Mix Examples of Some Basic Concept of Chemistry Questions in English

(C) $1$. Initial moles of $H_2SO_4 = \text{Molarity} \times \text{Volume (in L)} = 1 \ M \times 0.1 \ L = 0.1 \ \text{mol}$.
$2$. Mass of initial solution $= \text{Density} \times \text{Volume} = 1.5 \ g/mL \times 100 \ mL = 150 \ g$.
$3$. Mass of added water $= \text{Density} \times \text{Volume} = 1 \ g/mL \times 400 \ mL = 400 \ g$.
$4$. Total mass of final solution $= 150 \ g + 400 \ g = 550 \ g$.
$5$. Final volume of solution $= \frac{\text{Total mass}}{\text{Final density}} = \frac{550 \ g}{1.25 \ g/mL} = 440 \ mL = 0.44 \ L$.
$6$. Molarity of final solution $= \frac{\text{Moles of solute}}{\text{Final volume in L}} = \frac{0.1 \ \text{mol}}{0.44 \ L} \approx 0.227 \ M \approx 0.23 \ M$.

(A) For $HCl$,normality $(N)$ = molarity $(M)$ $\times$ n-factor = $6 \times 1 = 6 \, N$.
For $HNO_3$,normality $(N)$ = molarity $(M)$ $\times$ n-factor = $8 \times 1 = 8 \, N$.
Total milliequivalents of acid = $(250 \times 6) + (350 \times 8) = 1500 + 2800 = 4300 \, meq$.
Let the volume of water added be $x \, mL$.
The total volume of the final solution = $(250 + 350 + x) \, mL = (600 + x) \, mL$.
Given the final concentration is $3 \, N$,we use the formula $N_1V_1 + N_2V_2 = N_{final}V_{final}$.
$4300 = 3 \times (600 + x)$.
$4300 = 1800 + 3x$.
$3x = 2500$.
$x = 833.3 \, mL$.

(D) The mean molecular weight $M_{avg}$ is given by $\frac{n_1 M_1 + n_2 M_2}{n_1 + n_2}$.
Given,$\frac{a \times 32 + b \times 80}{a + b} = 40$.
$32a + 80b = 40a + 40b$.
$40b = 8a$,which simplifies to $a = 5b$.
Now,for the ratio $b : a$,the mean molecular weight is $\frac{b \times 32 + a \times 80}{b + a}$.
Substituting $a = 5b$ into the expression:
$= \frac{32b + (5b) \times 80}{b + 5b} = \frac{32b + 400b}{6b}$.
$= \frac{432b}{6b} = 72$.

(D) The number of gram equivalents is calculated as: $\text{Number of gram equivalents} = \text{Normality} \times \text{Volume in L} = \text{Molarity} \times n\text{-factor} \times \text{Volume in L}$.
For $H_2SO_4$,the $n\text{-factor} = 2$.
Option $A$: $0.5 \ M \times 2 \times 1 \ L = 1 \ g$ equivalent.
Option $B$: $\text{Number of equivalents} = \frac{\text{Mass}}{\text{Equivalent mass}} = \frac{49 \ g}{49 \ g/eq} = 1 \ g$ equivalent (since equivalent mass of $H_2SO_4 = \frac{98}{2} = 49$).
Option $C$: For $H_2 \ \text{gas}$,$n\text{-factor} = 2$. $\text{Equivalents} = \text{moles} \times n\text{-factor} = 0.5 \ mol \times 2 = 1 \ g$ equivalent.
Since all options result in $1 \ g$ equivalent,the correct answer is $D$.

(D) The number of moles of $AgCl$ is calculated as:
$n_{AgCl} = M \times V(L) = 0.1 \times 0.5 = 0.05 \ mole$.
Since $AgCl$ dissociates as $AgCl \to Ag^{+} + Cl^{-}$,one mole of $AgCl$ produces two moles of ions ($1 \ Ag^{+}$ and $1 \ Cl^{-}$).
Thus,$0.05 \ mole$ of $AgCl$ produces $0.05 \times 2 = 0.1 \ mole$ of total ions.
The number of $Cl^{-}$ ions is $0.05 \times N_A$.
Therefore,all the given options are correct.

