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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307031 2.76 g of silver carbonate on being strongly heated yield a residue weighing

1 2.16 g

2 2.48 g

3 2.64 g

4 2.32 g

Explanation:

(I) Mole method:
$2A g_{2} C O_{3} 4Ag + 2C O_{2} + O_{2}$
$2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$ $2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$
$2 \times 276 g o f A g_{2} C o_{3} g i v e s 4 \times 108 g o f A g$
$2 .76 g A g_{2} C O_{3} w i l l g i v e$
$(\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(II) Principle of Atomic Conservation (POAC) method:
Moles of $A g_{2} C O_{3}$ = Total moles of Ag as product
2 moles of $A g_{2} C O_{3}$ = 4 moles of Ag
$2 \times 276 g$ of $A g_{2} C O_{3}$ = $4 \times 108 g$ of Ag
2.76 g of $A g_{2} C O_{3}$ $= (\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(III) Equivalent Method:
Number of gram - equivalents of $A g_{2} C O_{3}$ = Number of gram-equivalents of Ag
$\frac{2 .76 g}{(\frac{2 \times 276 g / m o l}{1})} = \frac{y g}{(\frac{4 \times 108 g / m o l}{1})}$
$\Rightarrow y = 2 .16 g$
[Change in oxidation state of $A g_{2} C O_{3}$ and Ag is 1, respectively]

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307032 1.12 mL of a gas is produced at STP by the action of 4.12 mg of alcohol, with methyl magnesium iodide. The molecular mass of alcohol is

1 16.0

2 41.2

3 82.4

4 156.0

Explanation:

$ROH + {CH}_{3} MgI \to {CH}_{4} + MgROI$
$∴ 1$ mole alcohol gives 1 mole ${CH}_{4}$ .
According to data,
1.12 L(at STP) ${CH}_{4}$ is obtained from 4.12 g alcohol.
$∴ 22.4 L$ (at STP) ${CH}_{4}$ is obtained from
$= \frac{22.4 \times 4.12}{1.12} = 82.4 g$
So, $82.4 g {mol}^{-}$ is the molar mass of alcohol.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307033 1.0 g of Mg is burnt with 0.28 g of $O_{2}$ in a closed vessel. Which reactant is left in excess and how much?

1 Mg, 5.8 g

2 Mg, 0.58 g

3

O_{2}

, 0.24 g

4

O_{2}

, 2.4 g

Explanation:

Balanced reaction can be given as
$\underset{24 \times 2 = 48}{2 M g} + \underset{32}{O_{2}} \to 2 M g O$
32 g of $O_{2}$ is required to burn 48 g Mg
So, $0 .28 g O_{2}$ will be required for $\frac{48}{32} \times 0 .25 g M g = 0 .42 g o f M g$
Thus, Mg will remain in excess
$= 1 - 0 .42 = 0 .58 g$

KCET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307034 250 mg of a compound react with $A g N O_{3}$ produces 141 mg of silver bromide. The percentage of bromine in the compound is:
(at. mass Ag = 108; Br = 80 )

1 48

2 60

3 24

4 36

Explanation:

Mass of substance = 250mg = 0.250g
Mass of AgBr = 141mg = 0.141g
1 mole of AgBr = 1g atom of Br
188 g of AgBr = 80g of Br
$∴$ 188 g of AgBr contain bromine = 80 g
0.141 g of AgBr contain bromine = $\frac{80}{188} \times 0 .141$
This much amount of bromine present in 0.250 g of compound
$∴ % b r o m i n e = \frac{80}{188} \times \frac{0 .414}{0 .250} \times 100 = 24 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307031 2.76 g of silver carbonate on being strongly heated yield a residue weighing

1 2.16 g

2 2.48 g

3 2.64 g

4 2.32 g

Explanation:

(I) Mole method:
$2A g_{2} C O_{3} 4Ag + 2C O_{2} + O_{2}$
$2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$ $2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$
$2 \times 276 g o f A g_{2} C o_{3} g i v e s 4 \times 108 g o f A g$
$2 .76 g A g_{2} C O_{3} w i l l g i v e$
$(\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(II) Principle of Atomic Conservation (POAC) method:
Moles of $A g_{2} C O_{3}$ = Total moles of Ag as product
2 moles of $A g_{2} C O_{3}$ = 4 moles of Ag
$2 \times 276 g$ of $A g_{2} C O_{3}$ = $4 \times 108 g$ of Ag
2.76 g of $A g_{2} C O_{3}$ $= (\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(III) Equivalent Method:
Number of gram - equivalents of $A g_{2} C O_{3}$ = Number of gram-equivalents of Ag
$\frac{2 .76 g}{(\frac{2 \times 276 g / m o l}{1})} = \frac{y g}{(\frac{4 \times 108 g / m o l}{1})}$
$\Rightarrow y = 2 .16 g$
[Change in oxidation state of $A g_{2} C O_{3}$ and Ag is 1, respectively]

