Mix Examples of Some Basic Concept of Chemistry Questions in English

(N/A) The law of multiple proportions is followed. It states that if two elements can combine to form more than one compound,the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.
Fixing the mass of dinitrogen at $28 \ g$,the masses of dioxygen are $32 \ g, 64 \ g, 32 \ g, 80 \ g$. The ratio is $32:64:32:80$,which simplifies to $2:4:2:5$.
$(b)$ $(i) \ 1 \ km = 10^6 \ mm = 10^{15} \ pm$
$(ii) \ 1 \ mg = 10^{-6} \ kg = 10^6 \ ng$
$(iii) \ 1 \ mL = 10^{-3} \ L = 10^{-3} \ dm^3$

(A) $1$. Calculate moles of each gas: $n(H_2) = 20/2 = 10 \ mol$,$n(CO_2) = 220/44 = 5 \ mol$,$n(N_2) = 140/28 = 5 \ mol$.
$2$. Total moles $n_{\text{total}} = 10 + 5 + 5 = 20 \ mol$.
$3$. Using ideal gas equation $PV = nRT$: $P = (nRT)/V = (20 \times 8.34 \times 10^{-2} \times 300) / 2 = 249.42 \ bar$.
$4$. To reduce pressure by $50 \%$,total moles must be reduced to $10 \ mol$.
$5$. Since $n(H_2) = 10 \ mol$,removing all $H_2$ gas will reduce the total moles to $5 + 5 = 10 \ mol$,thereby reducing the pressure by $50 \%$.

(N/A) $(1)$ Mercury $(II)$ chloride: $HgCl_2$
$(2)$ Nickel $(II)$ sulphate: $NiSO_4$
$(3)$ Tin $(IV)$ oxide: $SnO_2$
$(4)$ Thallium $(I)$ sulphate: $Tl_2SO_4$

(D) $1$. Calculate percentage of $C$: $\text{Mass of } C = (12/44) \times 0.792 = 0.216 \ g$. $\% C = (0.216 / 0.45) \times 100 = 48\%$.
$2$. Calculate percentage of $H$: $\text{Mass of } H = (2/18) \times 0.324 = 0.036 \ g$. $\% H = (0.036 / 0.45) \times 100 = 8\%$.
$3$. Calculate percentage of $N$: $\text{Total } H_2SO_4 = 50 \times 0.25 = 12.5 \ \text{meq}$. $\text{Acid used for } NH_3 = 12.5 - (77.0 \times 0.25 / 2) = 12.5 - 9.625 = 2.875 \ \text{meq}$. $\text{Mass of } N = (2.875 \times 14) / 1000 = 0.04025 \ g$. $\% N = (0.04025 / 0.24) \times 100 \approx 16.77\%$.
$4$. Calculate percentage of $O$: $\% O = 100 - (48 + 8 + 16.77) = 27.23\%$.
$5$. Empirical formula calculation: $C: 48/12 = 4, H: 8/1 = 8, N: 16.77/14 \approx 1.2, O: 27.23/16 \approx 1.7$. Dividing by smallest $(1.2)$: $C: 3.33, H: 6.66, N: 1, O: 1.4$. Multiplying by $3$: $C: 10, H: 20, N: 3, O: 4$. The empirical formula is $C_{10}H_{20}N_3O_4$.

(N/A) $(i)$ Molar mass
$(ii)$ Molality
$(iii)$ $32^{\circ}F$ to $212^{\circ}F$
$(iv)$ $6$

(A) $(i)$ False. For ${H_3PO_3}$,the basicity is $2$. Therefore,$\text{Normality} = \text{Molarity} \times \text{Basicity}$. $0.6 = M \times 2$,so $M = 0.3 \ M$.
$(ii)$ False. The correct relation is $\text{Molecular formula} = n \times \text{Empirical formula}$.
$(iii)$ True. In ${Cl_2O}$,$2 \times 35.5 \ g$ of ${Cl}$ combines with $16 \ g$ of ${O}$. Thus,$35.5 \ g$ of ${Cl}$ combines with $8 \ g$ of ${O}$.

(N/A) $(i)$ True: This is the standard conversion formula between Fahrenheit and Celsius scales.
$(ii)$ False: John Dalton published his work in the book titled "$A$ New System of Chemical Philosophy" in $1808$,which laid the foundation for modern atomic theory.
$(iii)$ False: For a dibasic acid,the equivalent mass is calculated as $\frac{\text{Molar Mass}}{\text{Basicity}} = \frac{100}{2} = 50$,not $200$.