(B) For $HCl$,normality $(N)$ is equal to molarity $(M)$ because the $n$-factor is $1$.
Let $V$ be the volume of $5 \ M \ HCl$ used to mix with the entire $500 \ mL$ of $2 \ M \ HCl$ to obtain a $3 \ M$ solution.
Using the formula $M_{mix} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$:
$3 = \frac{2 \times 500 + 5 \times V}{500 + V}$
$3(500 + V) = 1000 + 5V$
$1500 + 3V = 1000 + 5V$
$500 = 2V$
$V = 250 \ mL$
Total volume $= 500 \ mL + 250 \ mL = 750 \ mL$.

(C) Volume of $1 \, drop = \frac{5 \, mL}{50} = 0.1 \, mL$.
Mass of $1 \, drop = \text{density} \times \text{volume} = 1.5 \, g/mL \times 0.1 \, mL = 0.15 \, g$.
Number of moles in $1 \, drop = \frac{\text{mass}}{\text{molecular weight}} = \frac{0.15 \, g}{100 \, g/mol} = 1.5 \times 10^{-3} \, mol$.
Number of molecules in $1 \, drop = \text{moles} \times N_A = 1.5 \times 10^{-3} \times 6.022 \times 10^{23} \approx 9.033 \times 10^{20} \approx 9 \times 10^{20}$.

(C) $NaOH$ is hygroscopic and reacts with $CO_2$ present in the atmosphere to form $Na_2CO_3$ according to the reaction: $2NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O$.
Since $NaOH$ is consumed in this reaction,the molarity (strength) of the $NaOH$ solution decreases.

(C) Let the volume of the $45\%$ acid solution be $V \ mL$. Then the volume of the $20\%$ acid solution is $(800 - V) \ mL$.
Using the principle of conservation of mass for the acid:
$\frac{V \times 45}{100} + \frac{(800 - V) \times 20}{100} = \frac{800 \times 29.875}{100}$
$0.45V + 160 - 0.2V = 239$
$0.25V = 79$
$V = \frac{79}{0.25} = 316 \ mL$

(A) $1$. Normality $(N)$ is calculated as $N = M \times n\text{-factor}$. For $H_2SO_4$,the $n\text{-factor}$ is $2$. Thus,$N = 2.05 \times 2 = 4.1 \ N$.
$2$. Molality $(m)$ is calculated using the formula $m = \frac{1000 \times M}{(1000 \times d - M \times M_B)}$,where $M$ is molarity,$d$ is density,and $M_B$ is the molar mass of $H_2SO_4$ $(98 \ g/mol)$.
$3$. Substituting the values: $m = \frac{1000 \times 2.05}{(1000 \times 1.08 - 2.05 \times 98)} = \frac{2050}{(1080 - 200.9)} = \frac{2050}{879.1} \approx 2.33 \ m$.

(B) Mass of $H_2O_2 = 5 \, g$
Molar mass of $H_2O_2 = 34 \, g/mol$
Number of moles of $H_2O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{5}{34} \approx 0.147 \, mol$
Volume of solution $= 100 \, mL = 0.1 \, L$
Molarity $(M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} = \frac{0.147 \, mol}{0.1 \, L} = 1.47 \, M \approx 1.5 \, M$

(B) $FeSO_4 \cdot 7H_2O$,$MgSO_4 \cdot 7H_2O$,and $ZnSO_4 \cdot 7H_2O$ are isomorphous compounds that crystallize with $7$ molecules of water of crystallization and possess similar crystal structures (heptahydrates).
$Na_2CO_3 \cdot 7H_2O$ does not share this structural similarity with the metal sulfate heptahydrates.

(B) $NaOH$ is a deliquescent substance,meaning it absorbs moisture from the atmosphere.
As it absorbs water,the total volume of the solution increases,which leads to a decrease in the molarity (strength) of the $NaOH$ solution.

(D) Baking powder is a mixture of a base $(NaHCO_3)$ and an acid source.
$Ca(H_2PO_4)_2$ acts as an acidic component. When moistened,it reacts with $NaHCO_3$ to release $CO_2$ gas,which helps the dough rise.
$NaAl(SO_4)_2$ (sodium aluminum sulfate) acts as a slow-acting acid. It reacts with $NaHCO_3$ upon heating or in the presence of moisture to release $CO_2$ gradually,ensuring a steady evolution of gas during the baking process.
Therefore,both statements $(A)$ and $(B)$ are correct.

