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

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]

0%

(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]

0%

(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]

0%

(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]

0%

(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

1

2

3

4

(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

1

2

3

4

(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]

0%

(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]

0%

(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]

0%

(11)

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

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

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