Quantitative Analysis Questions in English

(C) In the Carius method for the estimation of sulphur,the organic compound is heated with fuming $HNO_3$ in a sealed tube.
The sulphur present in the organic compound is oxidized to sulphuric acid $(H_2SO_4)$.
The reaction is: $S + 2HNO_3 \rightarrow H_2SO_4 + 2NO_2$.

(B) The percentage of chlorine is calculated using the formula:
$\% \text{ of chlorine} = \frac{35.5}{143.5} \times \frac{\text{Mass of } AgCl}{\text{Mass of organic compound}} \times 100$
Given:
Mass of organic compound = $0.099 \, g$
Mass of $AgCl$ = $0.287 \, g$
Molar mass of $Cl = 35.5 \, g/mol$
Molar mass of $AgCl = 143.5 \, g/mol$
Substituting the values:
$\% \text{ of chlorine} = \frac{35.5}{143.5} \times \frac{0.287}{0.099} \times 100$
$\% \text{ of chlorine} = 0.24738 \times 2.8989 \times 100 \approx 71.71 \%$
Thus,the correct option is $(B)$.

(D) The correct formula for the estimation of nitrogen by Kjeldahl's method is $\% N = \frac{1.4 \times V \times N}{W}$.
Here,$V$ represents the volume of standard acid consumed in $mL$.
$N$ represents the normality of the standard acid.
$W$ represents the weight of the organic substance taken in $g$.

(A) The percentage of sulphur is calculated using the formula: $\% \text{ of } S = \frac{32}{233} \times \frac{\text{mass of } BaSO_4}{\text{mass of organic compound}} \times 100$.
Substituting the given values: $\% \text{ of } S = \frac{32}{233} \times \frac{0.35}{0.2595} \times 100$.
Calculating the result: $\% \text{ of } S = 0.13734 \times 1.3487 \times 100 \approx 18.52 \%$.

(D) . Kjeldahl's and Dumas's methods are used for the quantitative estimation of nitrogen in an organic compound.
In the Kjeldahl method,the nitrogen element of the organic compound is converted into ammonia $(NH_3)$.

(B) Dumas' method involves the determination of nitrogen content in the organic compound in the form of nitrogen gas $(N_2)$.
In this method,the organic compound is heated with copper$(II)$ oxide $(CuO)$ in an atmosphere of carbon dioxide $(CO_2)$.
This process oxidizes the nitrogen present in the compound to free nitrogen gas $(N_2)$,along with the formation of carbon dioxide $(CO_2)$ and water $(H_2O)$.

(D) The percentage of sulphur is calculated using the formula: $\text{Percentage of sulphur} = \frac{32}{233} \times \frac{\text{mass of } BaSO_4}{\text{mass of organic compound}} \times 100$
Substituting the given values: $\text{Percentage of sulphur} = \frac{32}{233} \times \frac{1.158}{0.53} \times 100$
$\text{Percentage of sulphur} = 0.137339 \times 2.1849 \times 100 \approx 30\%$
Thus,the percentage of sulphur is $30\%$.

(D) The correct answer is $D$. In Kjeldahl’s method,the organic compound is heated with concentrated $H_2SO_4$ to convert nitrogen into $(NH_4)_2SO_4$.
Subsequently,this is treated with excess $NaOH$ to liberate ammonia gas $(NH_3)$.
The reaction for the liberation of ammonia is:
$CH_3CONH_2 + NaOH \xrightarrow{\Delta} CH_3COONa + H_2O + NH_3$
Thus,the nitrogen is quantitatively estimated in the form of ammonia.

(C) The Kjeldahl method is used for the estimation of nitrogen in organic compounds. However,it is not applicable to compounds containing nitrogen in nitro groups $(-NO_2)$,azo groups $(-N=N-)$,or nitrogen present in the ring (e.g.,pyridine,quinoline,or nitrogen in nucleic acids like $DNA$ and $RNA$).
In $DNA$ and $RNA$,nitrogen is present in the heterocyclic rings of the nitrogenous bases (purines and pyrimidines). Therefore,the Kjeldahl method cannot be used for their analysis.