(A) The correct matches are as follows:
$(1)$ Roald Hoffmann stated that chemistry is not the science of $100$ elements $(C)$.
$(2)$ Lavoisier proposed the Law of Conservation of Mass,stating that matter can neither be created nor destroyed $(D)$.
$(3)$ Joseph Proust proposed the Law of Definite Proportions,stating that the proportion of the mass of elements in any compound is fixed $(A)$.
$(4)$ Dalton proposed the Atomic Theory,stating that the smallest indivisible particle of matter is the atom $(B)$.
Therefore,the correct sequence is $(1-C, 2-D, 3-A, 4-B)$.

(A) $(A)-(2), (B)-(3), (C)-(1), (D)-(5), (E)-(4)$
Step-by-step calculation:
$(A)$ $88 \ g$ of $CO_2$: Molar mass of $CO_2 = 12 + 2(16) = 44 \ g/mol$. Moles $= 88/44 = 2 \ mol$.
$(B)$ $6.022 \times 10^{23}$ molecules of water: $1 \ mol$ contains $6.022 \times 10^{23}$ particles.
$(C)$ $5.6 \ L$ of $O_2$ at $STP$: Moles $= 5.6/22.4 = 0.25 \ mol$ (Note: The provided option $0.2$ is the closest approximation for $5.6 \ L$ if $22.4$ is used,though $5.6/22.4 = 0.25$. Given the options,$0.25$ is the standard value).
$(D)$ $96 \ g$ of $O_2$: Molar mass of $O_2 = 32 \ g/mol$. Moles $= 96/32 = 3 \ mol$.
$(E)$ One mole of any gas at $STP$ contains $6.022 \times 10^{23}$ molecules.

(N/A) From Avogadro's law,the volume of $1 \ mol$ of any gas at $STP$ is $22.4 \ L$.
The volume occupied by $1 \ g$ of a gas is given by $\frac{22.4 \ L}{\text{Molar mass in } g \ mol^{-1}}$.
$1$. For $CO$ (Molar mass $= 28 \ g \ mol^{-1}$): Volume $= \frac{22.4}{28} = 0.80 \ L$.
$2$. For $H_2O$ (Molar mass $= 18 \ g \ mol^{-1}$): Volume $= \frac{22.4}{18} \approx 1.24 \ L$.
$3$. For $CH_4$ (Molar mass $= 16 \ g \ mol^{-1}$): Volume $= \frac{22.4}{16} = 1.40 \ L$.
$4$. For $NO$ (Molar mass $= 30 \ g \ mol^{-1}$): Volume $= \frac{22.4}{30} \approx 0.75 \ L$.
Comparing the volumes,$CH_4$ occupies the greatest volume $(1.40 \ L)$ and $NO$ occupies the smallest volume $(0.75 \ L)$.

(B) The relationship between enthalpy of solution,lattice enthalpy,and hydration enthalpy is given by the Born-Haber cycle equation:
$\Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration}$
Given:
$\Delta H_{lattice} = 788 \ kJ \ mol^{-1}$
$\Delta H_{sol} = 4 \ kJ \ mol^{-1}$
Substituting these values into the equation:
$4 = 788 + \Delta H_{hydration}$
$\Delta H_{hydration} = 4 - 788 = -784 \ kJ \ mol^{-1}$

(C) To find the lightest,we calculate the mass for each option:
$1$. Mass of $0.2 \ mol$ of $H_2 = 0.2 \ mol \times 2 \ g/mol = 0.4 \ g$.
$2$. $6.023 \times 10^{23}$ molecules of $N_2$ corresponds to $1 \ mol$. Mass of $1 \ mol$ of $N_2 = 1 \ mol \times 28 \ g/mol = 28 \ g$.
$3$. Mass of $0.1 \ g$ of silver is given as $0.1 \ g$.
$4$. Mass of $0.1 \ mol$ of $O_2 = 0.1 \ mol \times 32 \ g/mol = 3.2 \ g$.
Comparing the masses: $0.1 \ g < 0.4 \ g < 2.8 \ g < 3.2 \ g$.
Therefore,$0.1 \ g$ of silver is the lightest.

(A) Mohr's salt is $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$. The number of water molecules is $6$.
Potash alum is $KAl(SO_4)_2 \cdot 12H_2O$. The number of water molecules is $12$.
The ratio of the number of water molecules in Mohr's salt to potash alum is $\frac{6}{12} = \frac{1}{2} = 0.5$.
Expressing $0.5$ in the form $.... \times 10^{-1}$,we get $5 \times 10^{-1}$.