(B) Similar to $K_2CO_3$ (potassium carbonate),where oxygen atoms are replaced by sulfur atoms,the compound $K_2CS_3$ is known as potassium thiocarbonate.

(A) The reaction of iron sulfide with oxygen is: $4FeS + 7O_2 \to 2Fe_2O_3 + 4SO_2$.
Here,the sulfur oxide $A$ is $SO_2$.
When $SO_2$ is dissolved in water,it forms sulfurous acid: $SO_2 + H_2O \to H_2SO_3$.
$H_2SO_3$ dissociates as: $H_2SO_3 \rightleftharpoons 2H^+ + SO_3^{2-}$.
Since the acid releases $2$ protons $(H^+)$ per molecule,its basicity is $2$.

(D) The chemical formula for calcium chlorite is derived from the calcium ion $(Ca^{2+})$ and the chlorite ion $(ClO_2^-)$.
To balance the charges,two chlorite ions are required for every one calcium ion.
Therefore,the formula is $Ca(ClO_2)_2$.

(D) To find the number of atoms,we calculate the moles of molecules and multiply by the number of atoms per molecule ($N_A$ is Avogadro's number).
$A) \ 1 \ mg \ C_4H_{10} = \frac{10^{-3} \ g}{58 \ g/mol} \times 14 \ atoms \times N_A \approx 0.24 \times 10^{-3} \ N_A \ atoms$
$B) \ 1 \ mg \ N_2 = \frac{10^{-3} \ g}{28 \ g/mol} \times 2 \ atoms \times N_A \approx 0.07 \times 10^{-3} \ N_A \ atoms$
$C) \ 1 \ mg \ Na = \frac{10^{-3} \ g}{23 \ g/mol} \times 1 \ atom \times N_A \approx 0.04 \times 10^{-3} \ N_A \ atoms$
$D) \ 1 \ mL \ H_2O = 1 \ g \ H_2O = \frac{1 \ g}{18 \ g/mol} \times 3 \ atoms \times N_A \approx 0.166 \ N_A \ atoms$
Comparing the values,$1 \ mL$ of $H_2O$ contains the largest number of atoms.

(D) To find the number of oxygen atoms,we calculate the moles of oxygen atoms in each option:
$A) \ 10 \ mL \ H_2O_{(l)}: \text{Density} \approx 1 \ g/mL, \text{so mass} = 10 \ g. \text{Moles of } H_2O = 10/18 \approx 0.556 \ mol. \text{Oxygen atoms} = 0.556 \times 1 = 0.556 \ mol.$
$B) \ 0.1 \ mol \ V_2O_5: \text{Oxygen atoms} = 0.1 \times 5 = 0.5 \ mol.$
$C) \ 12 \ g \ O_3: \text{Moles of } O_3 = 12/48 = 0.25 \ mol. \text{Oxygen atoms} = 0.25 \times 3 = 0.75 \ mol.$
$D) \ 12.04 \times 10^{22} \text{ molecules of } CO_2: \text{Moles} = (12.04 \times 10^{22}) / (6.022 \times 10^{23}) \approx 0.2 \ mol. \text{Oxygen atoms} = 0.2 \times 2 = 0.4 \ mol.$
Comparing the values: $0.556, 0.5, 0.75, 0.4$. The minimum is $0.4 \ mol$ from option $D$.

(A) Let the total number of moles be $n_{total} = 1 \ mol$.
Given mole fraction $X_{O_2} = 0.25$,so $n_{O_2} = 0.25 \ mol$.
Then $n_{O_3} = 1 - 0.25 = 0.75 \ mol$.
Mass of $O_2 = n_{O_2} \times M_{O_2} = 0.25 \ mol \times 32 \ g/mol = 8 \ g$.
Mass of $O_3 = n_{O_3} \times M_{O_3} = 0.75 \ mol \times 48 \ g/mol = 36 \ g$.
Total mass of mixture $= 8 \ g + 36 \ g = 44 \ g$.
Percentage concentration $\left( \frac{w}{W} \% \right)$ of $O_2 = \frac{\text{mass of } O_2}{\text{total mass}} \times 100 = \frac{8}{44} \times 100 = \frac{2}{11} \times 100 \approx 18.18 \%$.