(D) In $Kjeldahl's$ method,the nitrogen present in an organic compound is converted into $NH_3$ (ammonia),which is then estimated quantitatively.

(C) $1$. Calculate the total millimoles of $H_2SO_4$ taken: $100 \ mL \times 0.1 \ M = 10 \ mmol$.
$2$. Calculate the millimoles of $NaOH$ used for neutralization: $20 \ mL \times 0.5 \ M = 10 \ mmol$.
$3$. Since $2 \ mol$ of $NaOH$ reacts with $1 \ mol$ of $H_2SO_4$,the millimoles of $H_2SO_4$ neutralized by $NaOH$ is $10 / 2 = 5 \ mmol$.
$4$. Millimoles of $H_2SO_4$ reacted with $NH_3$ = Total $H_2SO_4$ - $H_2SO_4$ neutralized by $NaOH$ = $10 - 5 = 5 \ mmol$.
$5$. Since $2 \ mol$ of $NH_3$ reacts with $1 \ mol$ of $H_2SO_4$,millimoles of $NH_3$ = $5 \times 2 = 10 \ mmol$.
$6$. Mass of nitrogen = $(14 \times 10 \times 10^{-3}) \ g = 0.14 \ g$.
$7$. Percentage of nitrogen = $(0.14 / 0.3) \times 100 = 46.67 \%$.
$8$. Calculating nitrogen percentage for options: Urea $(NH_2CONH_2)$ has $28 \ g$ nitrogen in $60 \ g$ molar mass,so $\%N = (28/60) \times 100 = 46.67 \%$. Thus,the compound is Urea.

(A) The Dumas method is used for the estimation of nitrogen in organic compounds. In this method,the organic compound is heated with copper oxide in an atmosphere of carbon dioxide,which yields free nitrogen. The nitrogen gas is collected over a potassium hydroxide solution in an instrument called a nitrometer.

(A) In the Dumas method,the organic compound is heated with copper oxide $(CuO)$ to produce nitrogen gas.
The volume of nitrogen gas collected is measured at standard temperature and pressure $(STP)$.
The percentage of nitrogen is calculated using the formula:
$\text{Percentage of } N = \frac{28}{22400} \times \frac{V}{m} \times 100$,where $V$ is the volume of $N_2$ gas in $mL$ at $STP$ and $m$ is the mass of the organic compound in grams.
Since the molar mass of $N_2$ is $28 \, g/mol$ and $1 \, mole$ of gas occupies $22400 \, mL$ at $STP$,the correct formula is $\frac{28 \times V \times 100}{22400 \times m}$.

(B) In the Kjeldahl method,the percentage of nitrogen is calculated using the formula:
$\% \,N = \frac{1.4 \times N \times V}{W}$
Where:
$N$ = Normality of the standard acid used.
$V$ = Volume of the standard acid used in $mL$.
$W$ = Mass of the organic substance taken in $g$.

(B) In the Carius method,a known mass of an organic compound is heated with fuming nitric acid $(HNO_3)$ in the presence of silver nitrate $(AgNO_3)$ in a hard glass tube known as a Carius tube.
Carbon and hydrogen are oxidized to carbon dioxide and water,respectively.
The halogen present in the organic compound reacts with silver nitrate to form a precipitate of silver halide ($AgX$,where $X = Cl, Br, I$).
Therefore,the halogen is converted into silver halide.

(A) In the $Dumas$ method,the nitrogen-containing organic compound is heated with copper$(II)$ oxide $(CuO)$ in an atmosphere of carbon dioxide $(CO_2)$.
This process ensures that nitrogen is released as $N_2$ gas,while carbon and hydrogen are oxidized to $CO_2$ and $H_2O$ respectively.
$CuO$ acts as an oxidizing agent to ensure complete combustion of the organic compound.