(A) Let the number of moles of $NaOH$ and $Na_{2}CO_{3}$ be $a$ each,as the mixture is equimolar.
The molar mass of $NaOH$ is $40 \ g/mol$ and the molar mass of $Na_{2}CO_{3}$ is $106 \ g/mol$.
The total mass of the mixture is given by: $W_{NaOH} + W_{Na_{2}CO_{3}} = 4 \ g$.
Substituting the masses in terms of moles: $(a \times 40) + (a \times 106) = 4$.
$146a = 4 \Rightarrow a = \frac{4}{146} \ mol$.
The mass of $NaOH$ $(x)$ is: $x = a \times 40 = \frac{4}{146} \times 40 = \frac{160}{146} \approx 1.095 \ g$.
Rounding to the nearest integer,$x = 1 \ g$.

(C) Initial amount of $H_{2}SO_{4} = 0.02 \, mmol$.
$50 \%$ of the solution is taken,so the amount of $H_{2}SO_{4}$ in this portion is $0.5 \times 0.02 \, mmol = 0.01 \, mmol$.
After adding $0.01 \, mmol$ of $H_{2}SO_{4}$ to solution $(A)$,the total amount is $0.01 \, mmol + 0.01 \, mmol = 0.02 \, mmol$.
To express $0.02 \, mmol$ in the form $x \times 10^{-3} \, mmol$,we write $0.02 = 20 \times 10^{-3}$.
Thus,the value is $20$.

(B) Concentration in $ppm = \frac{\text{Mass of solute (in } g)}{\text{Volume of solution (in } mL)} \times 10^{6}$.
Given $ppm = 48$,Volume $= 2 \, L = 2000 \, mL$.
Mass of $Mg = \frac{48 \times 2000}{10^{6}} = 96 \times 10^{-3} \, g = 0.096 \, g$.
Moles of $Mg = \frac{\text{Mass}}{\text{Atomic mass}} = \frac{0.096}{24} = 0.004 \, mol = 4 \times 10^{-3} \, mol$.
Number of $Mg$ atoms $= \text{Moles} \times N_{A} = 4 \times 10^{-3} \times 6.02 \times 10^{23} = 24.08 \times 10^{20}$.
Comparing with $x \times 10^{20}$,we get $x = 24.08$.
The nearest integer value of $x$ is $24$.

(D) Initial moles of $HNO_3$ = $Molarity \times Volume(L) = 0.5 \times 0.8 = 0.4 \ mol$.
Moles of $HNO_3$ evaporated = $\frac{Mass}{Molar \ mass} = \frac{11.5 \ g}{63 \ g \ mol^{-1}} \approx 0.1825 \ mol$.
Moles of $HNO_3$ remaining = $0.4 - 0.1825 = 0.2175 \ mol$.
New volume of solution = $\frac{800 \ mL}{2} = 400 \ mL = 0.4 \ L$.
New molarity = $\frac{Moles \ remaining}{Volume(L)} = \frac{0.2175}{0.4} = 0.54375 \ M$.
Expressing in $x \times 10^{-2} \ M$,we get $54.375 \times 10^{-2} \ M$.
Rounding to the nearest integer,$x = 54$.

(A) The reaction is: $H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$.
Number of milliequivalents of $H_{2}SO_{4} = \text{Molarity} \times \text{n-factor} \times \text{Volume (mL)} = 0.1 \times 2 \times 100 = 20 \ mEq$.
Number of milliequivalents of $NaOH = 0.1 \times 1 \times 50 = 5 \ mEq$.
Since $H_{2}SO_{4}$ is in excess,the remaining milliequivalents of $H_{2}SO_{4} = 20 - 5 = 15 \ mEq$.
Total volume of the solution $= 100 \ mL + 50 \ mL = 150 \ mL$.
Normality of the resulting solution $= \frac{\text{Remaining mEq}}{\text{Total Volume (mL)}} = \frac{15}{150} = 0.1 \ N = 1 \times 10^{-1} \ N$.
Thus,the nearest integer is $1$.

(C) Mass of $AgX = 2.21 \ g$.
Mass of $X$ in $2.00 \ g$ of the compound $= \frac{46.78}{100} \times 2.00 \ g = 0.9356 \ g \approx 0.94 \ g$.
Mass of $Ag$ in $AgX = \text{Total mass of } AgX - \text{Mass of } X = 2.21 \ g - 0.94 \ g = 1.27 \ g$.
Number of moles of $Ag = \frac{1.27 \ g}{108 \ g/mol} \approx 0.01176 \ mol$.
Since the stoichiometry of $AgX$ is $1:1$,the number of moles of $X$ is also $0.01176 \ mol$.
Atomic mass of $X = \frac{\text{Mass of } X}{\text{Moles of } X} = \frac{0.94 \ g}{0.01176 \ mol} \approx 79.93 \ g/mol$.
This corresponds to the atomic mass of Bromine $(Br = 80)$.
Therefore,the halogen $X$ is $Br$.

