jee main chemistry important questions | Purification And Characterization Of Organic Compounds

Q 1 :

In Kjeldahl’s method for estimation of nitrogen, ${CuSO}_{4}$ acts as: [2024]

reducing agent
catalytic agent
hydrolysis agent
oxidizing agent

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(2)

In Kjeldahl’s method ${CuSO}_{4}$ acts as catalyst.

Q 2 :

Given below are two statements:

Statement (I): Kjeldahl method is applicable to estimate nitrogen in pyridine.
Statement (II): The nitrogen present in pyridine can easily be converted into ammonium sulphate in Kjeldahl method.

In the light of the above statements, choose the correct answer from the options given below: [2024]

Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Both Statement I and Statement II are false

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(4)

Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Q 3 :

Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]

(56)

$Millimoles of sulphuric acid (n_{H_{2} {SO}_{4}}) = Molarity \times volume in mL$

$= 2 M \times 10 mL = 20 mmol$

$Ammonia is a base, it reacts with sulphuric acid to form salt:$

$2 {NH}_{3} + H_{2} {SO}_{4} \to {({NH}_{4})}_{2} {SO}_{4}$

$By stoichiometry:$

$\frac{n_{N H_{3}}}{2} = \frac{n_{H_{2} S O_{4}}}{1}$

$\frac{n_{N H_{3}}}{2} = \frac{20 mmol}{1}$

$n_{N H_{3}} = 40 mmol = 0.04 mol$

$Moles of N (n_{N}) = Moles of ammonia (n_{N H_{3}}) = 0.04 mol$

$Since amount of nitrogen is conserved, number of moles of N in organic compound (n_{N}) = number of moles of N in ammonia$

$(n_{N H_{3}}) = 0.04 mol$

$Weight of N in organic compound (W_{N}) = n_{N} \times Molar mass of N$

$= 0.04 mol \times 14 {g mol}^{- 1} = 0.56 g$

$Percentage of nitrogen in organic compound:$

$= \frac{W_{N}}{W_{compound}} \times 100 = \frac{0.56}{1} \times 100 = 56 %$

Q 4 :

Given below are two statements I and II. [2025]

Statement I: Dumas method is used for estimation of “nitrogen” in an organic compound.

Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc. $H_{2} {SO}_{4}$ .

In the light of the above statements, choose the correct answer from the options given below.

Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Both Statement I and Statement II are true

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(1)

Nitrogen present is converted to nitrogen gas in Duma’s method.

Q 5 :

In Dumas’ method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is [2025]

1.257
20.87
18.67
12.57

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(4)

Pressure of nitrogen $(P_{N_{2}})$ = Total pressure - aqueous tension

$= (715 - 15) mmHg = 700 mmHg = \frac{700}{760} atm$

Volume of nitrogen $(V_{N_{2}})$ = 60 mL = 0.06 L

$n_{N_{2}} = \frac{P_{N_{2}} V_{N_{2}}}{R T} = \frac{700 \times 0.06}{760 \times 0.0821 \times 300} = 2.24 \times 10^{- 3} mol$

Mass of nitrogen = $n_{N_{2}} \times {molar mass of N}_{2}$

$= 2.24 \times 10^{- 3} \times 28 g = 0.06282 g$

Percentage of nitrogen in organic compound:

$= \frac{Mass of nitrogen}{Mass of organic compound} \times 100$

$= \frac{0.06282}{0.5} \times 100 \approx 12.57 %$

Q 6 :

In Dumas’ method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is

(Given: Aqueous tension at 300 K = 15 mm Hg) [2025]

15.71%
20.95%
17.46%
7.85%

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(1)

Pressure of nitrogen $(P_{N_{2}})$ = Total pressure - aqueous tension

$P_{N_{2}} = (715 - 15) mmHg = 700 mmHg = \frac{700}{760} atm$

Volume of nitrogen $(V_{N_{2}})$ = 60 mL = 0.06 L

Moles of nitrogen $(n_{N_{2}}) = \frac{P_{N_{2}} V_{N_{2}}}{R T} = \frac{\frac{700}{760} \times 0.06}{0.0821 \times 300}$