(C) The estimation of nitrogen in organic compounds is primarily carried out by two methods:
$1$. Dumas method: This is a general method applicable to all nitrogenous compounds.
$2$. Kjeldahl's method: This method is applicable to compounds containing nitrogen in the form of amino or nitro groups,but it is not applicable to compounds where nitrogen is linked to oxygen (like nitro or azo groups) or present in a ring (like pyridine).
Since both are standard methods for nitrogen estimation,the correct answer is $C$.

(A) In the Liebig's method for the quantitative estimation of carbon and hydrogen,the organic compound is burnt in the presence of excess oxygen and copper$(II)$ oxide $(CuO)$.
The carbon present in the compound is oxidized to $CO_2$,and the hydrogen is oxidized to $H_2O$.
The $CO_2$ gas produced is absorbed by a concentrated solution of potassium hydroxide $(KOH)$,while the $H_2O$ vapor is absorbed by anhydrous calcium chloride $(CaCl_2)$.

(D) The Carius method is a standard analytical technique used for the quantitative estimation of halogens (chlorine,bromine,and iodine) in organic compounds.
In this method,a known mass of the organic compound is heated with fuming nitric acid in the presence of silver nitrate in a hard glass tube known as a Carius tube.
The halogen present in the compound reacts with silver nitrate to form a precipitate of silver halide $(AgX)$,which is then filtered,washed,dried,and weighed to calculate the percentage of the halogen.

(B) In the quantitative analysis of carbon and hydrogen (Liebig's method),the organic compound is heated with excess of dry oxygen.
Carbon is oxidized to $CO_2$ and hydrogen is oxidized to $H_2O$.
The $H_2O$ vapors produced are passed through a $U$-tube containing anhydrous $CaCl_2$,which absorbs the water.
The $CO_2$ gas is then absorbed in a separate $U$-tube containing $KOH$ solution.

(A) In the estimation of phosphorus,the organic compound is heated with fuming $HNO_3$. Phosphorus is oxidized to phosphoric acid $(H_3PO_4)$.
It is precipitated as ammonium phosphomolybdate by adding ammonia and ammonium molybdate.
Finally,it is precipitated as magnesium ammonium phosphate $(MgNH_4PO_4)$ by adding magnesia mixture $(MgCl_2 + NH_4Cl + NH_4OH)$,which on ignition gives magnesium pyrophosphate $(Mg_2P_2O_7)$.
Thus,phosphorus is weighed as $Mg_2P_2O_7$. It is not weighed as phosphoric acid,ammonium molybdate,or ammonium phosphomolybdate.

(B) In the Carius method for the estimation of sulfur,a known mass of an organic compound is heated with fuming nitric acid $(HNO_3)$ in a hard glass tube known as a Carius tube.
Sulfur present in the compound is oxidized to sulfuric acid $(H_2SO_4)$.
It is then precipitated as barium sulfate $(BaSO_4)$ by adding an excess of barium chloride $(BaCl_2)$ solution.
The precipitate of $BaSO_4$ is filtered,washed,dried,and weighed.
From the mass of $BaSO_4$ obtained,the percentage of sulfur is calculated.

(A) In the $Kjeldahl$ method,the organic compound containing nitrogen is heated with concentrated $H_2SO_4$ in a $Kjeldahl$ flask.
During this process,the nitrogen present in the organic compound is quantitatively converted into ammonium sulphate,$(NH_4)_2SO_4$.

