Volumetric Analysis Questions in English

(B) In the titration of potassium permanganate $(KMnO_4)$ with ferrous ammonium sulphate (Mohr's salt),the reaction occurs in an acidic medium.
$KMnO_4$ acts as an oxidizing agent and $Fe^{2+}$ acts as a reducing agent.
The reduction half-reaction for $KMnO_4$ in acidic medium is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
The change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$.
The equivalent weight of an oxidizing agent is defined as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{n\text{-factor}}$.
Therefore,the equivalent weight of $KMnO_4$ is $\frac{\text{Molecular weight}}{5}$.

(B) Mohr's salt is a double salt with the formula $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$.
In aqueous solution,the $Fe^{2+}$ ions undergo hydrolysis to form $Fe(OH)_2$ and $H^+$ ions.
This hydrolysis makes the solution turbid.
To prevent this cationic hydrolysis of $Fe^{2+}$ ions,dilute $H_2SO_4$ is added,which provides an acidic medium and suppresses the hydrolysis reaction.

(A) According to the law of equivalence for neutralization:
$N_1 V_1 = N_2 V_2$
Here,$V_1 = 30 \ mL$,$V_2 = 15 \ mL$,and $N_2 = 0.2 \ N$.
Substituting the values:
$N_1 \times 30 = 0.2 \times 15$
$N_1 = \frac{0.2 \times 15}{30}$
$N_1 = 0.1 \ N$
Therefore,the strength of the acid solution is $0.1 \ N$.

(D) . All the given statements are true.
$1$. $H_2SO_4$ is preferred because it is neither an oxidising agent nor a reducing agent under these conditions.
$2$. $HCl$ is not used because it acts as a reducing agent and reacts with $KMnO_4$ to produce $Cl_2$ gas.
$3$. $HNO_3$ is not used because it is a strong oxidising agent and will interfere with the redox reaction by reacting with the reducing agent.

(B) The disodium salt of $EDTA$ (Ethylenediaminetetraacetic acid) is commonly used in complexometric titrations to determine the hardness of water.
It forms stable,water-soluble complexes with $Ca^{2+}$ and $Mg^{2+}$ ions,which are responsible for water hardness.

(C) The reaction between iodine $(I_2)$ and sodium thiosulfate ($Na_2S_2O_3$,commonly known as hypo) is a standard redox titration reaction.
The balanced chemical equation is:
$2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$
Thus,the products formed are sodium tetrathionate $(Na_2S_4O_6)$ and sodium iodide $(NaI)$.

(A) The correct answer is $(A)$. Alkaline solution of pyrogallol is a well-known reagent used in gas analysis to quickly and quantitatively absorb oxygen from a gas mixture.

(C) In iodometric titrations,an oxidizing agent reacts with iodide ions $(I^-)$ to liberate iodine $(I_2)$,which is then titrated against a standard sodium thiosulfate solution.
For a species to participate in an iodometric titration,it must be capable of oxidizing $I^-$ to $I_2$.
$Fe^{3+}$,$Cu^{2+}$,and $Ag^{2+}$ act as oxidizing agents and can oxidize $I^-$ to $I_2$.
$Pb^{2+}$ is in its stable oxidation state and does not act as an oxidizing agent towards iodide ions; therefore,it cannot undergo iodometric titration.

(B) The formula for normality is given by: $N = \frac{W \times 1000}{E \times V(mL)}$
Given:
Normality $(N)$ = $N/10 = 0.1 \, N$
Equivalent weight $(E)$ = $63$
Volume $(V)$ = $250 \, mL$
Substituting the values into the formula:
$0.1 = \frac{W \times 1000}{63 \times 250}$
$0.1 = \frac{W \times 4}{63}$
$W = \frac{0.1 \times 63}{4} = \frac{6.3}{4} = 1.575 \, g$
Therefore,the required amount is $1.575 \, g$.

(B) In the iodometric titration,$K_2Cr_2O_7$ acts as an oxidizing agent.
The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$.
Here,the change in oxidation state of $Cr$ is from $+6$ to $+3$ for each $Cr$ atom. Since there are two $Cr$ atoms in $K_2Cr_2O_7$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,the $n$-factor for $K_2Cr_2O_7$ is $6$.
Equivalent weight = $\frac{Molecular \text{ } weight}{n\text{-}factor} = \frac{Molecular \text{ } weight}{6}$.

