Chemical stoichiometry Questions in English

(C) Active mass is defined as the molar concentration,which is given by the formula: $\text{Active mass} = \frac{\text{Density} \times 1000}{\text{Molar mass}}$.
Given: $\text{Density} = 1.56 \, g/mL$,$\text{Molar mass} = 78 \, g/mol$.
$\text{Active mass} = \frac{1.56 \times 1000}{78} = \frac{1560}{78} = 20 \, mol/L$.

(A) The dissociation of the compound $AB_5$ in an aqueous solution is given by the equation:
$AB_5 \rightarrow A^{5+} + 5B^{-}$
Given the concentration of $AB_5$ is $10^{-3} \text{ mol/L}$.
According to the stoichiometry of the reaction,$1 \text{ mole}$ of $AB_5$ produces $5 \text{ moles}$ of $B^{-}$.
Therefore,the concentration of $B^{-}$ is:
$[B^{-}] = 5 \times [AB_5] = 5 \times 10^{-3} \text{ mol/L}$.

(C) The dissociation reaction is: $A_2B_3 \rightleftharpoons 2A^{+3} + 3B^{-2}$.
From the stoichiometry of the reaction,$2$ moles of $A^{+3}$ are produced for every $3$ moles of $B^{-2}$.
Therefore,the concentration relationship is given by $\frac{[A^{+3}]}{2} = \frac{[B^{-2}]}{3}$.
Rearranging this,we get $[A^{+3}] = \frac{2}{3} [B^{-2}]$.

(D) Active mass is defined as molar concentration,which is $\text{moles} / \text{volume} (V)$.
$(i)$ Moles of $CO_2 = \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$. Active mass $[CO_2] = \frac{0.5}{V}$.
$(ii)$ Moles of $H_2 = \frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$. Active mass $[H_2] = \frac{1.5}{V}$.
$(iii)$ Moles of $N_2 = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$. Active mass $[N_2] = \frac{0.25}{V}$.
Ratio $[CO_2] : [H_2] : [N_2] = \frac{0.5}{V} : \frac{1.5}{V} : \frac{0.25}{V} = 0.5 : 1.5 : 0.25$.
Multiplying by $4$,we get $2 : 6 : 1$. However,checking the provided options,the ratio $0.5 : 1.5 : 0.25$ simplifies to $1 : 3 : 0.5$.

(A) Given: Mass of $NaOH = 0.4 \ g$,Volume of solution = $40 \ mL$,Molecular weight of $NaOH = 40 \ g/mol$.
$1$. Molarity $(M) = \frac{\text{mass of solute}}{\text{molecular weight} \times \text{volume in L}} = \frac{0.4}{40 \times (40/1000)} = \frac{0.4 \times 1000}{40 \times 40} = 0.25 \ M$.
$2$. Since $NaOH$ is a monoacidic base,its $n$-factor is $1$.
$3$. Normality $(N) = M \times n\text{-factor} = 0.25 \times 1 = 0.25 \ N$.

(C) The relationship between normality $(N)$ and molarity $(M)$ is given by the formula: $N = M \times \text{n-factor}$.
For $Na_2CO_3$,the total positive charge on the cation is $2 \times (+1) = +2$,so the n-factor is $2$.
Substituting the values: $0.2 = M \times 2$.
Therefore,$M = \frac{0.2}{2} = 0.1 \ M$.

(A) The relationship between Normality $(N)$ and Molarity $(M)$ is given by: $N = M \times n$-factor.
For $HCl$,the $n$-factor is $1$. Therefore,for $2N$ $HCl$,the Molarity is $M = \frac{2}{1} = 2 \ M$.
Now,check the Molarity for the options:
For $4N$ $H_2SO_4$,the $n$-factor is $2$. Therefore,$M = \frac{4}{2} = 2 \ M$.
Since the Molarity of $4N$ $H_2SO_4$ is $2 \ M$,which is equal to the Molarity of $2N$ $HCl$,the correct option is $A$.