(D) $1$. Calculate the total millimoles of $HCl$ taken: $20 \, mL \times 0.1 \, M = 2.0 \, mmol$.
$2$. Calculate the millimoles of $NaOH$ used for neutralization of excess $HCl$: $15 \, mL \times 0.1 \, M = 1.5 \, mmol$.
$3$. Millimoles of $HCl$ reacted with $NH_3 = 2.0 - 1.5 = 0.5 \, mmol$.
$4$. Since $1 \, mol$ of $NH_3$ reacts with $1 \, mol$ of $HCl$,millimoles of $N = 0.5 \, mmol$.
$5$. Mass of nitrogen $= 0.5 \times 10^{-3} \, mol \times 14 \, g/mol = 7 \times 10^{-3} \, g = 7 \, mg$.
$6$. Percentage of nitrogen $= (\text{Mass of } N / \text{Mass of compound}) \times 100 = (7 \, mg / 29.5 \, mg) \times 100 \approx 23.7\%$.

(C) The formula for the percentage of nitrogen in the Kjeldahl method is: $\% \ N = \frac{1.4 \times \text{Normality of acid} \times \text{Volume of acid in mL}}{\text{Mass of organic compound in g}}$.
Given:
Normality of $H_2SO_4 = \frac{1}{5} \ N = 0.2 \ N$.
Volume of $H_2SO_4 = 58 \ mL$.
Mass of organic compound = $1 \ g$.
Substituting the values:
$\% \ N = \frac{1.4 \times 0.2 \times 58}{1} = 16.24 \%$.
Rounding to the nearest option,the percentage of nitrogen is $16.2 \%$.

(A) Mass of organic compound $= 0.25 \, g$
Pressure of dry nitrogen gas $(P_1) = 725 \, mm - 25 \, mm = 700 \, mm$
Volume of nitrogen $(V_1) = 40 \, mL$
Temperature $(T_1) = 300 \, K$
Using the ideal gas equation at $STP$ $(P_2 = 760 \, mm, T_2 = 273 \, K)$:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{700 \times 40 \times 273}{300 \times 760} = 33.52 \, mL$
Mass of $N_2$ at $STP = \frac{28 \times 33.52}{22400} = 0.0419 \, g$
Percentage of nitrogen $= \frac{0.0419}{0.25} \times 100 = 16.76 \%$

(A) The reaction between ammonia and sulfuric acid is: $2NH_3 + H_2SO_4 \longrightarrow (NH_4)_2SO_4$.
Number of millimoles of $H_2SO_4$ used = $M \times V(mL) = 1 \ M \times 10 \ mL = 10 \ mmol$.
Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,the number of millimoles of $NH_3$ produced = $2 \times 10 \ mmol = 20 \ mmol = 20 \times 10^{-3} \ mol$.
Mass of nitrogen = $20 \times 10^{-3} \ mol \times 14 \ g/mol = 0.28 \ g$.
Percentage of nitrogen = $\frac{\text{Mass of Nitrogen}}{\text{Mass of Sample}} \times 100 = \frac{0.28}{0.75} \times 100 = 37.33 \ \%$.

(B) The pressure of dry nitrogen gas is $p_1 = p - p_{aq} = 715 \ mm - 15 \ mm = 700 \ mm$.
Using the ideal gas equation $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$ to find the volume of $N_2$ at $STP$ $(p_2 = 760 \ mm, T_2 = 273 \ K)$:
$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2} = \frac{700 \times 55 \times 273}{300 \times 760} = 48.098 \ mL$.
The percentage of nitrogen is calculated as:
$\% \ N = \frac{28}{22400} \times \frac{V_2 \ (mL)}{W \ (g)} \times 100$
$\% \ N = \frac{28}{22400} \times \frac{48.098}{0.35} \times 100 = 16.45 \ \%$.

(B) The formula for the percentage of nitrogen is: $\% \ N = \frac{1.4 \times \text{meq. of acid consumed}}{\text{mass of organic compound}}$
Step $1$: Calculate milliequivalents $(meq)$ of $H_2SO_4$ taken:
$meq \text{ of } H_2SO_4 = 60 \ mL \times \frac{1}{10} \ M \times 2 \text{ (basicity)} = 12 \ meq$
Step $2$: Calculate $meq$ of $NaOH$ used for back titration:
$meq \text{ of } NaOH = 20 \ mL \times \frac{1}{10} \ M \times 1 \text{ (acidity)} = 2 \ meq$
Step $3$: Calculate $meq$ of acid consumed by ammonia:
$meq \text{ of acid consumed} = 12 - 2 = 10 \ meq$
Step $4$: Calculate percentage of nitrogen:
$\% \ N = \frac{1.4 \times 10}{1.4} = 10 \%$