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307032 1.12 mL of a gas is produced at STP by the action of 4.12 mg of alcohol, with methyl magnesium iodide. The molecular mass of alcohol is

1 16.0

2 41.2

3 82.4

4 156.0

Explanation:

$ROH + {CH}_{3} MgI \to {CH}_{4} + MgROI$
$∴ 1$ mole alcohol gives 1 mole ${CH}_{4}$ .
According to data,
1.12 L(at STP) ${CH}_{4}$ is obtained from 4.12 g alcohol.
$∴ 22.4 L$ (at STP) ${CH}_{4}$ is obtained from
$= \frac{22.4 \times 4.12}{1.12} = 82.4 g$
So, $82.4 g {mol}^{-}$ is the molar mass of alcohol.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307033 1.0 g of Mg is burnt with 0.28 g of $O_{2}$ in a closed vessel. Which reactant is left in excess and how much?

1 Mg, 5.8 g

2 Mg, 0.58 g

3

O_{2}

, 0.24 g

4

O_{2}

, 2.4 g

Explanation:

Balanced reaction can be given as
$\underset{24 \times 2 = 48}{2 M g} + \underset{32}{O_{2}} \to 2 M g O$
32 g of $O_{2}$ is required to burn 48 g Mg
So, $0 .28 g O_{2}$ will be required for $\frac{48}{32} \times 0 .25 g M g = 0 .42 g o f M g$
Thus, Mg will remain in excess
$= 1 - 0 .42 = 0 .58 g$

KCET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307034 250 mg of a compound react with $A g N O_{3}$ produces 141 mg of silver bromide. The percentage of bromine in the compound is:
(at. mass Ag = 108; Br = 80 )

1 48

2 60

3 24

4 36

Explanation:

Mass of substance = 250mg = 0.250g
Mass of AgBr = 141mg = 0.141g
1 mole of AgBr = 1g atom of Br
188 g of AgBr = 80g of Br
$∴$ 188 g of AgBr contain bromine = 80 g
0.141 g of AgBr contain bromine = $\frac{80}{188} \times 0 .141$
This much amount of bromine present in 0.250 g of compound
$∴ % b r o m i n e = \frac{80}{188} \times \frac{0 .414}{0 .250} \times 100 = 24 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307031 2.76 g of silver carbonate on being strongly heated yield a residue weighing

1 2.16 g

2 2.48 g

3 2.64 g

4 2.32 g

Explanation:

(I) Mole method:
$2A g_{2} C O_{3} 4Ag + 2C O_{2} + O_{2}$
$2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$ $2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$
$2 \times 276 g o f A g_{2} C o_{3} g i v e s 4 \times 108 g o f A g$
$2 .76 g A g_{2} C O_{3} w i l l g i v e$
$(\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(II) Principle of Atomic Conservation (POAC) method:
Moles of $A g_{2} C O_{3}$ = Total moles of Ag as product
2 moles of $A g_{2} C O_{3}$ = 4 moles of Ag
$2 \times 276 g$ of $A g_{2} C O_{3}$ = $4 \times 108 g$ of Ag
2.76 g of $A g_{2} C O_{3}$ $= (\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(III) Equivalent Method:
Number of gram - equivalents of $A g_{2} C O_{3}$ = Number of gram-equivalents of Ag
$\frac{2 .76 g}{(\frac{2 \times 276 g / m o l}{1})} = \frac{y g}{(\frac{4 \times 108 g / m o l}{1})}$
$\Rightarrow y = 2 .16 g$
[Change in oxidation state of $A g_{2} C O_{3}$ and Ag is 1, respectively]

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307032 1.12 mL of a gas is produced at STP by the action of 4.12 mg of alcohol, with methyl magnesium iodide. The molecular mass of alcohol is

1 16.0

2 41.2

3 82.4

4 156.0

Explanation:

$ROH + {CH}_{3} MgI \to {CH}_{4} + MgROI$
$∴ 1$ mole alcohol gives 1 mole ${CH}_{4}$ .
According to data,
1.12 L(at STP) ${CH}_{4}$ is obtained from 4.12 g alcohol.
$∴ 22.4 L$ (at STP) ${CH}_{4}$ is obtained from
$= \frac{22.4 \times 4.12}{1.12} = 82.4 g$
So, $82.4 g {mol}^{-}$ is the molar mass of alcohol.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307033 1.0 g of Mg is burnt with 0.28 g of $O_{2}$ in a closed vessel. Which reactant is left in excess and how much?