(B) The molar mass of $NaOH$ is $40 \ g/mol$. Since the valency factor ($n$-factor) of $NaOH$ is $1$,the equivalent mass is also $40 \ g/eq$.
Number of gram equivalents of $NaOH = \frac{\text{mass}}{\text{equivalent mass}} = \frac{2 \ g}{40 \ g/eq} = \frac{1}{20} \ eq$.
Volume of solution = $100 \ cm^3 = 0.1 \ L$.
Normality $(N)$ = $\frac{\text{Number of gram equivalents}}{\text{Volume of solution in } L} = \frac{1/20 \ eq}{0.1 \ L} = \frac{1}{2} \ N$ or $N/2$.

(A) The normality $(N)$ formula is given by: $N = \frac{w}{E \times V_{(L)}}$
For a dibasic acid,the equivalent mass $(E)$ is: $E = \frac{\text{Molar Mass}}{2} = \frac{200}{2} = 100 \, g/eq$.
Given: $N = 0.1 \, N$,$V = 100 \, mL = 0.1 \, L$.
Substituting the values: $0.1 = \frac{w}{100 \times 0.1}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1 \, g$.

(D) The relationship between normality $(N)$ and molarity $(M)$ is given by: $N = M \times \text{n-factor}$.
For phosphoric acid $(H_3PO_4)$,the basicity (n-factor) is $3$.
Therefore,$N = 1 \ M \times 3 = 3 \ N$.

(A) The formula for molarity $(M)$ given percentage by weight $(\% w/W)$ and density $(d)$ is:
$M = \frac{\% (w/W) \times d \times 10}{\text{Molar mass of } H_2SO_4}$
Given:
$\% (w/W) = 95$
$d = 1.834 \ g \ cm^{-3}$
$\text{Molar mass of } H_2SO_4 = 98 \ g \ mol^{-1}$
Calculation:
$M = \frac{95 \times 1.834 \times 10}{98}$
$M = \frac{1742.3}{98}$
$M \approx 17.778 \approx 17.8 \ M$

(C) Molarity is defined as the number of moles of solute per liter of solution.
First,calculate the molar mass of $NaOH$: $23 + 16 + 1 = 40 \ g/mol$.
Number of moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \ g}{40 \ g/mol} = 0.2 \ mol$.
Since the volume of the solution is $1 \ L$,the molarity $= \frac{0.2 \ mol}{1 \ L} = 0.2 \ M$.

(D) The total number of milliequivalents $(N_1V_1 + N_2V_2 + N_3V_3)$ is calculated as follows:
$N_1V_1 = 5 \, mL \times 1 \, N = 5 \, meq$
$N_2V_2 = 20 \, mL \times 0.5 \, N = 10 \, meq$
$N_3V_3 = 30 \, mL \times (1/3) \, N = 10 \, meq$
Total milliequivalents = $5 + 10 + 10 = 25 \, meq$.
The final volume of the solution is $1 \, L = 1000 \, mL$.
Normality $(N)$ = $\frac{\text{Total milliequivalents}}{\text{Total volume in mL}} = \frac{25}{1000} = \frac{1}{40} \, N$ or $N/40$.

(C) For complete neutralization,the number of equivalents of acid must be equal to the number of equivalents of base.
According to the law of equivalence: $N_1V_1 = N_2V_2$.
Given: $N_1 = N/10 = 0.1 \ N$,$V_1 = 25 \ mL$.
We need $N_2V_2 = 0.1 \times 25 = 2.5 \ meq$.
Checking option $C$: $N_2 = N/10 = 0.1 \ N$,$V_2 = 25 \ mL$.
$N_2V_2 = 0.1 \times 25 = 2.5 \ meq$.
Since the equivalents of $HCl$ match the equivalents of $NaOH$,option $C$ is correct.