Mass of nitrogen $(W_{N_{2}}) = n_{N_{2}} \times$ molar mass of nitrogen

$= \frac{\frac{700}{760} \times 0.06}{0.0821 \times 300} \times 28 g$

Mass percentage of nitrogen

$= \frac{W_{N_{2}}}{Mass of organic compound} \times 100$

$= \frac{\frac{700}{760} \times 0.06 \times 28}{0.0821 \times 300 \times 0.4} \times 100 = 15.71 %$

Q 7 :

In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ______ %.

(Given: molar mass in g ${mol}^{- 1}$ of Ag : 108, Cl : 35.5) [2025]

(20)

$Moles of Cl = Moles of AgCl = \frac{Mass of AgCl}{Molar mass of AgCl}$

$= \frac{143.5 mg}{143.5 g} = \frac{143.5 \times 10^{- 3}}{143.5} = 10^{- 3} mol$

$Mass of Cl = Moles of Cl \times Molar mass of Cl$

$= 10^{- 3} mol \times 35.5 {g mol}^{- 1} = 0.0355 g$

$Mass of organic compound = 180 mg = 0.180 g$

$Mass percent of Cl = \frac{Mass of Cl}{Mass of organic compound} \times 100$

$= \frac{0.0355}{0.180} \times 100 = 19.72 % \approx 20 %$

Q 8 :

During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %.

(Given molar mass in g ${mol}^{- 1}$ of Ba : 137, S : 32, O : 16) [2025]

(40)

${Moles of BaSO}_{4} = \frac{{Mass of BaSO}_{4}}{{Molar mass of BaSO}_{4}}$

$= \frac{466 mg}{233 {g mol}^{- 1}} = \frac{466 \times 10^{- 3} g}{233 {g mol}^{- 1}} = 0.002 mol$

$Moles of S = {moles of BaSO}_{4} = 0.002 mol$

$Mass of S = moles of S \times molar mass of S$

$= 0.002 \times 32g = 0.064 g = 64 mg$

$% of S = \frac{Mass of S}{Mass of organic compound} \times 100$

$= \frac{64 mg}{160 mg} \times 100 = 40 %$

Q 9 :

In Carius method of examination of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is _____ $\times 10^{- 1}$ % (Nearest integer).

(Given: Molar mass of Ag is 108 and Br is 80 g ${mol}^{- 1}$ ) [2025]

(255)

$Molar mass of AgBr = (108 + 80) {g mol}^{- 1} = 188 {g mol}^{- 1}$

$Number of moles of AgBr (n_{AgBr}) = \frac{Mass of AgBr}{Molar mass of AgBr} = \frac{0.15}{188} mol$

$Moles of Br = Moles of AgBr = \frac{0.15}{188} mol$

$Mass of Br = Moles of Br \times Molar mass of Br = \frac{0.15}{188} \times 80 g$

$Mass % of Br = \frac{Mass of Br}{Mass of organic compound} \times 100$

$= \frac{\frac{0.15}{188} \times 80}{0.25} \times 100 = 25.53 % = 255.3 \times 10^{- 1} %$

Q 10 :

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ______ $\times 10^{- 1}$ %. (Molar mass: O = 16, S = 32, Ba = 137 in g ${mol}^{- 1}$ ) [2025]

(275)

Molar mass of ${BaSO}_{4} (M_{{BaSO}_{4}})$ $= [1 \times 137 + 1 \times 32 + 4 \times 16] {g mol}^{- 1} = 233 {g mol}^{- 1}$

Moles of ${BaSO}_{4} (n_{{BaSO}_{4}})$ $= \frac{{Mass of BaSO}_{4}}{{Molar mass of BaSO}_{4}} = \frac{0.4}{233} mol$

Moles of S = Moles of ${BaSO}_{4} = \frac{0.4}{233} mol$

$Mass of S = Moles of S \times Molar mass of S = \frac{0.4}{233} \times 32 g$

Mass % of S in compound $= \frac{Mass of S}{Mass of organic compound} \times 100$