(A) Let $N_1$ be the normality of the weak monobasic acid and $N_2$ be the normality of the $NaOH$ solution.
For the acid-base neutralization: $N_1 \times 20 = N_2 \times 22.18$ $(i)$
For the $HCl$ and $NaOH$ neutralization: $N_{HCl} \times V_{HCl} = N_{NaOH} \times V_{NaOH}$
$\frac{1}{10} \times 20 = N_2 \times 21.5$
$N_2 = \frac{2}{21.5} \approx 0.09302 \, N$ $(ii)$
Substitute $N_2$ into equation $(i)$:
$N_1 = \frac{N_2 \times 22.18}{20} = \frac{2 \times 22.18}{21.5 \times 20} = \frac{22.18}{215} \approx 0.103 \, N$
The normality of the acid is nearly $0.10 \, N$.

(A) Iodometric titration involves the liberation of iodine $(I_2)$ by the reaction of an oxidizing agent with excess potassium iodide $(KI)$ in an acidic medium.
The liberated iodine is then titrated against a standard sodium thiosulfate $(Na_2S_2O_3)$ solution.
Option $A$ represents the reaction of dichromate $(Cr_2O_7^{2-})$ with iodide $(I^{-})$ in an acidic medium to produce $I_2$,followed by the titration of $I_2$ with thiosulfate $(S_2O_3^{2-})$,which is a standard iodometric procedure.

(C) For complete neutralization,the number of milli-equivalents of base must equal the number of milli-equivalents of acid.
Using the formula: $N_1V_1 = N_2V_2$
Where:
$N_1 = \frac{1}{10} \ N$ (Normality of $NaOH$)
$V_1 = ?$ (Volume of $NaOH$ in $mL$)
$N_2 = \frac{1}{25} \ N$ (Normality of $HCl$)
$V_2 = 100 \ mL$ (Volume of $HCl$)
Substituting the values:
$\frac{1}{10} \times V_1 = \frac{1}{25} \times 100$
$V_1 = \frac{100}{25} \times 10$
$V_1 = 4 \times 10 = 40 \ mL$
Therefore,the required volume of $NaOH$ is $40 \ mL$.

(A) The estimation of $Na_2S_2O_3$ (sodium thiosulfate) is typically performed via iodometric titration.
In this process,a standard solution of $I_2$ (iodine) is used as the titrant.
$I_2$ reacts with $Na_2S_2O_3$ to form $Na_2S_4O_6$ (sodium tetrathionate) and $NaI$ (sodium iodide).
Therefore,the correct option is $A$.

(C) Let the normality of the diluted $HCl$ solution be $N_1$.
Using the neutralization formula $N_1V_1 = N_2V_2$ for $HCl$ and $NaOH$:
$N_1 \times 20 \, mL = 0.1 \, N \times 25 \, mL$
$N_1 = \frac{0.1 \times 25}{20} = 0.125 \, N$.
The dilution factor for the concentrated $HCl$ is $\frac{1000 \, mL}{10 \, mL} = 100$.
Therefore,the normality of the concentrated $HCl$ is $0.125 \times 100 = 12.5 \, N$.

(A) $1$. Calculate the milliequivalents of $H_2SO_4$ taken: $100 \, mL \times 0.1 \, M \times 2 \, (\text{basicity}) = 20 \, meq$.
$2$. Calculate the milliequivalents of $NaOH$ used for back titration: $20 \, mL \times 0.5 \, M \times 1 \, (\text{acidity}) = 10 \, meq$.
$3$. Milliequivalents of $NH_3$ evolved = $20 - 10 = 10 \, meq = 0.01 \, eq$.
$4$. Percentage of nitrogen = $\frac{1.4 \times \text{meq of } NH_3}{\text{mass of compound}} = \frac{1.4 \times 10}{0.30} = 46.66 \%$.
$5$. Nitrogen percentage in Urea $(NH_2)_2CO$: $\frac{28}{60} \times 100 = 46.66 \%$.
$6$. Since the percentage matches,the compound is Urea.

(B) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
The equivalent mass of oxalic acid is $\frac{\text{Molar mass}}{\text{n-factor}} = \frac{126}{2} = 63 \ g/eq$.
Given mass of oxalic acid = $63 \ g$.
Number of equivalents = $\frac{\text{Given mass}}{\text{Equivalent mass}} = \frac{63}{63} = 1 \ eq$.
The normality $(N)$ of the solution is given as $\frac{N}{10} = 0.1 \ N$.
Using the formula $N = \frac{\text{Number of equivalents}}{\text{Volume in litres (V)}}$,we get $0.1 = \frac{1}{V}$.
Therefore,$V = \frac{1}{0.1} = 10 \ L$.