(A) $1$. Calculate the molarity of the concentrated $H_2SO_4$ solution:
Molarity $(M) = \frac{\text{density} \times \% \text{by weight} \times 10}{\text{molar mass}} = \frac{1.80 \times 98 \times 10}{98} = 18 \ M$.
$2$. Use the dilution formula $M_1V_1 = M_2V_2$:
$18 \ M \times V_1 = 0.1 \ M \times 1000 \ mL$.
$3$. Solve for $V_1$:
$V_1 = \frac{0.1 \times 1000}{18} = \frac{100}{18} \approx 5.55 \ mL$.

(A) The density of pure water is approximately $1000 \, g/L$.
The molar mass of water $(H_2O)$ is $18 \, g/mol$.
Molarity $(M)$ = $\frac{\text{mass of solute in } g}{\text{molar mass} \times \text{volume of solution in } L}$.
Molarity = $\frac{1000 \, g}{18 \, g/mol \times 1 \, L} = 55.56 \, M \approx 55.6 \, M$.

(C) The basicity (n-factor) of $H_3PO_4$ is $3$.
We know the relationship between Normality $(N)$ and Molarity $(M)$ is given by:
$N = M \times \text{n-factor}$
Substituting the given values:
$N = 1.5 \times 3 = 4.5 \ N$

(C) The balanced chemical equation for the reaction is:
$Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2$
From the stoichiometry of the reaction, $1 \, mole$ of $H_2SO_4$ produces $1 \, mole$ of $CO_2$.
First, calculate the number of moles of $H_2SO_4$:
$Moles = Molarity \times Volume \, (in \, L) = 0.1 \, M \times 0.1 \, L = 0.01 \, moles$.
Since $1 \, mole$ of $H_2SO_4$ produces $1 \, mole$ of $CO_2$, $0.01 \, moles$ of $H_2SO_4$ will produce $0.01 \, moles$ of $CO_2$.
At $STP$, $1 \, mole$ of any gas occupies $22.4 \, L$.
Therefore, the volume of $CO_2$ produced is:
$Volume = 0.01 \, moles \times 22.4 \, L/mole = 0.224 \, L$.

(C) The formula for molarity $(M)$ is given by: $M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}}$.
Given:
Mass of $NaCl = 5.85 \ g$
Molar mass of $NaCl = 58.5 \ g/mol$
Volume of solution $= 500 \ mL = 0.5 \ L$.
Substituting the values:
$M = \frac{5.85}{58.5 \times 0.5} = \frac{0.1}{0.5} = 0.2 \ M$.
Therefore,the correct option is $C$.

(A) The formula for Normality $(N)$ is: $N = \frac{W \times 1000}{E \times V(mL)}$
Where $W$ is the mass of the solute in grams,$E$ is the equivalent weight,and $V$ is the volume in $mL$.
For a dibasic acid,the equivalent weight $E = \frac{\text{Molar Mass}}{2} = \frac{200}{2} = 100 \ g/eq$.
Substituting the given values: $0.1 = \frac{W \times 1000}{100 \times 100}$.
$0.1 = \frac{W \times 1000}{10000} = \frac{W}{10}$.
$W = 0.1 \times 10 = 1 \ g$.

(B) To calculate the volume required for dilution,we use the dilution equation:
$M_1V_1 = M_2V_2$
Where:
$M_1 = 10 \, M$ (initial concentration)
$V_1 = ?$ (volume to be determined)
$M_2 = 5 \, M$ (final concentration)
$V_2 = 2.00 \, L$ (final volume)
Substituting the values into the equation:
$10 \, M \times V_1 = 5 \, M \times 2.00 \, L$
$V_1 = \frac{5 \times 2.00}{10} \, L$
$V_1 = 1.00 \, L$

(D) The dilution formula is $N_1V_1 = N_2V_2$.
Here,$N_1 = 36 \ N$ and $V_1 = 8.3 \ mL$.
The total volume after dilution is $V_2 = V_1 + V_{\text{water}} = 8.3 \ mL + 991.7 \ mL = 1000 \ mL$.
Substituting the values: $36 \times 8.3 = N_2 \times 1000$.
$N_2 = \frac{36 \times 8.3}{1000} = \frac{298.8}{1000} = 0.2988 \ N$.
Rounding to the nearest value,we get $N_2 \approx 0.3 \ N$.

