ions.
Therefore,$5 \times 10^{-4} \, mol$ of $Al_{2}O_{3}$ contains $2 \times 5 \times 10^{-4} = 10^{-3} \, mol$ of $Al^{3+}$ ions.
$4$. Calculate the number of ions:
Number of ions = $\text{Moles} \times \text{Avogadro's number} = 10^{-3} \times 6.022 \times 10^{23} = 6.022 \times 10^{20}$ ions.

(A) The molar mass of water $(H_2O)$ is $(2 \times 1) + 16 = 18 \,g/mol$.
Number of moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{360 \,g}{18 \,g/mol} = 20 \,moles$.
Thus,statement $(ii)$ is correct.
Number of molecules = $\text{Number of moles} \times \text{Avogadro's number} (N_A)$.
Number of molecules = $20 \times 6.022 \times 10^{23} = 1.2044 \times 10^{25}$ molecules.
Thus,statement $(iv)$ is correct.
Therefore,the correct options are $(ii)$ and $(iv)$.

(A) To determine which weighs the most,we calculate the mass of each substance using the formula: $\text{Mass} = \text{Number of moles} \times \text{Molar mass}$.
$(a)$ Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \, g/mol$. Mass of $2$ moles $= 2 \times 100 = 200 \, g$.
$(b)$ Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \, g/mol$. Mass of $2$ moles $= 2 \times 44 = 88 \, g$.
$(c)$ Molar mass of sucrose $(C_{12}H_{22}O_{11}) = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \, g/mol$. Mass of $0.2$ moles $= 0.2 \times 342 = 68.4 \, g$.
$(d)$ Molar mass of $H_2O = (2 \times 1) + 16 = 18 \, g/mol$. Mass of $10$ moles $= 10 \times 18 = 180 \, g$.
Comparing the values: $200 \, g > 180 \, g > 88 \, g > 68.4 \, g$. Therefore,$2$ moles of $CaCO_3$ weigh the highest.

(B) The number of atoms is calculated using the formula: $\text{Number of atoms} = \frac{\text{mass of substance}}{\text{molar mass}} \times \text{number of atoms in a molecule} \times N_A$.
$(a)$ For $18 \, g$ of $H_2O$: $\text{Molar mass} = 18 \, g/mol$. Atoms per molecule $= 3$. $\text{Atoms} = \frac{18}{18} \times 3 \times N_A = 3 N_A$.
$(b)$ For $18 \, g$ of $CH_4$: $\text{Molar mass} = 16 \, g/mol$. Atoms per molecule $= 5$. $\text{Atoms} = \frac{18}{16} \times 5 \times N_A = 5.625 N_A$.
$(c)$ For $18 \, g$ of $CO_2$: $\text{Molar mass} = 44 \, g/mol$. Atoms per molecule $= 3$. $\text{Atoms} = \frac{18}{44} \times 3 \times N_A \approx 1.23 N_A$.
$(d)$ For $18 \, g$ of $O_2$: $\text{Molar mass} = 32 \, g/mol$. Atoms per molecule $= 2$. $\text{Atoms} = \frac{18}{32} \times 2 \times N_A = 1.125 N_A$.
Comparing the values,$5.625 N_A$ is the maximum. Thus,$18 \, g$ of $CH_4$ has the maximum number of atoms.

(C) The number of molecules is calculated using the formula: $\text{Number of molecules} = \frac{\text{mass of substance}}{\text{molar mass}} \times N_A$.
For $1\,g$ of $H_2$: $\text{Molar mass} = 2\,g/mol$. $\text{Number of molecules} = \frac{1}{2} \times N_A = 0.5 N_A$.
For $1\,g$ of $CO_2$: $\text{Molar mass} = 44\,g/mol$. $\text{Number of molecules} = \frac{1}{44} \times N_A \approx 0.0227 N_A$.
For $1\,g$ of $N_2$: $\text{Molar mass} = 28\,g/mol$. $\text{Number of molecules} = \frac{1}{28} \times N_A \approx 0.0357 N_A$.
For $1\,g$ of $CH_4$: $\text{Molar mass} = 16\,g/mol$. $\text{Number of molecules} = \frac{1}{16} \times N_A = 0.0625 N_A$.
Comparing the values,$0.5 N_A$ is the largest. Therefore,$1\,g$ of $H_2$ contains the maximum number of molecules.

