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]

0%

(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]

0%

(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]

0%

(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]

0%

(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]

0%

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

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

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

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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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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]

0%

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]

0%

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]

0%

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]

0%

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]

0%

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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]