(A) $H_3PO_3$ is a dibasic acid,meaning it has two replaceable hydrogen atoms.
The neutralization reaction is: $H_3PO_3 + 2KOH \rightarrow K_2HPO_3 + 2H_2O$.
Using the equivalence relation: $n_{acid} \times \text{basicity} = n_{base} \times \text{acidity}$.
$M_1 \times V_1 \times n_f(acid) = M_2 \times V_2 \times n_f(base)$.
$0.1 \times 20 \times 2 = 0.1 \times V_2 \times 1$.
$V_2 = \frac{0.1 \times 20 \times 2}{0.1 \times 1} = 40 \, mL$.

(B) Let the ratio of volumes be $x : y$.
Using the formula for the concentration of a mixture: $M_{mix} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$
$0.7 = \frac{0.4x + 0.9y}{x + y}$
$0.7(x + y) = 0.4x + 0.9y$
$0.7x + 0.7y = 0.4x + 0.9y$
$0.3x = 0.2y$
$\frac{x}{y} = \frac{0.2}{0.3} = \frac{2}{3}$
Thus,the ratio is $2 : 3$.

(C) According to Avogadro's Law,at constant temperature and pressure,equal volumes of gases contain an equal number of moles.
Since the volume of the vessel is constant,$n_{O_2} = n_{SO_2}$.
We know that $n = \frac{w}{M}$,where $w$ is the weight and $M$ is the molar mass.
Therefore,$\frac{w_{O_2}}{M_{O_2}} = \frac{w_{SO_2}}{M_{SO_2}}$.
The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $SO_2$ is $64 \ g/mol$.
Substituting the values: $\frac{w_{O_2}}{32} = \frac{w_{SO_2}}{64}$.
$w_{O_2} = w_{SO_2} \times \frac{32}{64} = \frac{1}{2} \times w_{SO_2}$.
Thus,the weight of $O_2$ is half the weight of $SO_2$.

(B) Given,molarity of $CO_2$ = $0.1 \, M$ (i.e.,$0.1 \, mol$ of $CO_2$ in $1000 \, mL$ of solution).
Volume of solution = $200 \, mL$.
Number of moles of $CO_2$ in $200 \, mL$ = $\frac{0.1}{1000} \times 200 = 0.02 \, mol$.
At $STP$,$1 \, mol$ of an ideal gas occupies $22.4 \, L$.
Therefore,volume of $0.02 \, mol$ of $CO_2$ = $0.02 \times 22.4 \, L = 0.448 \, L$.

(C) At constant temperature and pressure,the volume is directly proportional to the number of moles: $\frac{V_1}{V_2} = \frac{n_1}{n_2}$.
Given $V_{N_2} = 1 \ L$ and $V_{O_2} = \frac{7}{8} \ L$.
Since $n = \frac{m}{M}$,where $m$ is mass and $M$ is molar mass:
$\frac{V_{N_2}}{V_{O_2}} = \frac{n_{N_2}}{n_{O_2}} = \frac{m_{N_2} / M_{N_2}}{m_{O_2} / M_{O_2}}$
$\frac{1}{7/8} = \frac{m_{N_2} / 28}{m_{O_2} / 32}$
$\frac{8}{7} = \frac{m_{N_2}}{m_{O_2}} \times \frac{32}{28}$
$\frac{8}{7} = \frac{m_{N_2}}{m_{O_2}} \times \frac{8}{7}$
Therefore,$m_{N_2} = m_{O_2}$.