(B) $1$. Molar mass of metal chloride $(MCl_x)$ = $2 \times$ vapour density = $2 \times 89 = 178 \ g/mol$.
$2$. The equation for the chloride is: $M + x(35.5) = 178$ ... $(i)$.
$3$. For the metal oxide $(M_2O_x)$,the mass percentage of metal is $75 \%$,so: $\frac{2M}{2M + 16x} = 0.75$.
$4$. Simplifying: $2M = 0.75(2M + 16x)$ $\Rightarrow 2M = 1.5M + 12x$ $\Rightarrow 0.5M = 12x$ $\Rightarrow M = 24x$ ... $(ii)$.
$5$. Substituting $(ii)$ into $(i)$: $24x + 35.5x = 178$ $\Rightarrow 59.5x = 178$ $\Rightarrow x \approx 3$.
$6$. Atomic weight $M = 24 \times 3 = 72 \ g/mol$.

(A) In the given reaction,$CuSO_4$ reacts with $NaOH$ to form $Cu(OH)_2$,which is an insoluble solid known as a precipitate.
Therefore,this is a precipitate formation reaction.
Thus,option $A$ is correct.

(A) The given reaction is $2AgF + MgCl_2 \longrightarrow MgF_2 \downarrow + 2AgCl \downarrow$.
In this reaction,both products $MgF_2$ and $AgCl$ are insoluble in water and form precipitates.
Since the reaction results in the formation of precipitates,it is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(A) The given reaction is $Pb(NO_3)_2(aq) + 2 KI(aq) \longrightarrow PbI_2(s) \downarrow + 2 KNO_3(aq)$.
In this reaction,lead$(II)$ nitrate reacts with potassium iodide to form a yellow precipitate of lead$(II)$ iodide $(PbI_2)$.
Since a solid precipitate is formed from the aqueous solutions of the reactants,this is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(D) The reaction $CaCl_2 + Na_2SO_4 \longrightarrow CaSO_4(s) + 2NaCl(aq)$ actually occurs,forming a white precipitate of $CaSO_4$.
However,if the question states '$No \ reaction$',it implies that the reactants do not undergo a chemical change under the given conditions or that the products are soluble.
Given the options,'$D$' corresponds to the condition where no reaction is observed.

(C) Let us analyze each reaction:
$1$. $SnSO_4 \xrightarrow{\Delta} SnO_2 + SO_2 \uparrow$. This reaction is correct as $SnSO_4$ decomposes to $SnO_2$ and $SO_2$.
$2$. $Ag_2C_2O_4 \xrightarrow{\Delta} 2Ag + 2CO_2 \uparrow$. This is a correct thermal decomposition reaction of silver oxalate.
$3$. $P_4O_{10(s)} + 6CaO_{(s)} \xrightarrow{\Delta} 2Ca_3(PO_4)_{2(s)}$. This is a correct acid-base reaction forming calcium phosphate. The option provided in the question incorrectly states $Ca_3(PO_4)_2 \uparrow$,implying it is a gas,which is incorrect as it is a solid.
$4$. $PbCl_4 \xrightarrow{\Delta} PbCl_2 + Cl_2 \uparrow$. This is a correct thermal decomposition reaction due to the inert pair effect.
Therefore,option $C$ is incorrect because $Ca_3(PO_4)_2$ is a solid,not a gas.

(D) Step $1$: Calculate the initial number of moles $(n_{initial})$ using the ideal gas equation $PV = nRT$.
Given: $P = 2 \ atm$,$V = 5.6 \ L$,$T = 273 \ K$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
$n_{initial} = \frac{PV}{RT} = \frac{2 \times 5.6}{0.0821 \times 273} \approx \frac{11.2}{22.4} = 0.5 \ mol$.
Step $2$: Calculate the number of moles removed $(n_{removed})$.
$n_{removed} = \frac{\text{Number of molecules}}{N_A} = \frac{10^{23}}{6.022 \times 10^{23}} \approx \frac{1}{6} \ mol$.
Step $3$: Calculate the remaining moles $(n_{remaining} = n_{initial} - n_{removed})$.
$n_{remaining} = 0.5 - \frac{1}{6} = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3} \ mol$.