(A) Given $1 \, kcal = 4.184 \, kJ$. Therefore,$2500 \, kcal = 2500 \times 4.184 \, kJ = 10460 \, kJ$. The energy requirement is $10460 \, kJ$.
$(b)$ Molar mass of sucrose $(C_{12}H_{22}O_{11})$ = $(12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \, g/mol$.
Energy released per mole of sucrose = $5.6 \times 10^6 \, J = 5600 \, kJ/mol$.
Moles of sucrose required = $\frac{10460 \, kJ}{5600 \, kJ/mol} \approx 1.8679 \, mol$.
Mass of sucrose = $1.8679 \, mol \times 342 \, g/mol \approx 638.82 \, g$.
From the equation,$1 \, mol$ of sucrose produces $12 \, mol$ of $CO_2$.
Moles of $CO_2$ produced = $1.8679 \times 12 = 22.4148 \, mol$.
Volume of $CO_2$ at $STP$ = $22.4148 \, mol \times 22.4 \, L/mol \approx 502.09 \, L$.

(B) Total energy provided by $100 \ g$ of glucose is $1800 \ kJ$.
Energy utilized for sports activities is $50 \%$ of $1800 \ kJ = 900 \ kJ$.
The remaining $900 \ kJ$ of energy must be dissipated to avoid storage,which is done through the evaporation of water.
Enthalpy of evaporation of water is $\Delta H_{vap} = 45 \ kJ \ mol^{-1}$.
Number of moles of water required to dissipate $900 \ kJ$ is $n = \frac{900 \ kJ}{45 \ kJ \ mol^{-1}} = 20 \ mol$.
Molar mass of water $(H_2O)$ is $(2 \times 1) + 16 = 18 \ g \ mol^{-1}$.
Mass of water to be perspired $= n \times M = 20 \ mol \times 18 \ g \ mol^{-1} = 360 \ g$.

(C) $1$. Calculate moles of $Na$: $n(Na) = \frac{0.69 \ g}{23 \ g \ mol^{-1}} = 0.03 \ mol$.
$2$. Reaction with water: $Na + H_2O \longrightarrow NaOH + \frac{1}{2} H_2$.
$3$. From the stoichiometry,$1 \ mol$ of $Na$ produces $1 \ mol$ of $NaOH$. Thus,$n(NaOH) = 0.03 \ mol$.
$4$. Neutralization reaction: $NaOH + HCl \longrightarrow NaCl + H_2O$.
$5$. Moles of $HCl$ required = $0.03 \ mol$.
$6$. Concentration of $HCl$ solution: Molarity $M = \frac{73 \ g \ L^{-1}}{36.5 \ g \ mol^{-1}} = 2 \ M$.
$7$. Volume of $HCl$ required: $V = \frac{n}{M} = \frac{0.03 \ mol}{2 \ mol \ L^{-1}} = 0.015 \ L = 15 \ mL$.

(A) . $16 \ g \ CH_{4} = 1 \ mol \ CH_{4}$. Total electrons in $1 \ molecule \ CH_{4} = 6 + 4 = 10$. Total electrons in $1 \ mol = 10 \times 6.02 \times 10^{23} = 60.2 \times 10^{23}$. Thus,$A-II$.
$B$. $1 \ g \ H_{2} = 0.5 \ mol \ H_{2}$. Volume at $STP$ = $0.5 \times 22.4 \ L = 11.2 \ L$. Thus,$B-IV$.
$C$. $1 \ mol \ N_{2} = 1 \times 28 \ g = 28 \ g$. Thus,$C-I$.
$D$. $0.5 \ mol \ SO_{2} = 0.5 \times 64 \ g = 32 \ g$. Thus,$D-III$.
Therefore,the correct match is $A-II, B-IV, C-I, D-III$.

(D) Volume of silver coating = $0.05 \ cm \times (0.05 \ m^2 \times 10^4 \ cm^2/m^2) = 0.05 \ cm \times 500 \ cm^2 = 25 \ cm^3$.
Mass of silver deposited = $\text{Volume} \times \text{Density} = 25 \ cm^3 \times 7.9 \ g \ cm^{-3} = 197.5 \ g$.
Moles of silver atoms = $\frac{\text{Mass}}{\text{Atomic mass}} = \frac{197.5}{108} \approx 1.8287 \ \text{mol}$.
Number of silver atoms = $\text{Moles} \times N_A = 1.8287 \times 6.022 \times 10^{23} \approx 11.01 \times 10^{23}$.
Thus,the number of silver atoms is approximately $11 \times 10^{23}$.