(B) Assume we have $100 \, g$ of air mixture.
Volume occupied by $23 \, g$ of $O_2 = \frac{22.4}{32} \times 23 = 16.1 \, L$ (since $32 \, g$ of $O_2$ occupies $22.4 \, L$ at $STP$).
Volume occupied by $77 \, g$ of $N_2 = \frac{22.4}{28} \times 77 = 61.6 \, L$ (since $28 \, g$ of $N_2$ occupies $22.4 \, L$ at $STP$).
Total volume of the mixture $= 16.1 + 61.6 = 77.7 \, L$.
Percentage by volume of $O_2 = \frac{16.1}{77.7} \times 100 \approx 20.72 \% \approx 20.8 \% $.

(C) Let the mass of each gas be $w \ g$.
The number of moles are: $n_{SO_2} = \frac{w}{64}$,$n_{CH_4} = \frac{w}{16}$,$n_{O_2} = \frac{w}{32}$.
The mole fraction of $CH_4$ is given by: $X_{CH_4} = \frac{n_{CH_4}}{n_{SO_2} + n_{CH_4} + n_{O_2}} = \frac{\frac{w}{16}}{\frac{w}{64} + \frac{w}{16} + \frac{w}{32}}$.
Simplifying the denominator: $\frac{w}{64} + \frac{4w}{64} + \frac{2w}{64} = \frac{7w}{64}$.
Thus,$X_{CH_4} = \frac{w/16}{7w/64} = \frac{64}{16 \times 7} = \frac{4}{7}$.
The partial pressure of $CH_4$ is: $P_{CH_4} = X_{CH_4} \times P_{total} = \frac{4}{7} \times 2.1 \ atm = 1.2 \ atm$.

(C) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in an acidic medium is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ react with $5$ moles of $H_2O_2$.
Therefore,$1$ mole of $KMnO_4$ will react with $\frac{5}{2}$ moles of $H_2O_2$.

(C) The balanced chemical equation for the reaction between acidic $KMnO_4$ and $H_2O_2$ is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
From the stoichiometry of the reaction,$2$ moles of $KMnO_4$ react with $5$ moles of $H_2O_2$.
Therefore,$1$ mole of $KMnO_4$ requires $\frac{5}{2} = 2.5$ moles of $H_2O_2$.

(C) The balanced chemical equation for the reaction of magnesium nitride with water is:
$Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3$
From the stoichiometry of the reaction,$1 \ mol$ of $Mg_3N_2$ reacts with $6 \ mol$ of $H_2O$ to produce $2 \ mol$ of ammonia $(NH_3)$.

(D) The chemical reaction for the Hall-Heroult process is: $2Al_2O_3 + 3C \rightarrow 4Al + 3CO_2$.
According to stoichiometry,$4 \times 27 \, g$ of $Al$ is produced by $3 \times 12 \, g$ of $C$.
So,$108 \, kg$ of $Al$ requires $36 \, kg$ of $C$.
For $270 \, kg$ of $Al$,the weight of $C$ required is: $\frac{36}{108} \times 270 = 90 \, kg$.

(A) The balanced chemical equation is: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$.
From the balanced equation,the coefficients of the products are:
$Na_2CrO_4 = 8$
$Fe_2O_3 = 2$
$CO_2 = 8$
Therefore,the ratio of the products $Na_2CrO_4 : Fe_2O_3 : CO_2$ is $8 : 2 : 8$.

(C) The balanced chemical equation for the oxidation of chromite ore is:
$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
From the balanced equation,the stoichiometric coefficients for the reactants are:
$FeCr_2O_4 = 4$
$Na_2CO_3 = 8$
$O_2 = 7$
Therefore,the ratio of the reactants is $4 : 8 : 7$.