$= \frac{\frac{0.4}{233} \times 32}{0.2} \times 100$

$= 27.5 % = 275 \times 10^{- 1} %$

Q 11 :

During estimation of nitrogen by Dumas’ method of compound X (0.42 g):

_____ mL of $N_{2}$ gas will be liberated at STP. (nearest integer)

(Given molar mass in g ${mol}^{- 1}$ : C : 12, H : 1, N : 14) [2025]

(111)

Molar mass of the compound $(C_{4} N_{2} H_{10})$ $= 4 \times 12 + 2 \times 14 + 10 \times 1 = 86 {g mol}^{- 1}$

$Moles of compound = \frac{Mass of compound}{Molar mass of compound}$ $= \frac{0.42}{86}$ mol

$Moles of N in compound = 2 \times Moles of compound$ $= 2 \times \frac{0.42}{86}$ mol

As all the N of organic compound is converted to $N_{2}$ gas in Duma’s method.

Moles of $N_{2} = \frac{Moles of N}{2}$
$= \frac{0.42}{86} mol$

Volume of $N_{2}$ at STP $= Moles of N_{2} \times 22.7 L = \frac{0.42}{86} \times 22.7 L$

$= 0.11086 L = 110.8 mL \approx 111 mL .$

Q 12 :

0.5 g of an organic compound on combustion gave 1.46 g of ${CO}_{2}$ and 0.9 g of $H_{2} O$ . The percentage of carbon in the compound is _____. (Nearest integer)

[Given: Molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16] [2025]

(80)

Moles of carbon in organic compound ( $n_{C}$ ) = moles of carbon in ${CO}_{2}$

$=Moles of C O_{2} = \frac{Mass of C O_{2}}{Molar mass of C O_{2}} = \frac{1.46}{44} mol$

Mass of carbon in organic compound ( $W_{C}$ ) $= n_{C} \times Molar mass of C = \frac{1.46}{44} \times 12 g$

Percentage of carbon in organic compound

$= \frac{Mass of carbon in organic compound}{Mass of organic compound} \times 100$

$= \frac{1.46 \times 12}{44 \times 0.5} \times 100 = 79.63 % \approx 80 %$

Q 13 :

In Dumas’ method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _____ % (nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

(20)

$Pressure of nitrogen gas (P_{N_{2}}) = Total pressure of gas - aqueous tension$
$= (900 - 15) mmHg = 885 mmHg = \frac{885}{760} atm$

$Volume of nitrogen (V_{N_{2}}) = 150 mL = 0.15 L$

By ideal gas equation, moles of nitrogen $(n_{N_{2}}) = \frac{P_{N_{2}} V_{N_{2}}}{R T}$

$= \frac{\frac{885}{760} \times 0.15}{0.0821 \times 300} = 0.0071 mol$

$Mass of nitrogen (W_{N_{2}}) = n_{N_{2}} \times Molar mass of N_{2}$

$= 0.0071 \times 28 g = 0.1988 g$

Percentage of nitrogen in organic compound $= \frac{Mass of nitrogen in organic compound}{Mass of organic compound} \times 100$

$= \frac{0.1988}{1} \times 100 = 19.88 %$

Q 14 :

In Dumas’ method 292 mg of an organic compound released 50 mL of nitrogen gas ( $N_{2}$ ) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ‘N’ in the organic compound is _____ % (Nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

(18)

Pressure of nitrogen $(P_{N_{2}})$ = Total pressure – aqueous tension

$= (715 - 15) mmHg = \frac{700}{760} atm$

Volume of nitrogen $(V_{N_{2}}) = 50 mL = 0.05 L$

Moles of nitrogen $(n_{N_{2}}) = \frac{P_{N_{2}} V_{N_{2}}}{R T} = \frac{\frac{700}{760} \times 0.05}{0.0821 \times 300} mol$