(D) Dalton's atomic theory states that atoms are indivisible,so statement $B$ (Matter consists of divisible atoms) is incorrect.
Dalton's theory states that all atoms of a given element are identical in all respects,including mass,so statement $D$ (All the atoms of given element have different properties including mass) is incorrect.
Statements $A$,$C$,and $E$ are consistent with the original postulates of Dalton's atomic theory.
Therefore,the incorrect statements are $B$ and $D$.

(C, A) $1.$ The reaction involved is $OCl^- + 2I^- + 2H^+ \rightarrow Cl^- + I_2 + H_2O$ and $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$.
From the stoichiometry,$1 \text{ mole of } OCl^- \equiv 1 \text{ mole of } I_2 \equiv 2 \text{ moles of } S_2O_3^{2-}$.
Millimoles of $Na_2S_2O_3 = 48 \ mL \times 0.25 \ N = 12 \text{ mmol}$.
Since $n$-factor for $S_2O_3^{2-}$ is $1$,millimoles of $S_2O_3^{2-} = 12 \text{ mmol}$.
Millimoles of $OCl^- = \frac{12}{2} = 6 \text{ mmol}$.
Molarity of bleach solution $= \frac{6 \text{ mmol}}{25 \ mL} = 0.24 \ M$.
$2.$ Bleaching powder is $CaOCl_2$,which is $Ca(OCl)Cl$.
The oxoacid is hypochlorous acid $(HOCl)$.
The anhydride of $HOCl$ is $Cl_2O$ $(2HOCl \rightarrow Cl_2O + H_2O)$.

(A) The formula for the final concentration $(M_F)$ of a mixture is $M_F = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$.
Given: $M_1 = 2 \ M$,$V_1 = 20 \ mL$,$M_2 = 0.5 \ M$,$V_2 = 400 \ mL$.
Substituting the values: $M_F = \frac{2 \times 20 + 0.5 \times 400}{20 + 400} = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.5714 \ M$.
Converting to the required format: $0.5714 \ M = 57.14 \times 10^{-2} \ M$.
Rounding to the nearest integer,we get $57 \times 10^{-2} \ M$.

(B) Step $1$: Reaction between $NaOH$ and $HCl$ in the beaker:
$NaOH + HCl \rightarrow NaCl + H_2O$
Initial moles of $NaOH = 2 \ M \times 0.01 \ L = 0.02 \ mol$.
Initial moles of $HCl = 1 \ M \times 0.02 \ L = 0.02 \ mol$.
Since the moles are equal,the resulting solution is neutral ($NaCl$ solution).
Step $2$: $10 \ mL$ of this neutral solution is added to a flask containing $2 \ moles$ of $HCl$.
The total volume of the flask is made up to $100 \ mL$ $(0.1 \ L)$.
Molarity of $HCl = \frac{\text{moles of } HCl}{\text{Volume of solution in } L} = \frac{2 \ mol}{0.1 \ L} = 20 \ M$.

(B) . $5.6 \ L$ of $SO_{2(g)}$ at $STP$ = $(5.6 / 22.4) \ mol = 0.25 \ mol$. Mass = $0.25 \times 64 \ g = 16 \ g$ $(IV)$.
$B$. $1.2 \times 10^{24}$ atoms of oxygen. If we assume $O_2$ molecule,$1.2 \times 10^{24} / 6.022 \times 10^{23} \approx 2 \ mol$ of $O$ atoms. Mass = $2 \times 16 \ g = 32 \ g$ $(I)$.
$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ = $(33.6 / 22.4) \ mol = 1.5 \ mol$. Mass = $1.5 \times 44 \ g = 66 \ g$ $(II)$.
$D$. $1.5 \ g$-molecule of $N_2$ = $1.5 \ mol$ of $N_2$. Mass = $1.5 \times 28 \ g = 42 \ g$ $(III)$.
Thus,the correct match is $A-IV, B-I, C-II, D-III$.

(B) The oxidation states are $X = +2$,$Y = +5$,and $Z = -2$.
For a neutral compound,the sum of oxidation states must be zero.
Checking option $B$: $X_3(YZ_4)_2$.
Total charge $= 3 \times (+2) + 2 \times (+5 + 4 \times (-2)) = 6 + 2 \times (5 - 8) = 6 + 2 \times (-3) = 6 - 6 = 0$.
Since the total charge is zero,the correct formula is $X_3(YZ_4)_2$.

(B) The mineral barite is chemically known as barium sulfate,which has the chemical formula $BaSO_4$.
Therefore,the elements present in barite are Barium $(Ba)$,Sulfur $(S)$,and Oxygen $(O)$.