1 Mg, 5.8 g

2 Mg, 0.58 g

3

O_{2}

, 0.24 g

4

O_{2}

, 2.4 g

Explanation:

Balanced reaction can be given as
$\underset{24 \times 2 = 48}{2 M g} + \underset{32}{O_{2}} \to 2 M g O$
32 g of $O_{2}$ is required to burn 48 g Mg
So, $0 .28 g O_{2}$ will be required for $\frac{48}{32} \times 0 .25 g M g = 0 .42 g o f M g$
Thus, Mg will remain in excess
$= 1 - 0 .42 = 0 .58 g$

KCET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307034 250 mg of a compound react with $A g N O_{3}$ produces 141 mg of silver bromide. The percentage of bromine in the compound is:
(at. mass Ag = 108; Br = 80 )

1 48

2 60

3 24

4 36

Explanation:

Mass of substance = 250mg = 0.250g
Mass of AgBr = 141mg = 0.141g
1 mole of AgBr = 1g atom of Br
188 g of AgBr = 80g of Br
$∴$ 188 g of AgBr contain bromine = 80 g
0.141 g of AgBr contain bromine = $\frac{80}{188} \times 0 .141$
This much amount of bromine present in 0.250 g of compound
$∴ % b r o m i n e = \frac{80}{188} \times \frac{0 .414}{0 .250} \times 100 = 24 %$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307031 2.76 g of silver carbonate on being strongly heated yield a residue weighing

1 2.16 g

2 2.48 g

3 2.64 g

4 2.32 g

Explanation:

(I) Mole method:
$2A g_{2} C O_{3} 4Ag + 2C O_{2} + O_{2}$
$2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$ $2 m o l 4 m o l$
$2 \times 276 g 4 \times 108 g$
$2 \times 276 g o f A g_{2} C o_{3} g i v e s 4 \times 108 g o f A g$
$2 .76 g A g_{2} C O_{3} w i l l g i v e$
$(\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(II) Principle of Atomic Conservation (POAC) method:
Moles of $A g_{2} C O_{3}$ = Total moles of Ag as product
2 moles of $A g_{2} C O_{3}$ = 4 moles of Ag
$2 \times 276 g$ of $A g_{2} C O_{3}$ = $4 \times 108 g$ of Ag
2.76 g of $A g_{2} C O_{3}$ $= (\frac{4 \times 108 \times 2 .76}{2 \times 276}) = 2 .16 g o f A g .$
(III) Equivalent Method:
Number of gram - equivalents of $A g_{2} C O_{3}$ = Number of gram-equivalents of Ag
$\frac{2 .76 g}{(\frac{2 \times 276 g / m o l}{1})} = \frac{y g}{(\frac{4 \times 108 g / m o l}{1})}$
$\Rightarrow y = 2 .16 g$
[Change in oxidation state of $A g_{2} C O_{3}$ and Ag is 1, respectively]

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307032 1.12 mL of a gas is produced at STP by the action of 4.12 mg of alcohol, with methyl magnesium iodide. The molecular mass of alcohol is

1 16.0

2 41.2

3 82.4

4 156.0

Explanation:

$ROH + {CH}_{3} MgI \to {CH}_{4} + MgROI$
$∴ 1$ mole alcohol gives 1 mole ${CH}_{4}$ .
According to data,
1.12 L(at STP) ${CH}_{4}$ is obtained from 4.12 g alcohol.
$∴ 22.4 L$ (at STP) ${CH}_{4}$ is obtained from
$= \frac{22.4 \times 4.12}{1.12} = 82.4 g$
So, $82.4 g {mol}^{-}$ is the molar mass of alcohol.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307033 1.0 g of Mg is burnt with 0.28 g of $O_{2}$ in a closed vessel. Which reactant is left in excess and how much?

1 Mg, 5.8 g

2 Mg, 0.58 g

3

O_{2}

, 0.24 g

4

O_{2}

, 2.4 g

Explanation:

Balanced reaction can be given as
$\underset{24 \times 2 = 48}{2 M g} + \underset{32}{O_{2}} \to 2 M g O$
32 g of $O_{2}$ is required to burn 48 g Mg
So, $0 .28 g O_{2}$ will be required for $\frac{48}{32} \times 0 .25 g M g = 0 .42 g o f M g$
Thus, Mg will remain in excess
$= 1 - 0 .42 = 0 .58 g$

KCET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307034 250 mg of a compound react with $A g N O_{3}$ produces 141 mg of silver bromide. The percentage of bromine in the compound is:
(at. mass Ag = 108; Br = 80 )

1 48

2 60

3 24

4 36

Explanation:

Mass of substance = 250mg = 0.250g
Mass of AgBr = 141mg = 0.141g
1 mole of AgBr = 1g atom of Br
188 g of AgBr = 80g of Br
$∴$ 188 g of AgBr contain bromine = 80 g
0.141 g of AgBr contain bromine = $\frac{80}{188} \times 0 .141$
This much amount of bromine present in 0.250 g of compound
$∴ % b r o m i n e = \frac{80}{188} \times \frac{0 .414}{0 .250} \times 100 = 24 %$

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