QBManager

PRACTICAL CHEMISTRY

37577 On passing a gas, ' $X$ ', through Nessler's reagent, a brown precipitate is obtained. The gas ' $X$ ' is

1

H_{2} S

2

{CO}_{2}

3

{NH}_{3}

4

{Cl}_{2}

Explanation:

Nessler's Reagent Reaction :
$\underset{(Nessler's Reagent)}{2 K_{2} {HgI}_{4}} + {NH}_{3} + 3 KOH \to \underset{\begin{matrix} (Iodine of Millon's base \\ Brorrn precipitate \end{matrix}}{HgO . Hg ({NH}_{2}) I} + 7 KI + 2 H_{2} O$

PRACTICAL CHEMISTRY

37578 The oxidation number of iron in the compound formed during brown ring test for ${NO}_{3}^{-}$ ion is___________

1

9

2

1

3

5

4

45

Explanation:

${[Fe {(H_{2} O)}_{5} (NO)]}^{2 +}$ ,
Oxidation no. of $Fe = + 1$

PRACTICAL CHEMISTRY

37579 If $50 mL$ of $0.5 M$ oxalic acid is required to neutralise $25 mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 mL$ of given $NaOH$ solution is__________ $g .$

1

4

2

5

3

7

4

9

Explanation:

Equivalent of Oxalic acid = Equivalents of $NaOH$
$50 \times 0.5 \times 2 = 25 \times M \times 1$
$M_{NaOH} = 2 M$
$W_{NaOH} in 50 ml = 2 \times 50 \times 40 \times 10^{- 3} g = 4 g$

PRACTICAL CHEMISTRY

37580 Given below are two statements:Statement- $I$ : The gas liberated on warming a salt with dil $H_{2} {SO}_{4}$ , turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.Statement- $I I$ : In statement-I the colour of paper turns black because of formation of lead sulphite.In the light of the above statements, choose the most appropriate answer from the options given below:

1 Both Statement-

I

and Statement-

I I

are false

2 Statement-

I

is false but Statement-

I I

is true

3 Statement-

I

is true but Statement-

I I

is false

4 Both Statement-

I

and Statement-

I I

are true.

Explanation:

$\begin{matrix} {Na}_{2} S + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + H_{2} S \\ {({CH}_{3} COO)}_{2} Pb + H_{2} S \to PbS + 2 {CH}_{3} COOH \\ Black lead \\ sulphide \end{matrix}$

PRACTICAL CHEMISTRY

37577 On passing a gas, ' $X$ ', through Nessler's reagent, a brown precipitate is obtained. The gas ' $X$ ' is

1

H_{2} S

2

{CO}_{2}

3

{NH}_{3}

4

{Cl}_{2}

Explanation:

Nessler's Reagent Reaction :
$\underset{(Nessler's Reagent)}{2 K_{2} {HgI}_{4}} + {NH}_{3} + 3 KOH \to \underset{\begin{matrix} (Iodine of Millon's base \\ Brorrn precipitate \end{matrix}}{HgO . Hg ({NH}_{2}) I} + 7 KI + 2 H_{2} O$

PRACTICAL CHEMISTRY

37578 The oxidation number of iron in the compound formed during brown ring test for ${NO}_{3}^{-}$ ion is___________

1

9

2

1

3

5

4

45

Explanation:

${[Fe {(H_{2} O)}_{5} (NO)]}^{2 +}$ ,
Oxidation no. of $Fe = + 1$

PRACTICAL CHEMISTRY

37579 If $50 mL$ of $0.5 M$ oxalic acid is required to neutralise $25 mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 mL$ of given $NaOH$ solution is__________ $g .$

1

4

2

5

3

7

4

9

Explanation:

Equivalent of Oxalic acid = Equivalents of $NaOH$
$50 \times 0.5 \times 2 = 25 \times M \times 1$
$M_{NaOH} = 2 M$
$W_{NaOH} in 50 ml = 2 \times 50 \times 40 \times 10^{- 3} g = 4 g$

PRACTICAL CHEMISTRY

37580 Given below are two statements:Statement- $I$ : The gas liberated on warming a salt with dil $H_{2} {SO}_{4}$ , turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.Statement- $I I$ : In statement-I the colour of paper turns black because of formation of lead sulphite.In the light of the above statements, choose the most appropriate answer from the options given below:

1 Both Statement-

I

and Statement-

I I

are false

2 Statement-

I

is false but Statement-

I I

is true

3 Statement-

I

is true but Statement-

I I

is false

4 Both Statement-

I

and Statement-

I I

are true.

