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CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317518 In Carius method of estimation of halogens, 355 mg of an organic compound gave 287 mg of AgCl . The percentage of chlorine in the compound is ____.
(At. Mass: $Ag = 108; Cl = 35.5 u$ )

1 10

2 20

3 30

4 40

Explanation:

$%$ of chlorine
$= \frac{35.5 \times Mass of AgCl}{143.5 \times Mass of organic compound} \times 100$
$= \frac{35.5 \times 0.287 \times 100}{143.5 \times 0.355} = 20 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317519 0.24 g of volatile liquid upon vapourisation gives 45 mL of vapours at STP. What will be the vapour density of the substance? (Density of $H_{2} = 0.089 g$ litre $^{- 1}$ )

1 2.4

2 40

3 30.3

4 59.7

Explanation:

It is an example of Victor Meyer's method.
$\begin{aligned} Mw & = \frac{Mass of compound \times 22400}{Volume of vapours at STP} \\ = \frac{0.24 \times 22400}{45} = 119.46 \end{aligned}$
Vapour density $= \frac{Mw}{2} = \frac{119.46}{2} = 59.7$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317520 Duma's method involves the determination of content of nitrogen in the organic compound in the form of

1 Gaseous

{NH}_{3}

2

N a C N

3 Gaseous

N_{2}

4

{({NH}_{4})}_{2} {SO}_{4}

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317521 In Duma's method for estimation of nitrogen $0.3 g$ of an organic compound gave $50 ml$ , of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. Calculate the $%$ composition of nitrogen in the compound: (Aqueous tension at $300 K = 15 mm$ )

1

11.46 %

2

17.46 %

3

22.06 %

4

39.65 %

Explanation:

Volume of nitrogen collected at $300 K$ and $715 mm$ pressure is $50 ml$ .
$P_{gas} = P_{Total} -$ Aqueous tension $= 715 - 15 = 700 mm$
Volume of nitrogen at STP
$\frac{273 \times 700 \times 50}{300 \times 760} = 41.9 ml$
$(from \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}})$
$% N = \frac{vol.of N_{2} (ml)}{mass of compound} \times \frac{28}{22400} \times 100$ (Duma's
method)
$= \frac{41.9}{0.3} \times \frac{28}{22400} \times 100 = 17.46 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317518 In Carius method of estimation of halogens, 355 mg of an organic compound gave 287 mg of AgCl . The percentage of chlorine in the compound is ____.
(At. Mass: $Ag = 108; Cl = 35.5 u$ )

1 10

2 20

3 30

4 40

Explanation:

$%$ of chlorine
$= \frac{35.5 \times Mass of AgCl}{143.5 \times Mass of organic compound} \times 100$
$= \frac{35.5 \times 0.287 \times 100}{143.5 \times 0.355} = 20 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317519 0.24 g of volatile liquid upon vapourisation gives 45 mL of vapours at STP. What will be the vapour density of the substance? (Density of $H_{2} = 0.089 g$ litre $^{- 1}$ )

1 2.4

2 40

3 30.3

4 59.7

Explanation:

It is an example of Victor Meyer's method.
$\begin{aligned} Mw & = \frac{Mass of compound \times 22400}{Volume of vapours at STP} \\ = \frac{0.24 \times 22400}{45} = 119.46 \end{aligned}$
Vapour density $= \frac{Mw}{2} = \frac{119.46}{2} = 59.7$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317520 Duma's method involves the determination of content of nitrogen in the organic compound in the form of

1 Gaseous

{NH}_{3}

2

N a C N

3 Gaseous

N_{2}

4

{({NH}_{4})}_{2} {SO}_{4}

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317521 In Duma's method for estimation of nitrogen $0.3 g$ of an organic compound gave $50 ml$ , of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. Calculate the $%$ composition of nitrogen in the compound: (Aqueous tension at $300 K = 15 mm$ )

1

11.46 %

2

17.46 %

3

22.06 %

4

39.65 %

Explanation:

Volume of nitrogen collected at $300 K$ and $715 mm$ pressure is $50 ml$ .
$P_{gas} = P_{Total} -$ Aqueous tension $= 715 - 15 = 700 mm$
Volume of nitrogen at STP
$\frac{273 \times 700 \times 50}{300 \times 760} = 41.9 ml$
$(from \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}})$
$% N = \frac{vol.of N_{2} (ml)}{mass of compound} \times \frac{28}{22400} \times 100$ (Duma's
method)
$= \frac{41.9}{0.3} \times \frac{28}{22400} \times 100 = 17.46 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317518 In Carius method of estimation of halogens, 355 mg of an organic compound gave 287 mg of AgCl . The percentage of chlorine in the compound is ____.
(At. Mass: $Ag = 108; Cl = 35.5 u$ )