(D) In Stock notation,the oxidation state of the central metal atom is represented by Roman numerals in parentheses.
For $SnCl_4$,the oxidation state of $Sn$ is $+4$,so it is correctly represented as $Sn(IV)Cl_4$.
For $FeCl_3$,the oxidation state of $Fe$ is $+3$,so it should be $Fe(III)Cl_3$.
For $MnO_2$,the oxidation state of $Mn$ is $+4$,so it should be $Mn(IV)O_2$.
For $AuCl_3$,the oxidation state of $Au$ is $+3$,so it should be $Au(III)Cl_3$.

(A) Mass of $1 \ g$ atom of carbon $= 12 \ g$.
Mass of $3.011 \times 10^{23}$ atoms of oxygen $= 0.5 \ mol \times 16 \ g/mol = 8 \ g$.
Mass of $10 \ mL$ of water $= 10 \ g$ (since density $\approx 1 \ g/mL$).
Mass of $\frac{1}{2} \ mol$ of $CH_4 = 0.5 \ mol \times 16 \ g/mol = 8 \ g$.
Comparing the masses: $12 \ g > 10 \ g > 8 \ g$.
Therefore,$1 \ g$ atom of carbon has the highest mass.

(A) The density of water is approximately $1 \ g/cm^3$. Thus,$20 \ cm^3$ of water has a mass of $20 \ g$.
The molar mass of water $(H_2O)$ is $18 \ g/mol$.
The number of moles of water is $n = \frac{20 \ g}{18 \ g/mol} = 1.11 \ mol$.
The number of molecules is $1.11 \times 6.022 \times 10^{23} \approx 6.69 \times 10^{23}$ molecules.
However,in liquid water,the molecules are closely packed. The actual volume occupied by the molecules themselves is negligible compared to the total volume of the liquid because the intermolecular space is very small. For standard calculations in this context,the volume of the molecules is considered to be the volume of the liquid itself,which is $20 \ cm^3$.

(D) The enthalpy of solution $(\Delta_{\text{soln}} H)$ is given by the sum of the lattice enthalpy $(\Delta_{L} H)$ and the hydration enthalpy $(\Delta_{\text{hyd}} H)$.
$\Delta_{\text{soln}} H = \Delta_{L} H + \Delta_{\text{hyd}} H$
Given: $\Delta_{L} H = 699 \ kJ \ mol^{-1}$ and $\Delta_{\text{hyd}} H = -681.8 \ kJ \ mol^{-1}$.
Substituting the values:
$\Delta_{\text{soln}} H = 699 \ kJ \ mol^{-1} + (-681.8 \ kJ \ mol^{-1})$
$\Delta_{\text{soln}} H = 17.2 \ kJ \ mol^{-1}$.

(A) The reaction of iron sulphide with oxygen is: $2FeS + 3O_2 \rightarrow 2FeO + 2SO_2$ $(A)$.
When $SO_2$ $(A)$ is dissolved in water,it forms sulphurous acid: $SO_2 + H_2O \rightarrow H_2SO_3$.
Sulphurous acid $(H_2SO_3)$ is a diprotic acid,meaning it has two replaceable hydrogen atoms.
Therefore,the basicity of $H_2SO_3$ is $2$.

(A) $I$. Moles of $NaOH = \text{Molarity} \times \text{Volume (L)} = 0.2 \times 0.5 = 0.1 \ mol$.
$II$. For $H_2SO_4$,$n$-factor $= 2$. Molarity $= \frac{\text{Normality}}{n-\text{factor}} = \frac{0.1}{2} = 0.05 \ M$.
Moles of $H_2SO_4 = 0.05 \times 0.2 = 0.01 \ mol$.
$III$. Molar mass of urea $(NH_2CONH_2) = 60 \ g/mol$.
Moles of urea $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6}{60} = 0.1 \ mol$.
Thus,the number of moles are $0.1, 0.01, 0.1$. Hence,option $(A)$ is correct.

(C) The pressure $P$ of a gas mixture is given by $P = \frac{nRT}{V}$. Since $R, T,$ and $V$ are constant,$P \propto n_{total}$.
In the first container,the total moles $n_1 = n_{H_2} + n_{He} + n_{O_2} = \frac{16}{2} + \frac{16}{4} + \frac{16}{32} = 8 + 4 + 0.5 = 12.5 \ mol$.
Given $P_1 = 10 \ atm$,so $10 \propto 12.5$.
In the second container,the total moles $n_2 = n_{He} + n_{O_2} = \frac{16}{4} + \frac{16}{32} = 4 + 0.5 = 4.5 \ mol$.
Therefore,the new pressure $P_2 = \frac{n_2}{n_1} \times P_1 = \frac{4.5}{12.5} \times 10 = 0.36 \times 10 = 3.6 \ atm$.