(B) The formula for normality is $N = \frac{w \times 1000}{Eq. wt. \times V(mL)}$.
For ferrous ammonium sulphate,the equivalent weight is equal to its molecular weight $(392)$ because the change in oxidation state of $Fe^{2+}$ to $Fe^{3+}$ involves $1$ electron.
Given: $N = 0.1 \, N$,$V = 100 \, mL$,$Eq. wt. = 392$.
Substituting the values: $0.1 = \frac{w \times 1000}{392 \times 100}$.
$w = \frac{0.1 \times 392 \times 100}{1000} = 3.92 \, g$.

(D) Mohr's salt is $FeSO_4(NH_4)_2SO_4 \cdot 6H_2O$.
In this compound,the iron is present as $Fe^{2+}$,meaning its oxidation state is $+2$.
Therefore,the statement that the oxidation state of iron is $+3$ is incorrect.

(A) The chemical formula $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ represents a double salt known as Mohr's salt.
It is a common laboratory reagent used in titrations as a primary standard for iron$(II)$ ions.

(B) In volumetric analysis,cerium$(IV)$ compounds are frequently used as oxidizing agents. Specifically,$Ce(IV)$ salts,such as ceric ammonium nitrate or ceric sulfate,are employed. Among the given options,$CeO_2$ (cerium dioxide) represents the oxidation state of cerium used in such analytical processes.

(C) $1$. Total millimoles of $H_2SO_4$ taken = $100 \ mL \times 0.1 \ M = 10 \ mmol$.
$2$. Millimoles of $H_2SO_4$ neutralized by $NaOH$ = $\frac{1}{2} \times (20 \ mL \times 0.5 \ M) = 5 \ mmol$ (since $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$).
$3$. Millimoles of $H_2SO_4$ consumed by $NH_3$ = $10 - 5 = 5 \ mmol$.
$4$. Since $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$,millimoles of $NH_3$ = $2 \times 5 = 10 \ mmol$.
$5$. Mass of nitrogen = $\frac{10 \times 10^{-3} \times 14}{0.30} \times 100 = 46.67\%$.
$6$. For Urea $(NH_2CONH_2)$,molar mass = $60 \ g/mol$. Nitrogen content = $\frac{28}{60} \times 100 = 46.67\%$.
$7$. Thus,the compound is Urea.

(A) The reaction of $Cu^{2+}$ with iodide ions is given by:
$2Cu^{2+} + 4I^- \to 2CuI + I_2$
The liberated $I_2$ is then titrated against sodium thiosulfate $(Na_2S_2O_3)$:
$I_2 + 2Na_2S_2O_3 \to 2NaI + Na_2S_4O_6$
From the stoichiometry,$2$ moles of $Cu^{2+}$ produce $1$ mole of $I_2$,which reacts with $2$ moles of $Na_2S_2O_3$.
Thus,$1$ mole of $Cu^{2+}$ reacts with $1$ mole of $Na_2S_2O_3$.
Since the $n$-factor for $Cu^{2+}$ in this reaction is $1$ (reduction from $+2$ to $+1$) and the $n$-factor for $Na_2S_2O_3$ is $1$ (oxidation from $+2$ to $+2.5$ average),the equivalent ratio is $1:1$.
However,the standard volumetric estimation of $Cu^{2+}$ involves the reaction $2Cu^{2+} + 4I^- \to 2CuI + I_2$ followed by $I_2 + 2S_2O_3^{2-} \to 2I^- + S_4O_6^{2-}$.
For $1$ equivalent of $Cu^{2+}$,$1$ equivalent of $Na_2S_2O_3$ is required.

(D) In complexometric titrations,the disodium salt of ethylene diamine tetraacetic acid $(EDTA)$ is widely used as a chelating agent.
It forms stable complexes with divalent metal ions such as $Ca^{2+}$ and $Mg^{2+}$.
Therefore,it is used to estimate the total hardness of water,which is due to the presence of both $Ca^{2+}$ and $Mg^{2+}$ ions.

(A) primary standard must be stable,non-hygroscopic,and available in a high state of purity.
$Na_2Cr_2O_7$ is highly hygroscopic,meaning it absorbs moisture from the atmosphere,which makes it difficult to weigh accurately.
In contrast,$K_2Cr_2O_7$ is non-hygroscopic and can be obtained in a very pure form,making it an ideal primary standard for volumetric analysis.