1 10

2 20

3 30

4 40

Explanation:

$%$ of chlorine
$= \frac{35.5 \times Mass of AgCl}{143.5 \times Mass of organic compound} \times 100$
$= \frac{35.5 \times 0.287 \times 100}{143.5 \times 0.355} = 20 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317519 0.24 g of volatile liquid upon vapourisation gives 45 mL of vapours at STP. What will be the vapour density of the substance? (Density of $H_{2} = 0.089 g$ litre $^{- 1}$ )

1 2.4

2 40

3 30.3

4 59.7

Explanation:

It is an example of Victor Meyer's method.
$\begin{aligned} Mw & = \frac{Mass of compound \times 22400}{Volume of vapours at STP} \\ = \frac{0.24 \times 22400}{45} = 119.46 \end{aligned}$
Vapour density $= \frac{Mw}{2} = \frac{119.46}{2} = 59.7$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317520 Duma's method involves the determination of content of nitrogen in the organic compound in the form of

1 Gaseous

{NH}_{3}

2

N a C N

3 Gaseous

N_{2}

4

{({NH}_{4})}_{2} {SO}_{4}

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317521 In Duma's method for estimation of nitrogen $0.3 g$ of an organic compound gave $50 ml$ , of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. Calculate the $%$ composition of nitrogen in the compound: (Aqueous tension at $300 K = 15 mm$ )

1

11.46 %

2

17.46 %

3

22.06 %

4

39.65 %

Explanation:

Volume of nitrogen collected at $300 K$ and $715 mm$ pressure is $50 ml$ .
$P_{gas} = P_{Total} -$ Aqueous tension $= 715 - 15 = 700 mm$
Volume of nitrogen at STP
$\frac{273 \times 700 \times 50}{300 \times 760} = 41.9 ml$
$(from \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}})$
$% N = \frac{vol.of N_{2} (ml)}{mass of compound} \times \frac{28}{22400} \times 100$ (Duma's
method)
$= \frac{41.9}{0.3} \times \frac{28}{22400} \times 100 = 17.46 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317518 In Carius method of estimation of halogens, 355 mg of an organic compound gave 287 mg of AgCl . The percentage of chlorine in the compound is ____.
(At. Mass: $Ag = 108; Cl = 35.5 u$ )

1 10

2 20

3 30

4 40

Explanation:

$%$ of chlorine
$= \frac{35.5 \times Mass of AgCl}{143.5 \times Mass of organic compound} \times 100$
$= \frac{35.5 \times 0.287 \times 100}{143.5 \times 0.355} = 20 %$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317519 0.24 g of volatile liquid upon vapourisation gives 45 mL of vapours at STP. What will be the vapour density of the substance? (Density of $H_{2} = 0.089 g$ litre $^{- 1}$ )

1 2.4

2 40

3 30.3

4 59.7

Explanation:

It is an example of Victor Meyer's method.
$\begin{aligned} Mw & = \frac{Mass of compound \times 22400}{Volume of vapours at STP} \\ = \frac{0.24 \times 22400}{45} = 119.46 \end{aligned}$
Vapour density $= \frac{Mw}{2} = \frac{119.46}{2} = 59.7$

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317520 Duma's method involves the determination of content of nitrogen in the organic compound in the form of

1 Gaseous

{NH}_{3}

2

N a C N

3 Gaseous

N_{2}

4

{({NH}_{4})}_{2} {SO}_{4}

CHXI12:ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES

317521 In Duma's method for estimation of nitrogen $0.3 g$ of an organic compound gave $50 ml$ , of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. Calculate the $%$ composition of nitrogen in the compound: (Aqueous tension at $300 K = 15 mm$ )

1

11.46 %

2

17.46 %

3

22.06 %

4

39.65 %

Explanation:

Volume of nitrogen collected at $300 K$ and $715 mm$ pressure is $50 ml$ .
$P_{gas} = P_{Total} -$ Aqueous tension $= 715 - 15 = 700 mm$
Volume of nitrogen at STP
$\frac{273 \times 700 \times 50}{300 \times 760} = 41.9 ml$
$(from \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}})$
$% N = \frac{vol.of N_{2} (ml)}{mass of compound} \times \frac{28}{22400} \times 100$ (Duma's
method)
$= \frac{41.9}{0.3} \times \frac{28}{22400} \times 100 = 17.46 %$

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