Mass of nitrogen $(W_{N_{2}}) = moles of nitrogen \times molar mass of nitrogen$

$= \frac{\frac{700}{760} \times 0.05}{0.0821 \times 300} \times 28 g$

Mass of organic compound $(W_{o . c .}) = 292 mg = 0.292 g$

Percentage of nitrogen in organic compound $= \frac{W_{N_{2}}}{W_{o . c .}} \times 100$

$= \frac{\frac{700}{760} \times 0.05 \times 28}{0.0821 \times 300 \times 0.292} \times 100 = 17.92 % \approx 18 %$

Q 15 :

In Dumas method for the estimation of $N_{2}$ , the sample is heated with copper oxide and the gas evolved is passed over: [2023]

Copper gauze
Pd
Ni
Copper oxide

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(1)

In Dumas method for the estimation of $N_{2}$ , the sample is heated with copper oxide and the gas evolved is passed over copper gauze to reduce traces of nitrogen oxides into nitrogen gas.

Q 16 :

In carius tube, an organic compound ‘X’ is treated with sodium peroxide to form a mineral acid ‘Y’.

The solution of ${BaCl}_{2}$ is added to ‘Y’ to form a precipitate ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be [2023]

Methionine
Chloroxylenol
Cytosine
A nucleotide

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(1)

Carius method is used for quantitative estimation of sulphur.

Q 17 :

In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ________ (Nearest integer)

(Given: Atomic mass Ba : 137u, S : 32u, O : 16u) [2023]

(42)

$% sulphur = \frac{32}{233} \times \frac{{Weight of BaSO}_{4} formed}{Weight of organic compound} \times 100$

$= \frac{32}{233} \times \frac{1.4439}{0.471} \times 100 = 42.10$

Q 18 :

0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is __________.

(Given: Molar mass AgBr = 188 g ${mol}^{- 1}$ ,

Br = 80 g ${mol}^{- 1}$ ) [2023]

(40)

$Organic compound (X) \overset{{Ag}^{+}}{\to} AgBr$

0.40 gram 0.376 gram

$Mole of AgBr = \frac{0.376}{188} = 0.002$

$Mole of Br = 0.002$

$Mass of Br = 0.002 \times 80 = 0.16 g$

$% of Br = \frac{0.16}{0.4} \times 100 = 40 %$

Q 19 :

0.5 g of an organic compound (X) with 60% carbon will produce ______ $\times 10^{- 1}$ g of ${CO}_{2}$ on complete combustion. [2023]

(11)

$60 = \frac{12}{44} \times \frac{{wt. of CO}_{2}}{0.5} \times 100$

${wt. of CO}_{2} = \frac{60 \times 44 \times 0.5}{12 \times 100} = 1.1 = 11 \times 10^{- 1} g$

Q 20 :

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur $(molar mass 32 g mol⁻¹).$ Molar mass of barium sulphate is $233 g mol⁻¹ .$ [2026]

4.55%
21.97%
16.48%
10.30%

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(2)

$\frac{n_{{BaSO}_{4}} \times 32}{W_{(unknown comp .)}} \times 100$

$= \frac{1.2 \times 32}{233} \times \frac{100}{0.75} = 21.97 %$

Q 21 :

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is _____ %.

(Aqueous tension at 300 K is 15 mm.) [2026]

(15)

$P_{N_{2}} = (715 - 15) mm = \frac{700}{760} atm$

$V_{N_{2}} = 70 ml = \frac{70}{1000} L$

$n_{N_{2}} = \frac{P V}{R T} = \frac{(\frac{700}{760}) (\frac{70}{1000})}{0.0821 \times 300}$

$W_{N_{2}} = \frac{700}{760} \times \frac{70}{1000} \times \frac{1}{0.0821 \times 300} \times 28$

$% N = \frac{W_{N_{2}}}{0.5} \times 100 = \frac{700}{760} \times \frac{70}{1000} \times \frac{1}{0.0821 \times 300} \times \frac{28}{0.5} \times 100$