(A) The balanced chemical equation for the disproportionation of potassium manganate in acidic medium is:
$3K_2MnO_4 + 2H_2SO_4 \rightarrow 2KMnO_4 + 2K_2SO_4 + MnO_2 + 2H_2O$.
In this reaction,the products are $KMnO_4$,$K_2SO_4$,$MnO_2$,and $H_2O$.
The stoichiometric coefficients of the products are $2$,$2$,$1$,and $2$ respectively.
However,looking at the provided options and the standard representation of this reaction,the ratio of the products $(KMnO_4 : K_2SO_4 : MnO_2 : H_2O)$ is $2 : 2 : 1 : 2$.

(B) The chemical equation for the reaction is: $C_3H_6 + Br_2 \rightarrow C_3H_6Br_2$
The molar mass of $C_3H_6 = (3 \times 12) + (6 \times 1) = 42 \ g/mol$.
The molar mass of $Br_2 = 2 \times 80 = 160 \ g/mol$.
According to the stoichiometry of the reaction,$42 \ g$ of $C_3H_6$ reacts with $160 \ g$ of $Br_2$.
Therefore,for $21 \ g$ of $C_3H_6$,the amount of $Br_2$ required is: $\frac{160}{42} \times 21 = 80 \ g$.

(A) $1$. Given: Purity = $70\%$ by weight,Specific gravity $(d)$ = $1.54 \ g/mL$.
$2$. Assume $100 \ g$ of the solution. The mass of $H_3PO_4$ is $70 \ g$.
$3$. Volume of the solution $(V)$ = $\frac{\text{Mass}}{\text{Density}} = \frac{100 \ g}{1.54 \ g/mL} = 64.935 \ mL$.
$4$. Equivalent mass of $H_3PO_4$ (tribasic acid) = $\frac{\text{Molar mass}}{3} = \frac{98}{3} = 32.67 \ g/eq$.
$5$. Normality $(N)$ = $\frac{\text{Mass of solute}}{\text{Equivalent mass} \times V(L)} = \frac{70}{32.67 \times (64.935 / 1000)} \approx 11 \ N$.

(B) The neutralization reaction is: $NaOH + HCl \rightarrow NaCl + H_2O$.
According to the law of equivalence,for a $1:1$ reaction,$M_1 V_1 = M_2 V_2$.
Here,$M_1 = 0.6\, M$,$V_1 = ?$,$M_2 = 0.4\, M$,and $V_2 = 30\, cm^3$.
Substituting the values: $0.6 \times V_1 = 0.4 \times 30$.
$V_1 = \frac{0.4 \times 30}{0.6} = \frac{12}{0.6} = 20\, cm^3$.

(A) Let the mass of both gases be $m \ g$.
Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{m}{16}$.
Number of moles of $H_2$ $(n_{H_2})$ = $\frac{m}{2}$.
The partial pressure of a gas is proportional to its mole fraction $(x_i = \frac{n_i}{n_{total}})$.
Mole fraction of $H_2$ $(x_{H_2})$ = $\frac{n_{H_2}}{n_{H_2} + n_{CH_4}} = \frac{\frac{m}{2}}{\frac{m}{2} + \frac{m}{16}}$.
$x_{H_2} = \frac{\frac{m}{2}}{\frac{8m + m}{16}} = \frac{\frac{m}{2}}{\frac{9m}{16}} = \frac{m}{2} \times \frac{16}{9m} = \frac{8}{9}$.
Therefore,the partial pressure caused by $H_2$ is $\frac{8}{9}$ of the total pressure.

(C) The active mass is defined as the molar concentration of the substance in the solution.
Active mass $= \frac{\text{moles of solute}}{\text{Volume of solution in } L}$
First,calculate the number of moles of urea $(NH_2CONH_2)$:
Molar mass of urea $= 12 + 16 + (14 + 2) \times 2 = 60 \ g/mol$
Moles of urea $= \frac{120 \ g}{60 \ g/mol} = 2 \ mol$
Now,calculate the active mass:
Active mass $= \frac{2 \ mol}{5 \ L} = 0.4 \ mol/L$
Therefore,the correct option is $C$.