(A) $K_2Cr_2O_7$ is preferred over $Na_2Cr_2O_7$ in volumetric analysis because $Na_2Cr_2O_7$ is hygroscopic in nature,which makes it difficult to weigh accurately for preparing a standard solution.
$K_2Cr_2O_7$ is non-hygroscopic,allowing for the preparation of a precise standard solution,which is a requirement for a primary standard.

(C) $K_2Cr_2O_7$ is used as a primary standard in volumetric analysis because it is available in a high state of purity,is stable,and its standard solution can be prepared accurately. While it is soluble in water,its suitability as a primary standard is due to its purity and stability,not just its solubility. Therefore,the Assertion is correct,but the Reason is incorrect.

(N/A) When a substance is precipitated,some ions that combine to form the precipitate get adsorbed on the surface of the precipitate.
Therefore,it becomes essential to wash the precipitate before estimating it quantitatively to remove these adsorbed ions or other such impurities.

(N/A) When a precipitate is formed,some ions required for precipitation are adsorbed on the surface of the precipitate. Thus,washing the precipitate before quantitative measurement removes these extra adsorbed ions and other impurities.

(A) Cation exchange resins contain acidic functional groups such as $-SO_3H$ or $-COOH$ which can release $H^+$ ions.
Anion exchange resins contain basic functional groups such as $-NH_2$ or $-NH_3^+OH^-$ which can release $OH^-$ ions.
Therefore,the correct pair for cation and anion exchange resins is $-SO_3H$ and $-NH_2$ respectively.

(D) Step $1$: Calculate the milliequivalents $(m_{eq})$ of $NaOH$ used for back titration: $m_{eq} = 30.0 \ mL \times 0.25 \ N = 7.5 \ m_{eq}$.
Step $2$: Calculate the total $m_{eq}$ of $H_2SO_4$ taken: $m_{eq} = 50.0 \ mL \times 0.5 \ N = 25.0 \ m_{eq}$.
Step $3$: Calculate the $m_{eq}$ of $H_2SO_4$ reacted with $NH_3$: $m_{eq} = 25.0 - 7.5 = 17.5 \ m_{eq}$.
Step $4$: Since $m_{eq}$ of $NH_3 = m_{eq}$ of $H_2SO_4$ reacted,$m_{eq}$ of $NH_3 = 17.5 \ m_{eq}$.
Step $5$: Calculate the mass of nitrogen: $\text{Mass of } N = 17.5 \times 10^{-3} \times 14 \ g = 0.245 \ g$.
Step $6$: Calculate the percentage of nitrogen: $\% N = (0.245 / 0.166) \times 100 \approx 147.59 \%$. Rounding to the nearest integer,we get $148 \%$.

(A) $MnO_2 + 4HCl \longrightarrow MnCl_2 + Cl_2 + 2H_2O$
$Cl_2 + 2KI \longrightarrow 2KCl + I_2$
$I_2 + 2Na_2S_2O_3 \longrightarrow 2NaI + Na_2S_4O_6$
Equivalents of $MnO_2 = \text{Equivalents of } Cl_2 = \text{Equivalents of } I_2 = \text{Equivalents of } Na_2S_2O_3$
Equivalents of $Na_2S_2O_3 = \text{Molarity} \times \text{Volume (L)} \times n\text{-factor} = 0.1 \times 0.060 \times 1 = 6 \times 10^{-3} \, eq$
Since the $n$-factor of $MnO_2$ in the reaction $MnO_2 \longrightarrow Mn^{2+}$ is $2$,moles of $MnO_2 = \frac{6 \times 10^{-3}}{2} = 3 \times 10^{-3} \, mol$
Molar mass of $MnO_2 = 55 + 2 \times 16 = 87 \, g/mol$
Mass of $MnO_2 = 3 \times 10^{-3} \times 87 = 0.261 \, g$
$\% \, MnO_2 = \frac{0.261}{2.0} \times 100 = 13.05 \, \%$
Nearest integer is $13$.

(A) $KMnO_4$ is a strong oxidizing agent. When $HCl$ is used in permanganate titrations,$KMnO_4$ oxidizes $HCl$ to $Cl_2$ gas according to the following reaction:
$2KMnO_4 + 16HCl \rightarrow 2MnCl_2 + 2KCl + 8H_2O + 5Cl_2$
Because $HCl$ is consumed in this side reaction,it interferes with the titration,leading to inaccurate results. Therefore,$HCl$ is not used. Both the assertion and the reason are correct,and the reason explains the assertion.