$= 14.65 % \approx 15$

Q 22 :

Sodium fusion extract of an organic compound (Y) with ${CHCl}_{3}$ and chlorine water gives violet color to the ${CHCl}_{3}$ layer. 0.15 g of (Y) gave 0.12 g of the silver halide precipitate in Carius method. Percentage of halogen in the compound (Y) is __________. (Nearest integer)

(Given: molar mass g ${mol}^{- 1}$ C : 12, H : 1, Cl : 35.5, Br : 80, I : 127) [2026]

(43)

$Iodine gives violet colour$

$% of I = \frac{Atomic weight of I}{Molecular weight of AgI} \times \frac{m}{W} \times 100$

$= \frac{127}{235} \times \frac{0.12}{0.15} \times 100$

$% of I = 43.23 % \approx 43 %$

Q 23 :

The cycloalkene (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C : Br ratio is 3 : 1. The percentage of bromine in the product (Y) is __________ %. (Nearest integer)

(Given: molar mass in g ${mol}^{- 1}$ H : 1, C : 12, O : 16, Br : 80) [2026]

(66)

$C_{6} H_{10} \overset{{Br}_{2}}{\to} C_{6} H_{10} {Br}_{2}$

$Molecular mass of C_{6} H_{10} {Br}_{2} is:$

$12 \times 6 + 10 + 160$

$72 + 10 + 160 = 242$

$% of Br = \frac{160}{242} \times 100$

$% of Br = 66.11 % \approx 66 %$

Q 24 :

A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is __________.
(Given Molar mass in g $mol ⁻ ¹ : S : 32, B a S O_{4} : 233$ ) [2026]

42.10%
48.24%
63.15%
21.05%

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(4)

$% S = \frac{32}{233} \times \frac{0.4813}{0.314} \times 100$

$= 21.052 %$

$Ans. (4) 21.05%$

Q 25 :

0.25 g of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of ${CaCl}_{2}$ tube and potash tube at the end of the experiment. The amount was found to be 0.15 g and 0.1837 g, respectively. The percentage of oxygen in compound A is ______ %. (Nearest integer)

[Given: molar mass in g ${mol}^{- 1}$ H : 1, C : 12, O : 16] [2026]

(73)

$\begin{matrix} C_{x} H_{y} O_{z} & + & O_{2} & \to & {CO}_{2} & + & H_{2} O \\ t = 0 & 0.25 gm \\ t_{e q} & - & 0.18 gm & 0.15 gm \end{matrix}$

$Mass of 'C' = \frac{0.18}{44} \times 12 = 0.049 \approx 0.05 gm$

$Mass of 'H' = \frac{0.15}{18} \times 2 = 0.016 \approx 0.017 gm$

$Mass of 'O' = 0.25 - 0.05 - 0.017 = 0.1833 gm$

$Mass % of 'O' = \frac{0.1833}{0.25} \times 100 = 73.32 %$

Q 26 :

In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is [2026]

34.79%
37.57%
87.65%
53.58%

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(4)

$Organic compound = 0.2425 gm$

$AgCl obtained = 0.5253 gm$

$In Carius method for estimation of halogen, amount of AgCl obtained is 0.5253 gm from 0.2425 gm of organic compound.$

$Hence percentage of Cl.$

$Percentage of Cl = \frac{35.5}{143.5} \times \frac{0.5253}{0.2425} \times 100$

$= 53.58 %$

Q 27 :

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer)

(Given, molar mass in g ${m o l}^{- 1}$ : O = 16, Mg = 24, P = 31) [2026]

50
30
40
20

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(1)

$% of P = \frac{n_{{Mg}_{2} P_{2} O_{7}} \times 2 \times 31}{W_{(unknown compound)}} \times 100$

$= \frac{(\frac{1.79}{222} \times 2 \times 31)}{1} \times 100$

$= 49.99 % \approx 50 % .$

Q 28 :

When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M $H_{2} {SO}_{4}$ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is : [2026]

0.42
42
0.21
21

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(2)