(C) Mass of organic compound $= 250 \ mg = 0.250 \ g$
Mass of $AgBr$ formed $= 141 \ mg = 0.141 \ g$
Molar mass of $AgBr = 108 + 80 = 188 \ g/mol$
$188 \ g$ of $AgBr$ contains $80 \ g$ of $Br$.
Mass of $Br$ in $0.141 \ g$ of $AgBr = \frac{80}{188} \times 0.141 \ g = 0.060 \ g$
Percentage of $Br = \frac{\text{Mass of } Br}{\text{Mass of organic compound}} \times 100$
Percentage of $Br = \frac{0.060}{0.250} \times 100 = 24 \%$

(C) Moles of $HCl$ taken $= 20 \times 0.1 \times 10^{-3} = 2 \times 10^{-3} \ mol$.
Moles of $HCl$ neutralized by $NaOH$ solution $= 15 \times 0.1 \times 10^{-3} = 1.5 \times 10^{-3} \ mol$.
Moles of $HCl$ neutralized by ammonia $= 2 \times 10^{-3} - 1.5 \times 10^{-3} = 0.5 \times 10^{-3} \ mol$.
Since $1 \ mol$ of $NH_3$ reacts with $1 \ mol$ of $HCl$,moles of $N = 0.5 \times 10^{-3} \ mol$.
Mass of nitrogen $= 0.5 \times 10^{-3} \ mol \times 14 \ g/mol = 7 \times 10^{-3} \ g = 7 \ mg$.
Percentage of nitrogen $= (\text{Mass of nitrogen} / \text{Mass of compound}) \times 100 = (7 \ mg / 29.5 \ mg) \times 100 \approx 23.7 \%$.

(A) The molar mass $(MM)$ of $BaSO_4 = 137 + 32 + (4 \times 16) = 233 \ g \ mol^{-1}$.
The formula for the percentage of sulphur is: $\% S = \frac{32}{233} \times \frac{\text{mass of } BaSO_4}{\text{mass of compound}} \times 100$.
Substituting the given values: $\% S = \frac{32}{233} \times \frac{0.12}{2.18} \times 100$.
$\% S = 0.1373 \times 0.05504 \times 100 \approx 7.56 \%$.

(A) The percentage of bromine is calculated using the formula: $\% \ Br = \frac{\text{Atomic mass of } Br}{\text{Molar mass of } AgBr} \times \frac{\text{Mass of } AgBr}{\text{Mass of organic compound}} \times 100$
Given:
Mass of organic compound = $0.15 \ g$
Mass of $AgBr$ = $0.12 \ g$
Atomic mass of $Br = 80 \ g/mol$
Molar mass of $AgBr = 108 + 80 = 188 \ g/mol$
Substituting the values:
$\% \ Br = \frac{80 \times 0.12 \times 100}{188 \times 0.15}$
$\% \ Br = \frac{960}{28.2} \approx 34.04 \%$

(B) The total milliequivalents $(meq)$ of $H_2SO_4$ is equal to the sum of $meq$ of $NH_3$ and $meq$ of $NaOH$ used for back titration.
$meq \text{ of } H_2SO_4 = meq \text{ of } NH_3 + meq \text{ of } NaOH$
$\frac{1}{5} \times 30 = meq \text{ of } NH_3 + \frac{1}{10} \times 20$
$6 = meq \text{ of } NH_3 + 2$
$meq \text{ of } NH_3 = 6 - 2 = 4$
The percentage of nitrogen is calculated using the formula:
$\% N = \frac{1.4 \times meq \text{ of } NH_3}{\text{mass of compound in } g}$
$\% N = \frac{1.4 \times 4}{0.1} = 56\%$