(C) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$. The equivalent weight is $126/2 = 63 \ g/eq$.
Initial normality $(N_1)$ of the $100 \ mL$ solution: $N_1 = \frac{2.52 \ g}{63 \ g/eq \times 0.1 \ L} = 0.4 \ N$.
Using the dilution formula $N_1V_1 = N_2V_2$: $0.4 \ N \times 10 \ mL = N_2 \times 500 \ mL$.
$N_2 = \frac{0.4 \times 10}{500} = 0.008 \ N$.
Amount of oxalic acid in $mg/mL$ in the final solution:
Mass of oxalic acid in $10 \ mL$ aliquot $= \frac{2.52 \ g}{100 \ mL} \times 10 \ mL = 0.252 \ g = 252 \ mg$.
Concentration $= \frac{252 \ mg}{500 \ mL} = 0.504 \ mg/mL$.

(A) The correct option is $A$.
The chemical reaction is: $2NH_3 + H_2SO_4 \longrightarrow (NH_4)_2SO_4$.
Equivalents of $H_2SO_4$ used = $Molarity \times Volume \times n-factor = 2 \times 10 \times 10^{-3} \times 2 = 0.04 \ eq$.
Since $1 \ mole$ of $H_2SO_4$ reacts with $2 \ moles$ of $NH_3$,the moles of $NH_3$ evolved = $2 \times (Molarity \times Volume) = 2 \times (2 \times 10 \times 10^{-3}) = 0.04 \ moles$.
Mass of nitrogen $(N)$ in $NH_3$ = $0.04 \times 14 \ g = 0.56 \ g$.
Weight percentage of nitrogen = $\frac{\text{Mass of } N}{\text{Mass of compound}} \times 100 = \frac{0.56}{2} \times 100 = 28 \ \%$.

(C) $KOH$ is a strong base that reacts with atmospheric $CO_2$ to form potassium carbonate $(K_2CO_3)$.
$2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$
Due to this reaction,the concentration of the $KOH$ solution decreases over time when exposed to air.
Therefore,it is necessary to standardize or check the concentration of $KOH$ solution before using it in volumetric analysis to ensure accuracy.
Thus,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A).$

(C) $K_2Cr_2O_7$ is preferred as a primary standard because it is non-deliquescent and can be obtained in a highly pure state.
$Na_2Cr_2O_7$ is deliquescent (absorbs moisture from the air),which makes it difficult to weigh accurately for preparing standard solutions.
$Na_2Cr_2O_7$ is much more soluble in water than $K_2Cr_2O_7$.
Therefore,Statement $I$ is true and Statement $II$ is false.

(C) The balanced redox reaction is given by:
$10[FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O] + 2KMnO_4 + 8H_2SO_4$ $\rightarrow 5Fe_2(SO_4)_3 + 2MnSO_4 + 10(NH_4)_2SO_4 + K_2SO_4 + 68H_2O$
In this reaction,for every $2$ molecules of $KMnO_4$ consumed,$68$ molecules of water are produced.

(D) primary standard must be stable,non-hygroscopic,and available in a high state of purity.
Statement $(A)$ is correct: It should be available in a pure,dry form.
Statement $(B)$ is correct: It should not react with atmospheric components like moisture or $CO_2$.
Statement $(C)$ is incorrect: $A$ primary standard must $NOT$ be hygroscopic.
Statement $(D)$ is correct: It should be readily soluble in water.
Statement $(E)$ is incorrect: $KMnO_4$ and $NaOH$ are secondary standards because they are not stable or pure enough to be primary standards.
Therefore,the incorrect statements are $(C)$ and $(E)$.

(C) During the preparation of Mohr's salt,dilute sulphuric acid is added to prevent the hydrolysis of $Fe^{2+}$ ion.
This is because $Fe^{2+}$ ions are prone to hydrolysis in water,forming basic salts,and the addition of dilute sulphuric acid provides an acidic medium that suppresses this hydrolysis.

(C) The estimation of carbon monoxide $(CO)$ is carried out by reacting it with iodine pentoxide $(I_2O_5)$.
The chemical reaction is as follows:
$I_2O_5(s) + 5CO(g) \rightarrow I_2(s) + 5CO_2(g)$
The amount of iodine $(I_2)$ produced is then determined by titration,which allows for the calculation of the amount of $CO$ present.