$eq. of H_{2} {SO}_{4} = eq. of Ammonia$

$\Rightarrow \frac{15 \times 1 \times 2}{1000} = moles of ammonia \times 1$

$\Rightarrow Moles of ammonia = moles of 'N'$

$\Rightarrow Weight of nitrogen = \frac{15 \times 1 \times 2}{1000} \times 14 = 0.42$

$% weight of 'N' = \frac{0.42}{1} \times 100 = 42 %$

Q 29 :

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M ${KMnO}_{4}$ solution in basic medium. The liberated iodine was titrated with standard 0.1 M ${Na}_{2} S_{2} O_{3}$ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of ${Na}_{2} S_{2} O_{3}$ consumed is _____. (Nearest integer) [2026]

(3)

${MnO}_{4}^{-} + I^{-} \to {MnO}_{2} + I_{2}$

$I_{2} + S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + I^{-}$

${gram eq of KMnO}_{4} = {gram eq of Na}_{2} S_{2} O_{3}$

$0.2 \times \frac{500}{1000} \times 3 = 0.1 \times V \times 1$

$V = 3 L$

Q 30 :

0.53 g of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave 0.75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of $C O_{2}$ gas on combustion. The percentage of hydrogen in the compound (x) is ____%. [Nearest Integer]

[Given: Molar mass in g ${mol}^{- 1}$ H : 1, C : 12, Br : 80, Ag : 108, O : 16 ;
Compound (x) : $C_{x} H_{y} B r_{z}$ ] [2026]

(4)

$\begin{matrix} C_{x} H_{y} {Br}_{z} & \to & {CO}_{2} \\ 1 g & 1.32 g \end{matrix}$

$% C = \frac{1.32 \times 12}{44 \times 1} \times 100 = 36 %$

$\begin{matrix} C_{x} H_{y} {Br}_{z} & \to & AgBr \\ 0.53 g & 0.75 g \end{matrix}$

$% Br = \frac{0.75 \times 80}{188 \times 0.53} \times 100 = 60.2 %$

$% H = 100 - (36 + 60.2)$

$% H \approx 4 %$

Topic Question Set

Q 1 :

In Kjeldahl’s method for estimation of nitrogen, ${CuSO}_{4}$ acts as: [2024]

Q 3 :

Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]

Q 5 :

In Dumas’ method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is [2025]

Q 6 :

In Dumas’ method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is

(Given: Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 7 :

In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ______ %.

(Given: molar mass in g ${mol}^{- 1}$ of Ag : 108, Cl : 35.5) [2025]

Q 8 :

During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %.

(Given molar mass in g ${mol}^{- 1}$ of Ba : 137, S : 32, O : 16) [2025]

Q 9 :

In Carius method of examination of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is _____ $\times 10^{- 1}$ % (Nearest integer).

(Given: Molar mass of Ag is 108 and Br is 80 g ${mol}^{- 1}$ ) [2025]

Q 10 :

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ______ $\times 10^{- 1}$ %. (Molar mass: O = 16, S = 32, Ba = 137 in g ${mol}^{- 1}$ ) [2025]

Q 11 :

During estimation of nitrogen by Dumas’ method of compound X (0.42 g):

_____ mL of $N_{2}$ gas will be liberated at STP. (nearest integer)

(Given molar mass in g ${mol}^{- 1}$ : C : 12, H : 1, N : 14) [2025]

Q 12 :

0.5 g of an organic compound on combustion gave 1.46 g of ${CO}_{2}$ and 0.9 g of $H_{2} O$ . The percentage of carbon in the compound is _____. (Nearest integer)

[Given: Molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16] [2025]

Q 13 :

In Dumas’ method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _____ % (nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 14 :

In Dumas’ method 292 mg of an organic compound released 50 mL of nitrogen gas ( $N_{2}$ ) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ‘N’ in the organic compound is _____ % (Nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 15 :

In Dumas method for the estimation of $N_{2}$ , the sample is heated with copper oxide and the gas evolved is passed over: [2023]

Q 16 :

In carius tube, an organic compound ‘X’ is treated with sodium peroxide to form a mineral acid ‘Y’.