(C) The molar mass of water $(H_2O)$ is $18 \, g/mol$.
The mass of one molecule of water is $\frac{18}{N_A} \, g$,where $N_A$ is the Avogadro number.
In one molecule of water $(H_2O)$,there are $2$ hydrogen atoms and $1$ oxygen atom.
Hydrogen $(^1H)$ has $0$ neutrons,while Oxygen $(^{16}O)$ has $8$ neutrons.
Therefore,the total number of neutrons in one molecule of water is $8$.
The mass of $8$ neutrons is $\frac{8}{N_A} \, g$.
The fraction of the mass of water due to neutrons is calculated as: $\frac{\text{Mass of neutrons}}{\text{Total mass of water}} = \frac{8/N_A}{18/N_A} = \frac{8}{18}$.

(D) Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \, g \, mol^{-1}$.
Mass of $5 \, mol$ of $CO_2 = 5 \times 44 = 220 \, g$.
Molar mass of $H_2O = (2 \times 1) + 16 = 18 \, g \, mol^{-1}$.
Mass of $5 \, mol$ of $H_2O = 5 \times 18 = 90 \, g$.
Since $220 \, g \neq 90 \, g$,they do not have the same mass.
$(b)$ Molar mass of $Ca = 40 \, g \, mol^{-1}$.
Moles of $Ca = \frac{240 \, g}{40 \, g \, mol^{-1}} = 6 \, mol$.
Molar mass of $Mg = 24 \, g \, mol^{-1}$.
Moles of $Mg = \frac{240 \, g}{24 \, g \, mol^{-1}} = 10 \, mol$.
Ratio of moles of $Ca$ to $Mg = 6:10 = 3:5$.

(B) The molar mass of $CaCl_{2} = 40 + (2 \times 35.5) = 111 \, g/mol$.
Given mass of $CaCl_{2} = 222 \, g$.
Number of moles of $CaCl_{2} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{222}{111} = 2 \, mol$.
According to the dissociation equation $CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}$,one formula unit of $CaCl_{2}$ produces $3$ ions ($1$ calcium ion and $2$ chloride ions).
Therefore,$1 \, mol$ of $CaCl_{2}$ produces $3 \, mol$ of ions.
$2 \, mol$ of $CaCl_{2}$ will produce $2 \times 3 = 6 \, mol$ of ions.
Total number of ions = $\text{Number of moles} \times \text{Avogadro number} (N_{A})$.
Total number of ions = $6 \times 6.022 \times 10^{23} = 36.132 \times 10^{23} = 3.6132 \times 10^{24}$ ions.

(D) $1$. Calculate the molar mass of $HgS$: $M(HgS) = 200.6 + 32 = 232.6 \,g \,mol^{-1}$.
$2$. Determine the mass fraction of mercury in $HgS$: $\text{Mass fraction of } Hg = \frac{200.6}{232.6}$.
$3$. Calculate the mass of mercury in $225 \,g$ of $HgS$: $\text{Mass of } Hg = \left( \frac{200.6}{232.6} \right) \times 225 \,g = 194.04 \,g$.