(B) Given: Mass of organic compound $(m)$ = $0.25 \ g$,Volume of $N_2$ $(V_1)$ = $40 \ mL$,Temperature $(T_1)$ = $300 \ K$,Pressure $(P_1)$ = $725 \ mm - 25 \ mm = 700 \ mm$ (corrected pressure).
Using the ideal gas equation at $STP$ $(P_0 = 760 \ mm, T_0 = 273 \ K)$:
$V_0 = \frac{P_1 V_1}{T_1} \times \frac{T_0}{P_0} = \frac{700 \ mm \times 40 \ mL}{300 \ K} \times \frac{273 \ K}{760 \ mm} = 33.53 \ mL$.
At $STP$,$22400 \ mL$ of $N_2$ weighs $28 \ g$.
Mass of $N_2 = \frac{28 \ g \times 33.53 \ mL}{22400 \ mL} = 0.0419 \ g$.
Percentage of nitrogen = $\frac{\text{Mass of } N_2}{\text{Mass of compound}} \times 100 = \frac{0.0419 \ g}{0.25 \ g} \times 100 = 16.76 \%$.
Thus,the correct option is $B$.

(B) Step $1$: Calculate the pressure of dry nitrogen gas. $P_{N_2} = P_{total} - P_{aqueous\ tension} = 715 \ mm - 15 \ mm = 700 \ mm$.
Step $2$: Convert the volume of nitrogen to $STP$ conditions using the ideal gas equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
$P_1 = 700 \ mm, V_1 = 50 \ mL, T_1 = 300 \ K$.
$P_2 = 760 \ mm, T_2 = 273 \ K$.
$V_2 = \frac{700 \times 50 \times 273}{760 \times 300} = 42.006 \ mL$.
Step $3$: Calculate the mass of nitrogen. $22400 \ mL$ of $N_2$ at $STP$ weighs $28 \ g$.
Mass of $N_2 = \frac{28 \times 42.006}{22400} = 0.0525 \ g$.
Step $4$: Calculate the percentage of nitrogen. $\% \ N = \frac{\text{mass of } N_2}{\text{mass of compound}} \times 100 = \frac{0.0525}{0.3} \times 100 = 17.5 \%$.
The closest option is $17.46 \%$.

(A) Kjeldahl's method is used for the estimation of nitrogen.
The organic compound is heated with conc. $H_2SO_4$ in the presence of $K_2SO_4$ and a little $CuSO_4$ to convert all the nitrogen into $(NH_4)_2SO_4$.
$K_2SO_4$ raises the boiling point of $H_2SO_4$ while $CuSO_4$ acts as a catalyst for the reaction.

(A) Milli-equivalents of $H_2SO_4$ taken $= 60 \ mL \times \frac{1}{10} \times 2 = 12 \ mEq$.
Milli-equivalents of $NaOH$ used for back titration $= 20 \ mL \times \frac{1}{10} = 2 \ mEq$.
Milli-equivalents of $NH_3$ evolved $= 12 - 2 = 10 \ mEq$.
Percentage of Nitrogen $= \frac{1.4 \times \text{milli-equivalents of } NH_3}{\text{mass of organic compound in } g} = \frac{1.4 \times 10}{1.4} = 10 \%$.

(C) The Kjeldahl method is not applicable to compounds containing nitrogen in nitro $(-NO_2)$ groups,azo $(-N=N-)$ groups,or nitrogen present in a heterocyclic ring (like pyridine).
In these cases,nitrogen is not quantitatively converted to ammonium sulphate under the reaction conditions.
Alanine $(CH_3CH(NH_2)COOH)$ is an amino acid where nitrogen is present as an amino group $(-NH_2)$,which is readily converted to ammonium sulphate.
Therefore,the Kjeldahl method is suitable for Alanine.