The solution of ${BaCl}_{2}$ is added to ‘Y’ to form a precipitate ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be [2023]

Q 17 :

In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ________ (Nearest integer)

(Given: Atomic mass Ba : 137u, S : 32u, O : 16u) [2023]

Q 18 :

0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is __________.

(Given: Molar mass AgBr = 188 g ${mol}^{- 1}$ ,

Br = 80 g ${mol}^{- 1}$ ) [2023]

Q 19 :

0.5 g of an organic compound (X) with 60% carbon will produce ______ $\times 10^{- 1}$ g of ${CO}_{2}$ on complete combustion. [2023]

Q 20 :

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur $(molar mass 32 g mol⁻¹).$ Molar mass of barium sulphate is $233 g mol⁻¹ .$ [2026]

Q 21 :

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is _____ %.

(Aqueous tension at 300 K is 15 mm.) [2026]

Q 26 :

In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is [2026]

Q 27 :

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer)

(Given, molar mass in g ${m o l}^{- 1}$ : O = 16, Mg = 24, P = 31) [2026]

Q 28 :

When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M $H_{2} {SO}_{4}$ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is : [2026]

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Topic Question Set

Q 1 : In Kjeldahl’s method for estimation of nitrogen, CuSO4 acts as: [2024]

Q 3 : Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M H2SO4. The percentage of nitrogen in the compound is ______%. [2024]

Q 5 : In Dumas’ method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is [2025]

Q 6 : In Dumas’ method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given: Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 7 : In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ______ %. (Given: molar mass in g mol-1 of Ag : 108, Cl : 35.5) [2025]

Q 8 : During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %. (Given molar mass in g mol-1 of Ba : 137, S : 32, O : 16) [2025]

Q 9 : In Carius method of examination of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is _____ ×10-1 % (Nearest integer). (Given: Molar mass of Ag is 108 and Br is 80 g mol-1) [2025]

Q 10 : In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ______ ×10-1 %. (Molar mass: O = 16, S = 32, Ba = 137 in g mol-1) [2025]

Q 11 : During estimation of nitrogen by Dumas’ method of compound X (0.42 g): _____ mL of N2 gas will be liberated at STP. (nearest integer) (Given molar mass in g mol-1: C : 12, H : 1, N : 14) [2025]

Q 12 : 0.5 g of an organic compound on combustion gave 1.46 g of CO2 and 0.9 g of H2O. The percentage of carbon in the compound is _____. (Nearest integer) [Given: Molar mass (in g mol-1) C : 12, H : 1, O : 16] [2025]

Q 13 : In Dumas’ method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _____ % (nearest integer). (Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 14 : In Dumas’ method 292 mg of an organic compound released 50 mL of nitrogen gas (N2) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ‘N’ in the organic compound is _____ % (Nearest integer). (Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 15 : In Dumas method for the estimation of N2, the sample is heated with copper oxide and the gas evolved is passed over: [2023]

Q 16 : In carius tube, an organic compound ‘X’ is treated with sodium peroxide to form a mineral acid ‘Y’. The solution of BaCl2 is added to ‘Y’ to form a precipitate ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be [2023]

Q 17 : In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ________ (Nearest integer) (Given: Atomic mass Ba : 137u, S : 32u, O : 16u) [2023]

Q 18 : 0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is __________. (Given: Molar mass AgBr = 188 g mol-1, Br = 80 g mol-1) [2023]

Q 19 : 0.5 g of an organic compound (X) with 60% carbon will produce ______ ×10-1 g of CO2 on complete combustion. [2023]

Q 20 : In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g mol⁻¹). Molar mass of barium sulphate is 233 g mol⁻¹. [2026]

Q 21 : In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is _____ %. (Aqueous tension at 300 K is 15 mm.) [2026]