(B) For constant $T$ and $P$,$V \propto n$.
Percentage of $O_2$ by volume $= \frac{n_{O_2}}{n_{O_2} + n_{N_2}} \times 100$.
$= \frac{W_{O_2} / M_{O_2}}{(W_{O_2} / M_{O_2}) + (W_{N_2} / M_{N_2})} \times 100$.
$= \frac{23 / 32}{(23 / 32) + (77 / 28)} \times 100$.
$= \frac{0.71875}{0.71875 + 2.75} \times 100$.
$= \frac{0.71875}{3.46875} \times 100 \approx 20.72 \% \approx 20.8 \%$.

(D) An equimolar mixture means equal number of moles of each gas.
Let the number of moles of $CH_4$ be $1 \, mol$ and $CO$ be $1 \, mol$.
Total mass of the mixture $= (1 \times 16) + (1 \times 28) = 44 \, g$.
Total volume of the mixture at $STP = 2 \times 22.4 \, L = 44.8 \, L$.
Density $= \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{44 \, g}{44.8 \, L} = 0.98 \, g \, L^{-1}$.

(B) $1$. Moles of $He = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \ g}{4 \ g/mol} = 0.25 \ mol$ $(1/4 \ mol)$. So,option $A$ is true.
$2$. Volume at $STP = \text{moles} \times 22.4 \ L = 0.25 \times 22.4 \ L = 5.6 \ L$. Thus,option $B$ is false.
$3$. Rate of diffusion $r \propto \frac{1}{\sqrt{M}}$. Ratio $\frac{r_{He}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{He}}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4$. So,option $C$ is true.
$4$. Atoms in $1 \ g \ He = 0.25 \times N_A$. Molecules in $4 \ g \ CH_4 = \frac{4}{16} \times N_A = 0.25 \times N_A$. So,option $D$ is true.

(D) The molar mass of $H_2SO_4 = 98 \, g/mol$.
The given mass of $H_2SO_4 = 392 \, mg = 392 \times 10^{-3} \, g = 0.392 \, g$.
The number of moles of $H_2SO_4 = \frac{0.392 \, g}{98 \, g/mol} = 0.004 \, mol = 4 \times 10^{-3} \, mol$.
The total number of molecules initially present $= 4 \times 10^{-3} \times 6.022 \times 10^{23} = 2.4088 \times 10^{21}$ molecules.
Since $2.4088 \times 10^{21}$ molecules are removed,the remaining molecules $= 2.4088 \times 10^{21} - 2.4088 \times 10^{21} = 0$.

(C) Initial mass of $CO_2 = 100 \, mg = 0.1 \, g$.
Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \, g/mol$.
Initial moles of $CO_2 = \frac{0.1 \, g}{44 \, g/mol} \approx 2.27 \times 10^{-3} \, mol$.
Moles of $CO_2$ removed = $\frac{10^{21}}{6.022 \times 10^{23}} \approx 1.66 \times 10^{-3} \, mol$.
Remaining moles of $CO_2 = (2.27 \times 10^{-3}) - (1.66 \times 10^{-3}) = 0.61 \times 10^{-3} \, mol = 6.1 \times 10^{-4} \, mol$.
Rounding to the nearest provided option,the answer is $6 \times 10^{-4} \, mol$.

(C) Molar mass of sucrose $(C_{12}H_{22}O_{11})$ = $342 \ g/mol$.
Moles of sucrose = $\frac{34.2 \ g}{342 \ g/mol} = 0.1 \ mol$.
Oxygen atoms from sucrose = $0.1 \ mol \times 11 = 1.1 \ mol$.
Molar mass of water $(H_2O)$ = $18 \ g/mol$.
Moles of water = $\frac{90 \ g}{18 \ g/mol} = 5 \ mol$.
Oxygen atoms from water = $5 \ mol \times 1 = 5 \ mol$.
Total moles of oxygen atoms = $1.1 + 5 = 6.1 \ mol$.
Number of oxygen atoms = $6.1 \times 6.022 \times 10^{23} \approx 3.67 \times 10^{24}$.