(A) The reaction sequence is: $P$ $\xrightarrow{HNO_3} H_3PO_4$ $\xrightarrow{Mg^{2+}, NH_4^+} MgNH_4PO_4$ $\xrightarrow{\Delta} Mg_2P_2O_7$.
$1$. Molecular weight of $Mg_2P_2O_7 = (2 \times 24.3) + (2 \times 31) + (7 \times 16) = 48.6 + 62 + 112 = 222.6 \ g/mol$ (using $222 \ g/mol$ as per standard approximation).
$2$. The mass of $P$ in $Mg_2P_2O_7$ is $62 \ g$ per $222 \ g$ of $Mg_2P_2O_7$.
$3$. Percentage of $P = \left( \frac{\text{Mass of } P}{\text{Mass of } Mg_2P_2O_7} \right) \times \left( \frac{\text{Mass of } Mg_2P_2O_7}{\text{Mass of organic compound}} \right) \times 100$.
$4$. Percentage of $P = \left( \frac{62}{222} \right) \times \left( \frac{1.0 \ g}{0.5 \ g} \right) \times 100 = \left( \frac{62}{222} \right) \times 2 \times 100 = 55.85 \%$.

(A) The reaction for the neutralisation of $NH_3$ by $H_2SO_4$ is: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$.
Given,$20 \ mmol$ of $H_2SO_4$ is used.
Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,$20 \ mmol$ of $H_2SO_4$ reacts with $40 \ mmol$ of $NH_3$.
$1 \ mmol$ of $NH_3$ contains $1 \ mmol$ of $N$ atoms,so $40 \ mmol$ of $NH_3$ contains $40 \ mmol$ of $N$ atoms.
Mass of $N = 40 \times 10^{-3} \ mol \times 14 \ g/mol = 0.56 \ g$.
Percentage of $N = \frac{\text{Mass of } N}{\text{Mass of compound}} \times 100 = \frac{0.56}{2.8} \times 100 = 20 \%$.

(A) Zeisel's method is used to estimate the methoxy group in an organic compound. In this method,an organic compound containing a methoxy group is treated with $HI$. The resulting alkyl iodide is then treated with $AgNO_3$ to form a precipitate of $AgI$.

(B) First,calculate the pressure of dry nitrogen gas by subtracting the aqueous tension from the total pressure: $P_{N_2} = 715 \ mm - 15 \ mm = 700 \ mm \ Hg$.
Next,convert the volume of nitrogen to $STP$ conditions using the ideal gas equation: $V_{STP} = \frac{V \times P \times 273}{T \times 760} = \frac{55 \times 700 \times 273}{300 \times 760} \approx 46.05 \ mL$.
Finally,calculate the percentage of nitrogen using the formula: $\text{Percentage of } N = \frac{28}{22400} \times \frac{V_{STP}}{\text{mass of compound}} \times 100$.
Substituting the values: $\text{Percentage of } N = \frac{28}{22400} \times \frac{46.05}{0.35} \times 100 \approx 16.45 \%$.
Thus,the percentage of nitrogen in the compound is $16.45 \%$.

(A) The molar mass of $Br = 80 \ g/mol$.
The molar mass of $AgBr = 108 + 80 = 188 \ g/mol$.
Weight of $AgBr$ formed $= 0.12 \ g$.
Weight of organic compound taken $= 0.15 \ g$.
The formula for the percentage of bromine is:
$\text{Percentage of } Br = \frac{\text{Atomic mass of } Br}{\text{Molar mass of } AgBr} \times \frac{\text{Mass of } AgBr \text{ formed}}{\text{Mass of organic compound}} \times 100$.
Substituting the values:
$\text{Percentage of } Br = \frac{80}{188} \times \frac{0.12}{0.15} \times 100$.
$\text{Percentage of } Br = 0.4255 \times 0.8 \times 100 = 34.04 \%$.
Therefore,the correct option is $A$.