Explanation:

$\begin{matrix} {Na}_{2} S + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + H_{2} S \\ {({CH}_{3} COO)}_{2} Pb + H_{2} S \to PbS + 2 {CH}_{3} COOH \\ Black lead \\ sulphide \end{matrix}$

PRACTICAL CHEMISTRY

37577 On passing a gas, ' $X$ ', through Nessler's reagent, a brown precipitate is obtained. The gas ' $X$ ' is

1

H_{2} S

2

{CO}_{2}

3

{NH}_{3}

4

{Cl}_{2}

Explanation:

Nessler's Reagent Reaction :
$\underset{(Nessler's Reagent)}{2 K_{2} {HgI}_{4}} + {NH}_{3} + 3 KOH \to \underset{\begin{matrix} (Iodine of Millon's base \\ Brorrn precipitate \end{matrix}}{HgO . Hg ({NH}_{2}) I} + 7 KI + 2 H_{2} O$

PRACTICAL CHEMISTRY

37578 The oxidation number of iron in the compound formed during brown ring test for ${NO}_{3}^{-}$ ion is___________

1

9

2

1

3

5

4

45

Explanation:

${[Fe {(H_{2} O)}_{5} (NO)]}^{2 +}$ ,
Oxidation no. of $Fe = + 1$

PRACTICAL CHEMISTRY

37579 If $50 mL$ of $0.5 M$ oxalic acid is required to neutralise $25 mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 mL$ of given $NaOH$ solution is__________ $g .$

1

4

2

5

3

7

4

9

Explanation:

Equivalent of Oxalic acid = Equivalents of $NaOH$
$50 \times 0.5 \times 2 = 25 \times M \times 1$
$M_{NaOH} = 2 M$
$W_{NaOH} in 50 ml = 2 \times 50 \times 40 \times 10^{- 3} g = 4 g$

PRACTICAL CHEMISTRY

37580 Given below are two statements:Statement- $I$ : The gas liberated on warming a salt with dil $H_{2} {SO}_{4}$ , turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.Statement- $I I$ : In statement-I the colour of paper turns black because of formation of lead sulphite.In the light of the above statements, choose the most appropriate answer from the options given below:

1 Both Statement-

I

and Statement-

I I

are false

2 Statement-

I

is false but Statement-

I I

is true

3 Statement-

I

is true but Statement-

I I

is false

4 Both Statement-

I

and Statement-

I I

are true.

Explanation:

$\begin{matrix} {Na}_{2} S + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + H_{2} S \\ {({CH}_{3} COO)}_{2} Pb + H_{2} S \to PbS + 2 {CH}_{3} COOH \\ Black lead \\ sulphide \end{matrix}$

PRACTICAL CHEMISTRY

37577 On passing a gas, ' $X$ ', through Nessler's reagent, a brown precipitate is obtained. The gas ' $X$ ' is

1

H_{2} S

2

{CO}_{2}

3

{NH}_{3}

4

{Cl}_{2}

Explanation:

Nessler's Reagent Reaction :
$\underset{(Nessler's Reagent)}{2 K_{2} {HgI}_{4}} + {NH}_{3} + 3 KOH \to \underset{\begin{matrix} (Iodine of Millon's base \\ Brorrn precipitate \end{matrix}}{HgO . Hg ({NH}_{2}) I} + 7 KI + 2 H_{2} O$

PRACTICAL CHEMISTRY

37578 The oxidation number of iron in the compound formed during brown ring test for ${NO}_{3}^{-}$ ion is___________

1

9

2

1

3

5

4

45

Explanation:

${[Fe {(H_{2} O)}_{5} (NO)]}^{2 +}$ ,
Oxidation no. of $Fe = + 1$

PRACTICAL CHEMISTRY

37579 If $50 mL$ of $0.5 M$ oxalic acid is required to neutralise $25 mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 mL$ of given $NaOH$ solution is__________ $g .$

1

4

2

5

3

7

4

9

Explanation:

Equivalent of Oxalic acid = Equivalents of $NaOH$
$50 \times 0.5 \times 2 = 25 \times M \times 1$
$M_{NaOH} = 2 M$
$W_{NaOH} in 50 ml = 2 \times 50 \times 40 \times 10^{- 3} g = 4 g$

PRACTICAL CHEMISTRY

37580 Given below are two statements:Statement- $I$ : The gas liberated on warming a salt with dil $H_{2} {SO}_{4}$ , turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.Statement- $I I$ : In statement-I the colour of paper turns black because of formation of lead sulphite.In the light of the above statements, choose the most appropriate answer from the options given below:

1 Both Statement-

I

and Statement-

I I

are false

2 Statement-

I

is false but Statement-

I I

is true

3 Statement-

I

is true but Statement-

I I

is false

4 Both Statement-

I

and Statement-

I I

are true.

Explanation:

$\begin{matrix} {Na}_{2} S + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + H_{2} S \\ {({CH}_{3} COO)}_{2} Pb + H_{2} S \to PbS + 2 {CH}_{3} COOH \\ Black lead \\ sulphide \end{matrix}$

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