(C) The neutralization reaction follows the principle of equivalence: $N_{1}V_{1} = N_{2}V_{2}$.
Given:
Volume of monobasic acid $(V_{1})$ = $10 \ cm^{3}$
Normality of monobasic acid $(N_{1})$ = $0.1 \ N$
Volume of $NaOH$ solution $(V_{2})$ = $15 \ cm^{3}$
Normality of $NaOH$ solution $(N_{2})$ = ?
Substituting the values in the formula:
$10 \ cm^{3} \times 0.1 \ N = 15 \ cm^{3} \times N_{2}$
$1 = 15 \times N_{2}$
$N_{2} = \frac{1}{15} \ N = 0.066 \ N$.

(C) The reaction between $Cu^{2+}$ and $I^-$ is: $2Cu^{2+} + 4I^- \rightarrow 2CuI(s) + I_2$.
Moles of $Cu^{2+} = 0.1 \ L \times 0.05 \ M = 0.005 \ mol$.
Moles of $I_2$ produced = $\frac{1}{2} \times \text{moles of } Cu^{2+} = 0.0025 \ mol$.
The titration reaction is: $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.
Moles of $Na_2S_2O_3$ required = $2 \times \text{moles of } I_2 = 2 \times 0.0025 = 0.005 \ mol$.
Volume of $Na_2S_2O_3 = \frac{\text{moles}}{\text{molarity}} = \frac{0.005 \ mol}{0.01 \ M} = 0.5 \ L = 500 \ mL$.

(D) $K_2Cr_2O_7$ acts as an oxidizing agent in an acidic medium. The half-reaction is:
$Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_2O$
Since the reaction involves $6$ electrons,the n-factor is $6$.
$\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{294}{6} = 49 \ g/eq$.
Normality $(N)$ is given by:
$N = \frac{\text{Weight}}{\text{Equivalent weight} \times \text{Volume in Liters}}$
$0.04 = \frac{\text{Weight}}{49 \times 0.250}$
$\text{Weight} = 0.04 \times 49 \times 0.250$
$\text{Weight} = 0.01 \times 49 = 0.49 \ g$.

(C) The formula for normality is $N = \frac{w \times 1000}{E \times V(mL)}$.
Here,$N = 1/20$,$V = 1000 \ mL$,and the equivalent weight $E$ of oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is $63 \ g/eq$.
Substituting the values: $w = \frac{N \times E \times V}{1000} = \frac{1}{20} \times 63 \times \frac{1000}{1000} = \frac{63}{20} \ g$.

(C) In basic medium,the reaction between $KMnO_4$ and $KI$ is: $2MnO_4^{-} + I^{-} + H_2O \rightarrow 2MnO_2 + I_2 + 2OH^{-}$.
Here,$n$-factor for $KMnO_4$ is $3$ (change in oxidation state of $Mn$ from $+7$ to $+4$).
Equivalents of $KMnO_4 = 0.2 \times 0.5 \times 3 = 0.3 \ \text{eq}$.
Since $I_2$ is produced from $KMnO_4$,the equivalents of $I_2$ produced are equal to the equivalents of $KMnO_4$ reacted,which is $0.3 \ \text{eq}$.
The titration reaction is $I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}$.
Equivalents of $I_2 = \text{Equivalents of } Na_2S_2O_3$.
$0.3 = 0.1 \times V \times 1$.
$V = 3 \ L$.

(B) primary standard is a reagent that is pure,stable,and has a high molar mass.
Statement $A$ is incorrect because hydrated salts often lose or gain water of crystallization,making them unstable.
Statement $B$ is correct; a primary standard must be stable and not react with air (e.g.,not hygroscopic or efflorescent).
Statement $C$ is correct; the reaction must be rapid and follow a definite stoichiometry.
Statement $D$ is incorrect; a primary standard must be readily soluble in water to prepare a standard solution.
Statement $E$ is incorrect; a primary standard should have a high relative molar mass to minimize weighing errors.
Therefore,only statement $B$ and $C$ are correct,but based on standard criteria,$B$ and $C$ are the most accurate. Given the options,$B$ is the best fit.

(D) $EDTA$,$Cupron$,and $DMG$ (Dimethylglyoxime) are standard analytical reagents used for complexometric or gravimetric analysis.
$D$-penicillamine is primarily used as a medication for Wilson's disease and is not a standard reagent for general quantitative chemical analysis.