(A) $1$. The reaction of metallic sodium with water is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$.
$2$. Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$.
$3$. Since $2 \ mol$ of $Na$ produce $2 \ mol$ of $NaOH$,$0.02 \ mol$ of $Na$ will produce $0.02 \ mol$ of $NaOH$.
$4$. The neutralization reaction is: $NaOH + HCl \rightarrow NaCl + H_2O$.
$5$. Moles of $HCl$ required = Moles of $NaOH = 0.02 \ mol$.
$6$. The concentration of $HCl$ solution is $73 \ g/L$. Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
$7$. Molarity of $HCl = \frac{73 \ g/L}{36.5 \ g/mol} = 2 \ M$.
$8$. Using $n = M \times V(L)$,we get $0.02 = 2 \times V$.
$9$. $V = 0.01 \ L = 10 \ mL$.

(A) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent as follows:
$MnO_4^- + e^- \to MnO_4^{2-}$
Here,the change in oxidation state of $Mn$ is from $+7$ to $+6$,so the $n$-factor is $1$.
Therefore,the equivalent weight of $KMnO_4 = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{158}{1} = 158 \ g/eq$.
Using the formula for normality: $\text{Normality} = \frac{\text{Mass}}{\text{Equivalent mass} \times \text{Volume in } L}$.
$0.1 = \frac{\text{Mass}}{158 \times (100/1000)}$.
$\text{Mass} = 0.1 \times 158 \times 0.1 = 1.58 \ g$.

(D) For complete neutralization,the number of equivalents of oxalic acid must equal the number of equivalents of $NaOH$.
$Equivalents = \frac{w}{Eq. \ wt.} = N \times V(L)$
For hydrated oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$,the molar mass is $126 \ g/mol$ and the $n$-factor is $2$,so the equivalent weight is $\frac{126}{2} = 63 \ g/eq$.
Given $V = 100 \ mL = 0.1 \ L$ and $N = 0.2 \ N$.
$\frac{w}{63} = 0.2 \times 0.1$
$\frac{w}{63} = 0.02$
$w = 0.02 \times 63 = 1.26 \ g$.
Thus,the correct option is $(D)$.

(B) $1$. Calculate the moles of $AgNO_3$: The molar mass of $AgNO_3 = 107.8 + 14 + (3 \times 16) = 169.8 \ g/mol$. Assuming the density of the solution is $1 \ g/mL$,mass of $50 \ mL$ solution = $50 \ g$. Moles of $AgNO_3 = (50 \times 0.169) / 169.8 \approx 0.05 \ mol$.
$2$. Calculate the moles of $NaCl$: The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. Moles of $NaCl = (50 \times 0.058) / 58.5 \approx 0.05 \ mol$.
$3$. The reaction is: $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$.
$4$. Since the stoichiometry is $1:1$,$0.05 \ mol$ of $AgNO_3$ reacts with $0.05 \ mol$ of $NaCl$ to produce $0.05 \ mol$ of $AgCl$ precipitate.
$5$. Molar mass of $AgCl = 107.8 + 35.5 = 143.3 \ g/mol$.
$6$. Mass of $AgCl = 0.05 \ mol \times 143.3 \ g/mol = 7.165 \ g \approx 7.17 \ g$.

(D) Let the mass of $H_2$ gas be $x \ g$ and the mass of $O_2$ gas be $4x \ g$.
The molar mass of $H_2$ is $2 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$.
The number of moles of $H_2$ $(n_{H_2})$ $= \frac{x}{2}$.
The number of moles of $O_2$ $(n_{O_2})$ $= \frac{4x}{32} = \frac{x}{8}$.
The molar ratio of $H_2$ to $O_2$ is $\frac{n_{H_2}}{n_{O_2}} = \frac{x/2}{x/8} = \frac{8}{2} = 4 : 1$.