(B) Given: Total pressure $P_{total} = 2 \ bar$,Partial pressure of $H_2$ $(P_{H_2})$ = $1.778 \ bar$.
Partial pressure of $O_2$ $(P_{O_2})$ = $P_{total} - P_{H_2} = 2 - 1.778 = 0.222 \ bar$.
According to Dalton's Law,the mole fraction $(x)$ is proportional to the partial pressure: $x_{H_2} = \frac{P_{H_2}}{P_{total}} = \frac{1.778}{2} = 0.889$ and $x_{O_2} = \frac{P_{O_2}}{P_{total}} = \frac{0.222}{2} = 0.111$.
Assuming $1 \ mole$ of the mixture,moles of $H_2$ $(n_{H_2})$ = $0.889 \ mol$ and moles of $O_2$ $(n_{O_2})$ = $0.111 \ mol$.
Mass of $H_2$ = $n_{H_2} \times M_{H_2} = 0.889 \times 2 = 1.778 \ g$.
Mass of $O_2$ = $n_{O_2} \times M_{O_2} = 0.111 \times 32 = 3.552 \ g$.
Total mass = $1.778 + 3.552 = 5.33 \ g$.
Weight percentage of $H_2$ = $(\frac{1.778}{5.33}) \times 100 \approx 33.33 \%$.

(A) According to the conditions of the question:
$2Fe_2S_3 + 9O_2 \rightarrow 2Fe_2O_3 + 6SO_2$ $(A)$
$SO_2 + H_2O \rightarrow H_2SO_3$
The acid formed is sulphurous acid $(H_2SO_3)$.
Since $H_2SO_3$ has two replaceable hydrogen atoms,its basicity is $2$.

(B) Step $1$: Calculate the concentration of the $HCl$ solution using the titration formula $N_1 V_1 = N_2 V_2$.
Given: $N_1 = 0.1 \ N$ $(NaOH)$,$V_1 = 25 \ mL$,$V_2 = 12.5 \ mL$ $(HCl)$.
$0.1 \times 25 = N_2 \times 12.5 \implies N_2 = \frac{2.5}{12.5} = 0.2 \ N$.
Step $2$: Calculate the volume of water to be added for dilution using $N_1 V_1 = N_2 V_2$.
Initial state: $N_1 = 0.2 \ N$,$V_1 = 500 \ mL$.
Final state: $N_2 = 0.1 \ N$,$V_2 = ?$.
$0.2 \times 500 = 0.1 \times V_2 \implies V_2 = 1000 \ mL$.
Step $3$: Calculate the volume of water to be added.
Volume of water = $V_2 - V_1 = 1000 \ mL - 500 \ mL = 500 \ mL$.

(C) $1$. When $Na_2O$ and $CaO$ are dissolved in water,they form their respective hydroxides: $Na_2O + H_2O \rightarrow 2NaOH$ and $CaO + H_2O \rightarrow Ca(OH)_2$.
$2$. When this solution is saturated with excess $CO_2$,the hydroxides react to form bicarbonates: $NaOH + CO_2 \rightarrow NaHCO_3$ and $Ca(OH)_2 + 2CO_2 \rightarrow Ca(HCO_3)_2$.
$3$. Both $NaHCO_3$ and $Ca(HCO_3)_2$ are soluble in water.
$4$. The resulting solution contains these bicarbonates,which make the solution slightly acidic due to the presence of excess dissolved $CO_2$ (forming carbonic acid) and the nature of the bicarbonate salts.

(A) The percent labeling of oleum is defined as $100 + x$,where $x$ is the mass of water required to convert all $SO_3$ in $100 \ g$ of oleum into $H_2SO_4$.
For $118\%$ oleum,$x = 18 \ g$ of $H_2O$.
The reaction is $SO_3 + H_2O \rightarrow H_2SO_4$.
Since $18 \ g$ of $H_2O$ is $1 \ mole$,it reacts with $1 \ mole$ of $SO_3$ $(80 \ g)$.
Thus,in $100 \ g$ of oleum,there is $80 \ g$ of $SO_3$ and $20 \ g$ of $H_2SO_4$.
Moles of $SO_3 = \frac{80 \ g}{80 \ g/mol} = 1 \ mol$.
Moles of $H_2SO_4 = \frac{20 \ g}{98 \ g/mol} \approx 0.204 \ mol$.
Neutralization reactions are:
$SO_3 + 2NaOH \rightarrow Na_2SO_4 + H_2O$ (requires $2 \ mol$ $NaOH$ per $mol$ $SO_3$)
$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ (requires $2 \ mol$ $NaOH$ per $mol$ $H_2SO_4$)
Total $NaOH$ moles $= 2(1) + 2(0.204) = 2 + 0.408 = 2.408 \ mol \approx 2.4 \ mol$.