(D) The mass is calculated using the formula: $\text{Mass} = \text{moles} \times \text{molar mass}$.
$A)$ $0.1 \, \text{mol} \times 12 \, \text{g/mol} = 1.2 \, \text{g}$
$B)$ $\text{Moles} = \frac{1120 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.05 \, \text{mol}$. $\text{Mass} = 0.05 \times 44 = 2.2 \, \text{g}$
$C)$ $0.1 \, \text{mol} \times 17 \, \text{g/mol} = 1.7 \, \text{g}$
$D)$ $\text{Moles} = \frac{6.022 \times 10^{22}}{6.022 \times 10^{23}} = 0.1 \, \text{mol}$. $\text{Mass} = 0.1 \times 44 = 4.4 \, \text{g}$
Comparing the values,$4.4 \, \text{g}$ is the maximum mass.

(A) For a compound $A_xB_y$,$1 \, mol$ of the compound contains $x \, mol$ of $A$ and $y \, mol$ of $B$ atoms.
Therefore,option $C$ is correct.
Option $D$ is also correct as it represents the definition of molecular mass.
Option $A$ is incorrect because $1 \, mol$ of $A_xB_y$ contains $x \, mol$ of $A$ and $y \, mol$ of $B$,not $1 \, mol$ each.
Option $B$ is also incorrect in a general sense as equivalent mass depends on the reaction stoichiometry (n-factor),but in the context of multiple-choice questions where $A$ is the primary error,$A$ is the most fundamentally incorrect statement regarding stoichiometry.

(D) At $STP$,$22.4 \, L$ of $H_2O_{(g)}$ corresponds to $1 \, \text{mole}$ of $H_2O_{(g)}$.
The molar mass of $H_2O$ is $18 \, g/mol$.
Therefore,the mass of $1 \, \text{mole}$ of $H_2O$ is $18 \, g$.
The density of liquid water is $1 \, g/mL$.
Using the formula $\text{Volume} = \frac{\text{Mass}}{\text{Density}}$,we get $\text{Volume} = \frac{18 \, g}{1 \, g/mL} = 18 \, mL$.

(B) The normality of $HCl$ is $N_1 = 0.1 \, N$ and volume is $V_1 = 40 \, mL$.
For $H_2SO_4$,the molarity is $M = 0.1 \, M$. The n-factor for $H_2SO_4$ is $2$,so its normality is $N_2 = M \times \text{n-factor} = 0.1 \times 2 = 0.2 \, N$. The volume is $V_2 = 20 \, mL$.
The formula for the normality of the mixture is $N_R = \frac{N_1 V_1 + N_2 V_2}{V_1 + V_2}$.
Substituting the values: $N_R = \frac{0.1 \times 40 + 0.2 \times 20}{40 + 20}$.
$N_R = \frac{4 + 4}{60} = \frac{8}{60} = \frac{2}{15} \, N$.

(C) $1 \ g-$ atom of phosphorous $(P)$ $= 31 \ g$
$2 \ moles$ of water $(H_2O)$ $= 2 \times 18 = 36 \ g$
$22.4 \ L$ of $CO_2$ gas at $NTP = 1 \ mole = 44 \ g$
$6.02 \times 10^{23}$ atoms of sulphur $(S)$ $= 1 \ mole = 32 \ g$
Comparing the masses: $44 \ g > 36 \ g > 32 \ g > 31 \ g$.
Thus,$22.4 \ L$ of $CO_2$ gas at $NTP$ has the highest mass.