(A) Kjeldahl's method is not applicable to compounds containing nitrogen in nitro groups $(-NO_2)$,azo groups $(-N=N-)$,or nitrogen present in the ring (like pyridine) because these compounds do not quantitatively convert to ammonium sulfate $( (NH_4)_2SO_4 )$ upon digestion with concentrated $H_2SO_4$.
In $C_6H_5NO_2$ (nitrobenzene),the nitrogen is in the nitro group,which resists conversion to ammonia under standard Kjeldahl conditions.

The percentage of carbon is calculated as: $\text{Percentage of carbon} = \frac{12 \times \text{mass of } CO_2 \times 100}{44 \times \text{mass of compound}} = \frac{12 \times 0.198 \times 100}{44 \times 0.246} = 21.95 \ \%$.
The percentage of hydrogen is calculated as: $\text{Percentage of hydrogen} = \frac{2 \times \text{mass of } H_2O \times 100}{18 \times \text{mass of compound}} = \frac{2 \times 0.1014 \times 100}{18 \times 0.246} = 4.58 \ \%$.

(N/A) The pressure of dry nitrogen is $P_{N_2} = P_{total} - P_{aqueous} = 715 \ mm - 15 \ mm = 700 \ mm$.
Using the combined gas law to convert the volume to $STP$ ($273 \ K$ and $760 \ mm$):
$V_{STP} = \frac{P_{N_2} \times V \times 273}{P_{STP} \times T} = \frac{700 \times 50 \times 273}{760 \times 300} \approx 41.91 \ mL$.
Since $22,400 \ mL$ of $N_2$ at $STP$ weighs $28 \ g$,the mass of $N_2$ is:
$Mass_{N_2} = \frac{28 \times 41.91}{22400} \approx 0.05239 \ g$.
Percentage of nitrogen $= \frac{Mass_{N_2}}{Mass_{compound}} \times 100 = \frac{0.05239}{0.3} \times 100 \approx 17.46 \%$.

(B) The reaction between $NH_{3}$ and $H_{2}SO_{4}$ is: $2NH_{3} + H_{2}SO_{4} \rightarrow (NH_{4})_{2}SO_{4}$.
$1 \ mol$ of $H_{2}SO_{4}$ neutralizes $2 \ mol$ of $NH_{3}$.
$10 \ mL$ of $1 \ M$ $H_{2}SO_{4}$ contains $10 \ mmol$ of $H_{2}SO_{4}$.
Therefore,$NH_{3}$ produced $= 2 \times 10 \ mmol = 20 \ mmol = 0.02 \ mol$.
Mass of nitrogen $= 0.02 \ mol \times 14 \ g/mol = 0.28 \ g$.
Percentage of nitrogen $= \frac{\text{Mass of Nitrogen}}{\text{Mass of Compound}} \times 100 = \frac{0.28}{0.5} \times 100 = 56 \ \%$.

(A) The molar mass of $AgBr = 108 + 80 = 188 \, g \, mol^{-1}$.
$188 \, g$ of $AgBr$ contains $80 \, g$ of bromine.
Therefore,$0.12 \, g$ of $AgBr$ contains $\frac{80 \times 0.12}{188} \, g$ of bromine.
Percentage of bromine $= \frac{\text{mass of bromine}}{\text{mass of organic compound}} \times 100$
$= \frac{80 \times 0.12 \times 100}{188 \times 0.15} = 34.04 \%$.

(42.10%) The molecular mass of $BaSO_4 = 137 + 32 + 4 \times 16 = 233 \ g/mol$.
The amount of sulphur in $233 \ g$ of $BaSO_4$ is $32 \ g$.
The amount of sulphur in $0.4813 \ g$ of $BaSO_4$ is $\frac{32 \times 0.4813}{233} \ g$.
The percentage of sulphur in the organic compound is calculated as:
$\text{Percentage of sulphur} = \frac{32 \times 0.4813 \times 100}{233 \times 0.157} \%$.
$\text{Percentage of sulphur} = 42.10 \%$.