Q 23 : The cycloalkene (X) on bromination consumes one mole of bromine per mole of (X) and gives the product (Y) in which C : Br ratio is 3 : 1. The percentage of bromine in the product (Y) is __________ %. (Nearest integer) (Given: molar mass in g mol-1 H : 1, C : 12, O : 16, Br : 80) [2026]

Q 24 : A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in the compound is __________. (Given Molar mass in g mol⁻¹: S:32, BaSO4:233) [2026]

Q 26 : In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is [2026]

Q 27 : By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer) (Given, molar mass in g mol-1: O = 16, Mg = 24, P = 31) [2026]

Q 28 : When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M H2SO4 was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is : [2026]

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Q 1 :

In Kjeldahl’s method for estimation of nitrogen, ${CuSO}_{4}$ acts as: [2024]

Q 3 :

Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2M $H_{2} {SO}_{4}$ . The percentage of nitrogen in the compound is ______%. [2024]

Q 5 :

In Dumas’ method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is [2025]

Q 6 :

In Dumas’ method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is

(Given: Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 7 :

In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl. The percentage composition of chlorine in the compound is ______ %.

(Given: molar mass in g ${mol}^{- 1}$ of Ag : 108, Cl : 35.5) [2025]

Q 8 :

During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %.

(Given molar mass in g ${mol}^{- 1}$ of Ba : 137, S : 32, O : 16) [2025]

Q 9 :

In Carius method of examination of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is _____ $\times 10^{- 1}$ % (Nearest integer).

(Given: Molar mass of Ag is 108 and Br is 80 g ${mol}^{- 1}$ ) [2025]

Q 10 :

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ______ $\times 10^{- 1}$ %. (Molar mass: O = 16, S = 32, Ba = 137 in g ${mol}^{- 1}$ ) [2025]

Q 11 :

During estimation of nitrogen by Dumas’ method of compound X (0.42 g):

_____ mL of $N_{2}$ gas will be liberated at STP. (nearest integer)

(Given molar mass in g ${mol}^{- 1}$ : C : 12, H : 1, N : 14) [2025]

Q 12 :

0.5 g of an organic compound on combustion gave 1.46 g of ${CO}_{2}$ and 0.9 g of $H_{2} O$ . The percentage of carbon in the compound is _____. (Nearest integer)

[Given: Molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16] [2025]

Q 13 :

In Dumas’ method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is _____ % (nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 14 :

In Dumas’ method 292 mg of an organic compound released 50 mL of nitrogen gas ( $N_{2}$ ) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ‘N’ in the organic compound is _____ % (Nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

Q 15 :

In Dumas method for the estimation of $N_{2}$ , the sample is heated with copper oxide and the gas evolved is passed over: [2023]

Q 16 :

In carius tube, an organic compound ‘X’ is treated with sodium peroxide to form a mineral acid ‘Y’.

The solution of ${BaCl}_{2}$ is added to ‘Y’ to form a precipitate ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be [2023]

Q 17 :

In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ________ (Nearest integer)

(Given: Atomic mass Ba : 137u, S : 32u, O : 16u) [2023]

Q 18 :

0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is __________.

(Given: Molar mass AgBr = 188 g ${mol}^{- 1}$ ,

Br = 80 g ${mol}^{- 1}$ ) [2023]

Q 19 :

0.5 g of an organic compound (X) with 60% carbon will produce ______ $\times 10^{- 1}$ g of ${CO}_{2}$ on complete combustion. [2023]

Q 20 :

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur $(molar mass 32 g mol⁻¹).$ Molar mass of barium sulphate is $233 g mol⁻¹ .$ [2026]

Q 21 :

In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is _____ %.

(Aqueous tension at 300 K is 15 mm.) [2026]

Q 26 :

In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride. The percentage of chlorine in the organic compound is [2026]

Q 27 :

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer)

(Given, molar mass in g ${m o l}^{- 1}$ : O = 16, Mg = 24, P = 31) [2026]

Q 28 :

When 1 g of compound (X) is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M $H_{2} {SO}_{4}$ was neutralized by ammonia evolved. The percentage of nitrogen in compound (X) is : [2026]