(B) $(i)$ For $100 \, g$ of $N_2$:
Number of moles $= \frac{100 \, g}{28 \, g/mol} \approx 3.57 \, mol$.
Number of molecules $= 3.57 \times 6.022 \times 10^{23} \approx 21.505 \times 10^{23}$.
Since each $N_2$ molecule has $2$ atoms,total atoms $= 2 \times 21.505 \times 10^{23} = 43.01 \times 10^{23}$.
$(ii)$ For $100 \, g$ of $NH_3$:
Number of moles $= \frac{100 \, g}{17 \, g/mol} \approx 5.88 \, mol$.
Number of molecules $= 5.88 \times 6.022 \times 10^{23} \approx 35.42 \times 10^{23}$.
Since each $NH_3$ molecule has $4$ atoms ($1$ Nitrogen $+ 3$ Hydrogen),total atoms $= 4 \times 35.42 \times 10^{23} = 141.69 \times 10^{23}$.
Comparing the two,$100 \, g$ of $NH_3$ has more atoms.

(B) The mass of the gold sample is $1 \, g$.
Since the sample contains $90\%$ gold, the mass of gold present is $0.9 \, g$.
The atomic mass of gold $(Au)$ is approximately $197 \, g/mol$.
The number of moles of gold is calculated as: $\text{moles} = \frac{\text{mass}}{\text{atomic mass}} = \frac{0.9 \, g}{197 \, g/mol} \approx 0.004568 \, mol$.
The number of atoms is calculated by multiplying the number of moles by Avogadro's number $(N_A = 6.022 \times 10^{23} \, atoms/mol)$: $\text{Number of atoms} = 0.004568 \times 6.022 \times 10^{23} \approx 2.75 \times 10^{21} \approx 2.77 \times 10^{21}$ atoms.

(B) Mass of silver $= m \, g$
Mass of gold $= \frac{m}{100} \, g$
Number of atoms of silver $= \frac{\text{mass}}{\text{atomic mass}} \times N_A = \frac{m}{108} \times N_A$
Number of atoms of gold $= \frac{\text{mass}}{\text{atomic mass}} \times N_A = \frac{m}{100 \times 197} \times N_A$
Ratio of number of atoms of gold to silver $= \frac{\text{Number of atoms of gold}}{\text{Number of atoms of silver}}$
$= \frac{\frac{m}{100 \times 197} \times N_A}{\frac{m}{108} \times N_A} = \frac{108}{100 \times 197}$
$= \frac{108}{19700} \approx \frac{1}{182.41}$
Thus,the ratio is $1: 182.41$.

(C) The molar mass of methane $(CH_4)$ is $16 \text{ g/mol}$ and ethane $(C_2H_6)$ is $30 \text{ g/mol}$.
The mass of $1.5 \times 10^{20}$ molecules of $CH_4$ is given by: $\text{Mass} = \frac{1.5 \times 10^{20}}{N_A} \times 16 \text{ g}$.
Let $x$ be the number of molecules of $C_2H_6$. The mass of $x$ molecules of $C_2H_6$ is: $\text{Mass} = \frac{x}{N_A} \times 30 \text{ g}$.
Since the masses are equal:
$\frac{1.5 \times 10^{20} \times 16}{N_A} = \frac{x \times 30}{N_A}$.
Solving for $x$:
$x = \frac{1.5 \times 10^{20} \times 16}{30} = \frac{24 \times 10^{20}}{30} = 0.8 \times 10^{20}$ molecules.

(D) The balanced chemical equation for photosynthesis is: $6CO_2 + 6H_2O \xrightarrow{\text{Chlorophyll, Sunlight}} C_6H_{12}O_6 + 6O_2$.
From the equation,$1$ mole of glucose $(C_6H_{12}O_6)$ is produced from $6$ moles of water $(H_2O)$.
The molar mass of glucose is $(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180\, g/mol$.
The molar mass of water is $(2 \times 1) + 16 = 18\, g/mol$.
Thus,$180\, g$ of glucose requires $6 \times 18 = 108\, g$ of water.
Therefore,$18\, g$ of glucose requires: $\frac{108}{180} \times 18 = 10.8\, g$ of water.
Given the density of water is $1\, g\, cm^{-3}$,the volume of water consumed is: $\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{10.8\, g}{1\, g\, cm^{-3}} = 10.8\, cm^3$.