(A) Let the total number of moles be $n_{total} = 1 \ mol$.
Given mole fraction $X_{O_2} = 0.25$,so $n_{O_2} = 0.25 \ mol$.
Consequently,$n_{O_3} = 1 - 0.25 = 0.75 \ mol$.
Mass of $O_2$ $(w_{O_2})$ = $n_{O_2} \times M_{O_2} = 0.25 \ mol \times 32 \ g/mol = 8 \ g$.
Mass of $O_3$ $(w_{O_3})$ = $n_{O_3} \times M_{O_3} = 0.75 \ mol \times 48 \ g/mol = 36 \ g$.
Percentage concentration $\left( \frac{w}{W}\% \right)$ of $O_2 = \frac{w_{O_2}}{w_{O_2} + w_{O_3}} \times 100 = \frac{8}{8 + 36} \times 100 = \frac{8}{44} \times 100 = \frac{2}{11} \times 100 \approx 18.18\%$.

(B) Let the fraction of $A_2B_{(g)}$ decomposing via the first reaction be $x$ and via the second reaction be $(1-x)$.
In the first reaction,$1 \, \text{mole}$ of $A_2B$ produces $1 \, \text{mole}$ of $A_2$.
In the second reaction,$1 \, \text{mole}$ of $A_2B$ produces $1 \, \text{mole}$ of $A$.
The molar ratio of $A_2$ to $A$ is given as $5:3$,so $\frac{x}{1-x} = \frac{5}{3}$.
Solving for $x$: $3x = 5 - 5x \implies 8x = 5 \implies x = \frac{5}{8}$.
Thus,the fraction of the second reaction is $1 - \frac{5}{8} = \frac{3}{8}$.
The total enthalpy change is $\Delta_r H = x(\Delta H_1) + (1-x)(\Delta H_2)$.
$\Delta_r H = \frac{5}{8} \times 40 + \frac{3}{8} \times 50 = 25 + 18.75 = 43.75 \, \text{kJ/mol}$.

(C) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Given $\Delta H_{vap} = 40.79 \ kJ \ mol^{-1}$.
For $18 \ g$ of water,the heat required for evaporation is $40.79 \ kJ$.
For $30 \ g$ of water,the heat required is calculated as:
$q = \frac{30 \ g}{18 \ g \ mol^{-1}} \times 40.79 \ kJ \ mol^{-1} = 1.666 \ mol \times 40.79 \ kJ \ mol^{-1} \approx 67.98 \ kJ \approx 68 \ kJ$.

(A) $1$. Calculate the moles of reactants:
$n(H_2SO_4) = 0.4 \ L \times 0.2 \ M = 0.08 \ mol$.
$n(NaOH) = 0.6 \ L \times 0.1 \ M = 0.06 \ mol$.
$2$. The reaction is $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Since $NaOH$ is the limiting reagent, the heat evolved corresponds to the neutralization of $0.06 \ mol$ of $NaOH$.
$3$. The total volume of the solution is $400 \ mL + 600 \ mL = 1000 \ mL = 1 \ L$.
Assuming the density of the solution is $1 \ g/mL$, the mass of the solution is $1000 \ g$.
$4$. Using the formula $q = m \times c \times \Delta T$:
$3430 \ J = 1000 \ g \times 4.18 \ J \ K^{-1} \ g^{-1} \times \Delta T$.
$\Delta T = \frac{3430}{4180} \approx 0.82 \ K$.

(D) In aqueous solution,$AlCl_3$ undergoes ionization because the high hydration energy released during the formation of $[Al(H_2O)_6]^{3+}$ ions is sufficient to overcome the high lattice energy and ionization energy required to break the $Al-Cl$ bonds.

(A) Mass of $6.022 \times 10^{23}$ atoms of oxygen $= 16 \ g$.
Mass of one atom of oxygen $= \frac{16}{6.022 \times 10^{23}} \approx 2.66 \times 10^{-23} \ g$.
Mass of $6.022 \times 10^{23}$ atoms of nitrogen $= 14 \ g$.
Mass of one atom of nitrogen $= \frac{14}{6.022 \times 10^{23}} \approx 2.32 \times 10^{-23} \ g$.
Mass of $1 \times 10^{-10}$ mole of oxygen $= 16 \times 10^{-10} \ g$.
Mass of $1 \times 10^{-10}$ mole of copper $= 63 \times 10^{-10} \ g$.
Comparing the values:
$2.32 \times 10^{-23} (II) < 2.66 \times 10^{-23} (I) < 16 \times 10^{-10} (III) < 63 \times 10^{-10} (IV)$.
Thus,the order of increasing mass is $II < I < III < IV$.