(B) The reaction for neutralization is:
$2NaOH + H_2SO_4 \longrightarrow Na_2SO_4 + 2H_2O$
Since $Na_2SO_4$ is neutral,only $NaOH$ reacts with $H_2SO_4$.
Number of millimoles of $H_2SO_4 = 100 \ mL \times 0.5 \ M = 50 \ mmol$.
From the stoichiometry,$1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$.
Therefore,moles of $NaOH = 2 \times 50 \ mmol = 100 \ mmol = 0.1 \ mol$.
Mass of $NaOH = 0.1 \ mol \times 40 \ g/mol = 4 \ g$.
Mass of $Na_2SO_4 = \text{Total mass} - \text{Mass of } NaOH = 100 \ g - 4 \ g = 96 \ g$.

(B) Given,molarity of $H_2SO_4$ solution $= 10.0 \ M$.
Volume $(V) = 50 \ mL$.
Density of solution $(d) = 1.4 \ g/cc$.
Molarity $10 \ M$ means $10 \ moles$ of $H_2SO_4$ are present in $1000 \ mL$ of solution.
Molar mass of $H_2SO_4$ $(M_B) = 98 \ g/mol$.
Mass of $H_2SO_4$ in $1000 \ mL = 10 \ mol \times 98 \ g/mol = 980 \ g$.
Mass of $H_2SO_4$ in $50 \ mL = (980 \ g / 1000 \ mL) \times 50 \ mL = 49 \ g$.
Mass of solution $= d \times V = 1.4 \ g/cc \times 50 \ mL = 70 \ g$.
Mass of solvent $(w_A) = \text{Mass of solution} - \text{Mass of solute} = 70 \ g - 49 \ g = 21 \ g$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{49 \ g / 98 \ g/mol}{21 \ g / 1000 \ g/kg} = 0.5 \ mol \times (1000 / 21) \ kg^{-1} \approx 23.8 \ m$.

(C) Let the weight of each gas be $x \ g$.
Number of moles of $CH_{4} = \frac{x}{16} \ mol$.
Number of moles of $H_{2} = \frac{x}{2} \ mol$.
Total moles = $\frac{x}{16} + \frac{x}{2} = \frac{x + 8x}{16} = \frac{9x}{16} \ mol$.
The mole fraction of $H_{2}$ is $X_{H_{2}} = \frac{n_{H_{2}}}{n_{total}} = \frac{x/2}{9x/16} = \frac{x}{2} \times \frac{16}{9x} = \frac{8}{9}$.
According to Dalton's Law,the partial pressure is proportional to the mole fraction.
Therefore,the fraction of total pressure exerted by $H_{2} = \frac{8}{9}$.

(A) $1$. Calculate initial moles:
$n_A = \frac{36.0 \ g}{60 \ g \ mol^{-1}} = 0.6 \ mol$
$n_B = \frac{56.0 \ g}{80 \ g \ mol^{-1}} = 0.7 \ mol$
$2$. Determine limiting reagent:
According to the reaction $A + 2B \rightarrow AB_{2}$,$1 \ mol$ of $A$ requires $2 \ mol$ of $B$.
For $0.6 \ mol$ of $A$,required $B = 0.6 \times 2 = 1.2 \ mol$.
Since we have only $0.7 \ mol$ of $B$,'$B$' is the limiting reagent.
Statement $(A)$ is incorrect.
$3$. Calculate product formed:
$0.7 \ mol$ of $B$ will produce $\frac{0.7}{2} = 0.35 \ mol$ of $AB_{2}$.
Molar mass of $AB_{2} = 60 + (2 \times 80) = 220 \ g \ mol^{-1}$.
Mass of $AB_{2} = 0.35 \ mol \times 220 \ g \ mol^{-1} = 77.0 \ g$.
Statement $(B)$ and $(C)$ are correct.
$4$. Calculate unreacted '$A$':
$B$ is limiting,so $0.7 \ mol$ of $B$ reacts with $\frac{0.7}{2} = 0.35 \ mol$ of $A$.
Remaining moles of $A = 0.6 - 0.35 = 0.25 \ mol$.
Mass of remaining $A = 0.25 \ mol \times 60 \ g \ mol^{-1} = 15.0 \ g$.
Statement $(D)$ is correct.
Thus,statements $(B)$,$(C)$,and $(D)$ are correct. Based on the options provided,$(C)$ and $(D)$ are correct.