(N/A) The formula unit mass of a substance is the sum of the atomic masses of all atoms in a formula unit of a compound. It is calculated in the same manner as molecular mass,with the distinction that the term 'formula unit mass' is used for substances whose constituent particles are ions,such as $NaCl$.
$(a)$ $Na_{2}CO_{3} \cdot 10H_{2}O = (2 \times \text{Mass of } Na) + (1 \times \text{Mass of } C) + (3 \times \text{Mass of } O) + 10 \times (2 \times \text{Mass of } H + 1 \times \text{Mass of } O)$
$= (2 \times 23) + (1 \times 12) + (3 \times 16) + 10 \times (2 \times 1 + 16)$
$= 46 + 12 + 48 + 10 \times (18)$
$= 106 + 180 = 286 \text{ u}$
$(b)$ $CuSO_{4} = (1 \times \text{Mass of } Cu) + (1 \times \text{Mass of } S) + (4 \times \text{Mass of } O)$
$= 63.5 + 32 + (4 \times 16)$
$= 63.5 + 32 + 64 = 159.5 \text{ u}$

$(i)$ Molar mass of $NH_3 = 14 + (3 \times 1) = 17 \, g/mol$.
$(ii)$ Number of moles $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{1.7 \, g}{17 \, g/mol} = 0.1 \, mol$.
$(iii)$ Number of molecules $= \text{Number of moles} \times \text{Avogadro's number} = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$ molecules.

(N/A) $(i)$ Mole: One mole of any species (atoms,molecules,ions,or particles) is the quantity in number equal to Avogadro's constant $(6.022 \times 10^{23})$ having a mass equal to its atomic or molecular mass in grams.
$(ii)$ Gram atomic mass: The mass of one mole of atoms of an element expressed in grams.
$(iii)$ Gram molecular mass: The mass of one mole of molecules of a substance expressed in grams,which is the sum of the atomic masses of all atoms in the molecule.
$(b)$ $(i)$ The molar mass of $CO_2$ is $(12 + 2 \times 16) = 44 \text{ g/mol}$. Therefore,the mass of $2$ moles of $CO_2$ is $2 \times 44 = 88 \text{ g}$.
$(ii)$ Since $6.022 \times 10^{23}$ molecules represent $1$ mole of $CO_2$,the mass is equal to its molar mass,which is $44 \text{ g}$.

(B) $(i)$ $5$ moles of $CO_{2} = 5 \text{ mol} \times \text{molar mass of } CO_{2}$
$= 5 \text{ mol} \times 44 \text{ g/mol} = 220 \text{ g}$
$(ii)$ $5$ moles of $H_{2}O = 5 \text{ mol} \times \text{molar mass of } H_{2}O$
$= 5 \text{ mol} \times 18 \text{ g/mol} = 90 \text{ g}$
Thus,it can be seen that both have different masses.

(N/A) $(i)$ $NH_{4}NO_{3} = 14 + (1 \times 4) + 14 + (16 \times 3) = 14 + 4 + 14 + 48 = 80 \ u$
$(ii)$ $H_{2}SO_{4} = (2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \ u$
$(iii)$ $Na_{2}CO_{3} = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \ u$
$(iv)$ $NaOH = 23 + 16 + 1 = 40 \ u$
$(v)$ $HNO_{3} = 1 + 14 + (3 \times 16) = 1 + 14 + 48 = 63 \ u$
$(vi)$ $Al_{2}(SO_{4})_{3} = (2 \times 27) + 3 \times [32 + (4 \times 16)] = 54 + 3 \times [32 + 64] = 54 + 3 \times 96 = 54 + 288 = 342 \ u$

(B) Given mass of water $= 90 \text{ g}$.
The molar mass of water $(H_2O)$ is calculated as follows:
$H_2O = (2 \times 1) + 16 = 18 \text{ g/mol}$.
We know that the number of moles is calculated by the formula:
$\text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}}$.
Substituting the values:
$\text{Number of moles} = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ moles}$.
Therefore,there are $5 \text{ moles}$ in $90 \text{ g}$ of water.