(A) Molar mass of methanol $(CH_3OH) = (1 \times 12) + (4 \times 1) + (1 \times 16) = 32 \, g \, mol^{-1} = 0.032 \, kg \, mol^{-1}$.
Molarity of pure methanol $= \frac{\text{Density}}{\text{Molar mass}} = \frac{0.793 \, kg \, L^{-1}}{0.032 \, kg \, mol^{-1}} = 24.78 \, mol \, L^{-1}$.
Using the dilution formula $M_1V_1 = M_2V_2$:
$24.78 \, mol \, L^{-1} \times V_1 = 0.25 \, mol \, L^{-1} \times 2.5 \, L$.
$V_1 = \frac{0.25 \times 2.5}{24.78} \, L = 0.02522 \, L$.
$V_1 = 0.02522 \times 1000 \, mL = 25.22 \, mL$.

(N/A) Chemistry is the branch of science that studies the composition,properties,and interactions of matter.
It plays a central role in science and is often intertwined with other disciplines like physics,biology,and geology.
Chemistry is essential for understanding chemical transformations in nature.
It has significant applications in daily life,such as understanding weather patterns,the functioning of the brain,and the operation of computers.
In medicine,chemistry has provided life-saving drugs like $Cisplatin$ and $Taxol$ for cancer therapy,and $AZT$ $(Azidothymidine)$ for $AIDS$ patients.
It also contributes to environmental protection by synthesizing safer alternatives to hazardous substances like $CFCs$ $(Chlorofluorocarbons)$.

$(i)$ Number of electrons in $1$ molecule of methane $(CH_4) = 6 + 4(1) = 10$.
Number of electrons in $1$ mole $(6.022 \times 10^{23} \text{ molecules}) = 6.022 \times 10^{23} \times 10 = 6.022 \times 10^{24}$.
$(ii)$ $(a)$ $1$ atom of $^{14}C$ contains $(14 - 6) = 8$ neutrons.
Number of neutrons in $14 \, g$ of $^{14}C = 6.022 \times 10^{23} \times 8$.
Number of neutrons in $7 \, mg$ $(0.007 \, g) = \frac{6.022 \times 10^{23} \times 8 \times 0.007}{14} = 2.4088 \times 10^{21}$.
$(b)$ Mass of neutrons $= (2.4088 \times 10^{21}) \times (1.675 \times 10^{-27} \, kg) = 4.0347 \times 10^{-6} \, kg$.
$(iii)$ $(a)$ $1$ mole of $NH_3 = 17 \, g = 6.022 \times 10^{23}$ molecules.
Protons in $1$ molecule of $NH_3 = 7 + 3(1) = 10$.
Number of protons in $34 \, mg$ $(0.034 \, g) = \frac{6.022 \times 10^{23} \times 10 \times 0.034}{17} = 1.2044 \times 10^{22}$.
$(b)$ Mass of protons $= (1.2044 \times 10^{22}) \times (1.6726 \times 10^{-27} \, kg) \approx 2.0145 \times 10^{-5} \, kg$.
The number of subatomic particles is independent of temperature and pressure; hence,the values remain unchanged.

(N/A) Molarity is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V(L)}$.
$(a)$ Molar mass of $Co(NO_3)_2 \cdot 6H_2O = 59 + 2(14 + 48) + 6(18) = 291 \, g \, mol^{-1}$.
Moles of $Co(NO_3)_2 \cdot 6H_2O = \frac{30 \, g}{291 \, g \, mol^{-1}} \approx 0.103 \, mol$.
Molarity $= \frac{0.103 \, mol}{4.3 \, L} \approx 0.0239 \, M$.
$(b)$ Using the dilution formula $M_1V_1 = M_2V_2$:
$0.5 \, M \times 30 \, mL = M_2 \times 500 \, mL$.
$M_2 = \frac{0.5 \times 30}{500} = 0.03 \, M$.

(N/A) The $Arthashastra$ by $Kautilya$ contains a long section on mines and minerals,which describes chemical methods. It includes ores of gold,silver,copper,lead,tin,and iron. $Kautilya$ describes various types of gold,which is known as $Rasavidya$.