(C) $(i)$ The molar mass of water $(H_2O)$ is $(2 \times 1) + 16 = 18 \text{ g/mol}$.
$(ii)$ Number of moles in $90 \text{ g}$ of water $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{90}{18} = 5 \text{ moles}$.
$(iii)$ Number of molecules $=$ Number of moles $\times$ Avogadro's number $(N_A)$.
$= 5 \times 6.022 \times 10^{23} \text{ molecules/mol}$.
$= 3.011 \times 10^{24} \text{ molecules}$.

(A) The molar mass (gram molecular mass) of water $(H_2O)$ is $18 \text{ g/mol}$.
One mole of water contains $6.022 \times 10^{23}$ molecules (Avogadro's number).
Therefore,the mass of $1$ water molecule is calculated as:
$\text{Mass of } 1 \text{ molecule} = \frac{\text{Molar mass}}{\text{Avogadro's number}} = \frac{18 \text{ g}}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23} \text{ g}$.
To find the mass of $20$ water molecules,we multiply the mass of one molecule by $20$:
$\text{Mass of } 20 \text{ molecules} = 20 \times 2.99 \times 10^{-23} \text{ g} = 5.98 \times 10^{-22} \text{ g}$.

(A) $1$. Molar mass of sucrose $(C_{12}H_{22}O_{11})$ = $(12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \, g/mol$.
$2$. Moles of sucrose = $\frac{3.42 \, g}{342 \, g/mol} = 0.01 \, mol$.
$3$. Oxygen atoms from sucrose = $0.01 \times 11 \times N_A = 0.11 \times 6.022 \times 10^{23} = 6.6242 \times 10^{22}$ atoms.
$4$. Moles of water $(H_2O)$ = $\frac{18 \, g}{18 \, g/mol} = 1 \, mol$.
$5$. Oxygen atoms from water = $1 \times 1 \times N_A = 6.022 \times 10^{23}$ atoms.
$6$. Total oxygen atoms = $(0.066242 \times 10^{23}) + (6.022 \times 10^{23}) = 6.088242 \times 10^{23} \approx 6.09 \times 10^{23}$ atoms. (Note: The provided option $A$ is $6.09 \times 10^{22}$,which appears to be a typo in the question's options; however,based on standard calculation,the result is $6.09 \times 10^{23}$).

(C) To find the substance with the highest weight, we calculate the mass of each option using the formula: $\text{Mass} = \text{Number of moles} \times \text{Molar mass}$.
$A$. $0.2$ mole of sucrose $(C_{12}H_{22}O_{11})$:
$\text{Molar mass} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \text{ g/mol}$.
$\text{Mass} = 0.2 \times 342 = 68.4 \text{ g}$.
$B$. $2$ moles of $CO_2$:
$\text{Molar mass} = 12 + (2 \times 16) = 44 \text{ g/mol}$.
$\text{Mass} = 2 \times 44 = 88 \text{ g}$.
$C$. $2$ moles of $CaCO_3$:
$\text{Molar mass} = 40 + 12 + (3 \times 16) = 100 \text{ g/mol}$.
$\text{Mass} = 2 \times 100 = 200 \text{ g}$.
$D$. $10$ moles of $H_2O$:
$\text{Molar mass} = (2 \times 1) + 16 = 18 \text{ g/mol}$.
$\text{Mass} = 10 \times 18 = 180 \text{ g}$.
Comparing the masses: $68.4 \text{ g} < 88 \text{ g} < 180 \text{ g} < 200 \text{ g}$.
Thus, $2$ moles of $CaCO_3$ weigh the highest.

(D) To find the number of atoms,we first calculate the number of moles $(n = \text{mass} / \text{molar mass})$ and then multiply by the number of atoms per molecule and Avogadro's number $(N_A)$.
$A) \, 18 \, g$ of $H_2O$: Molar mass = $18 \, g/mol$. Moles = $18/18 = 1 \, mol$. Atoms per molecule = $3$ $(2H + 1O)$. Total atoms = $1 \times 3 \times N_A = 3N_A$.
$B) \, 18 \, g$ of $O_2$: Molar mass = $32 \, g/mol$. Moles = $18/32 = 0.5625 \, mol$. Atoms per molecule = $2$. Total atoms = $0.5625 \times 2 \times N_A = 1.125N_A$.
$C) \, 18 \, g$ of $CO_2$: Molar mass = $44 \, g/mol$. Moles = $18/44 \approx 0.409 \, mol$. Atoms per molecule = $3$ $(1C + 2O)$. Total atoms = $0.409 \times 3 \times N_A \approx 1.227N_A$.
$D) \, 18 \, g$ of $CH_4$: Molar mass = $16 \, g/mol$. Moles = $18/16 = 1.125 \, mol$. Atoms per molecule = $5$ $(1C + 4H)$. Total atoms = $1.125 \times 5 \times N_A = 5.625N_A$.
Comparing the values,$CH_4$ has the maximum number of atoms.

(A) The number of molecules in a substance is calculated by the formula: $\text{Number of molecules} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_A$,where $N_A$ is Avogadro's number $(6.022 \times 10^{23})$.
Since the given mass is $1\, g$ for all options,the number of molecules is inversely proportional to the molar mass.
$1$. Molar mass of $H_2 = 2 \times 1 = 2\, g/mol$. Number of molecules $\propto 1/2 = 0.5$.
$2$. Molar mass of $CO_2 = 12 + (2 \times 16) = 44\, g/mol$. Number of molecules $\propto 1/44 \approx 0.0227$.
$3$. Molar mass of $N_2 = 2 \times 14 = 28\, g/mol$. Number of molecules $\propto 1/28 \approx 0.0357$.
$4$. Molar mass of $CH_4 = 12 + (4 \times 1) = 16\, g/mol$. Number of molecules $\propto 1/16 = 0.0625$.
Comparing the values,$H_2$ has the smallest molar mass,therefore it contains the maximum number of molecules.

(D) The molar mass of water $(H_{2}O)$ is calculated as: $(2 \times 1) + 16 = 18 \text{ g/mol}$.
By definition,one mole of any substance contains $6.022 \times 10^{23}$ particles (Avogadro's number) and its mass is equal to its molar mass in grams.
Therefore,$1 \text{ mole}$ of $H_{2}O = 18 \text{ g}$ of $H_{2}O$ and contains $6.022 \times 10^{23}$ molecules of $H_{2}O$.
Thus,statements $(i)$ and $(ii)$ are correct.

(B) To find the maximum mass,we calculate the mass for each option:
$1$. $10^{22}$ molecules of $H_2O$: Mass = $(10^{22} / 6.022 \times 10^{23}) \times 18 \approx 0.0166 \times 18 \approx 0.299 \text{ g}$.
$2$. $2$ moles of $H_2O$: Mass = $2 \times 18 = 36 \text{ g}$.
$3$. $10^{22}$ molecules of $NH_3$: Mass = $(10^{22} / 6.022 \times 10^{23}) \times 17 \approx 0.0166 \times 17 \approx 0.282 \text{ g}$.
$4$. $2$ moles of $NH_3$: Mass = $2 \times 17 = 34 \text{ g}$.
Comparing these values,$36 \text{ g}$ is the maximum mass. Thus,$2$ moles of $H_2O$ has the maximum mass.

(B) The statement is False.
Molar mass is defined as the mass of one mole of a substance (atoms or molecules) expressed in grams.
It is the mass of $6.022 \times 10^{23}$ particles of that substance.
Therefore,the mass of two moles would be twice the molar mass,not the molar mass itself.