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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306846 The equivalent weight of $H_{2} O_{2}$ in
$2 {MnO}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 {Mn}^{2 +} + 5 O_{2} + 8 H_{2} O$ is

1 17

2 34

3 68

4 85

Explanation:

${2 M n O}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$
$H_{2} \overset{- 1}{O_{2}} \to 2 H_{2} \overset{- 2}{O_{2}} + 2 H^{+}$
Total change in oxidation number
$= - 4 - (- 2) = - 2$
The equivalent of $H_{2} O_{2}$
$= \frac{M . w t o f H_{2} O_{2}}{C h a n g e i n o x i d a t i o n n u m b e r} = \frac{34}{2} = 17$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306847 If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of $A g N O_{3}$ , yields 0.50 g of AgCl, the equivalent mass of the metal is

1 20.04

2

43 .80

3 40.08

4

21 .90

Explanation:

$\frac{E q u i v a l e n t m a s s o f m e t a l + E q u i v a l e n t m a s s o f C l^{-}}{E q u i v a l e n t m a s s o f A g + E q u i v a l e n t m a s s o f C l^{-}}$
$= \frac{0 .2}{0 .5} \Rightarrow \frac{E + 35 .5}{108 + 35 .5} = \frac{0 .2}{0 .5} \Rightarrow E = 21 .90$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306848 $50 c m^{3} o f 0 .04 M K_{2} C r_{2} O_{7}$ in acidic medium oxidizes a sample of $H_{2} S$ gas to sulphur. Volume of $0 .03 M K M n O_{4}$ required to oxidize the same amount of $H_{2} S$ gas to sulphur, in acidic medium is

1

60 c m^{3}

2

80 c m^{3}

3

90 c m^{3}

4

120 c m^{3}

Explanation:

$K_{2} C r_{2} O_{7} a n d K M n O_{4}$
both are good oxidizing agents in acidic medium. The reaction is as follows:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Normality of $C r_{2} O_{7}^{2 -} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .04 \times 6 = 0 .24 N$
Normality of ${M n O}_{4}^{-} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .03 \times 5 = 0 .15 N$
$N_{1} V_{1} = N_{2} V_{2}$
$(K_{2} C r_{2} O_{7}) (K M n O_{4})$
$\Rightarrow 0 .24 \times 50 = 0 .15 \times V_{2}$
$V_{2} = \frac{0 .24 \times 50}{0 .15} = 80 c m^{3}$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306849 Assertion :
Equivalent weight of $Cu$ in $CuO$ is 31.8 and in ${Cu}_{2} O$ is 63.6.
Reason :
Equivalent weight of an element
$= \frac{Atomic weight}{x - factor}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The equivalent weight (eq. wt.) of copper $(Cu)$ in copper oxide $(CuO)$ is calculated as follows:The atomic weight (At.wt) of copper $(Cu)$ is $63.6 g / mol$ , and its valency is 2 .
Equivalent weight of $Cu$ in $CuO = \frac{63.6}{2} = 31.8$
Equivalent weight of $Cu$ in ${Cu}_{2} O = \frac{63.6}{1} = 63.6$
So, the option (1) is correct

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306846 The equivalent weight of $H_{2} O_{2}$ in
$2 {MnO}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 {Mn}^{2 +} + 5 O_{2} + 8 H_{2} O$ is

1 17

2 34

3 68

4 85

Explanation:

${2 M n O}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$
$H_{2} \overset{- 1}{O_{2}} \to 2 H_{2} \overset{- 2}{O_{2}} + 2 H^{+}$
Total change in oxidation number
$= - 4 - (- 2) = - 2$
The equivalent of $H_{2} O_{2}$
$= \frac{M . w t o f H_{2} O_{2}}{C h a n g e i n o x i d a t i o n n u m b e r} = \frac{34}{2} = 17$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306847 If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of $A g N O_{3}$ , yields 0.50 g of AgCl, the equivalent mass of the metal is

1 20.04

2

43 .80

3 40.08

4

21 .90

Explanation:

$\frac{E q u i v a l e n t m a s s o f m e t a l + E q u i v a l e n t m a s s o f C l^{-}}{E q u i v a l e n t m a s s o f A g + E q u i v a l e n t m a s s o f C l^{-}}$
$= \frac{0 .2}{0 .5} \Rightarrow \frac{E + 35 .5}{108 + 35 .5} = \frac{0 .2}{0 .5} \Rightarrow E = 21 .90$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306848 $50 c m^{3} o f 0 .04 M K_{2} C r_{2} O_{7}$ in acidic medium oxidizes a sample of $H_{2} S$ gas to sulphur. Volume of $0 .03 M K M n O_{4}$ required to oxidize the same amount of $H_{2} S$ gas to sulphur, in acidic medium is

1

60 c m^{3}

2

80 c m^{3}

3

90 c m^{3}

4

120 c m^{3}

Explanation:

$K_{2} C r_{2} O_{7} a n d K M n O_{4}$
both are good oxidizing agents in acidic medium. The reaction is as follows:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Normality of $C r_{2} O_{7}^{2 -} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .04 \times 6 = 0 .24 N$
Normality of ${M n O}_{4}^{-} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .03 \times 5 = 0 .15 N$
$N_{1} V_{1} = N_{2} V_{2}$
$(K_{2} C r_{2} O_{7}) (K M n O_{4})$
$\Rightarrow 0 .24 \times 50 = 0 .15 \times V_{2}$
$V_{2} = \frac{0 .24 \times 50}{0 .15} = 80 c m^{3}$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306849 Assertion :
Equivalent weight of $Cu$ in $CuO$ is 31.8 and in ${Cu}_{2} O$ is 63.6.
Reason :
Equivalent weight of an element
$= \frac{Atomic weight}{x - factor}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The equivalent weight (eq. wt.) of copper $(Cu)$ in copper oxide $(CuO)$ is calculated as follows:The atomic weight (At.wt) of copper $(Cu)$ is $63.6 g / mol$ , and its valency is 2 .
Equivalent weight of $Cu$ in $CuO = \frac{63.6}{2} = 31.8$
Equivalent weight of $Cu$ in ${Cu}_{2} O = \frac{63.6}{1} = 63.6$
So, the option (1) is correct

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306846 The equivalent weight of $H_{2} O_{2}$ in
$2 {MnO}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 {Mn}^{2 +} + 5 O_{2} + 8 H_{2} O$ is

1 17

2 34

3 68

4 85

Explanation:

${2 M n O}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$
$H_{2} \overset{- 1}{O_{2}} \to 2 H_{2} \overset{- 2}{O_{2}} + 2 H^{+}$
Total change in oxidation number
$= - 4 - (- 2) = - 2$
The equivalent of $H_{2} O_{2}$
$= \frac{M . w t o f H_{2} O_{2}}{C h a n g e i n o x i d a t i o n n u m b e r} = \frac{34}{2} = 17$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306847 If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of $A g N O_{3}$ , yields 0.50 g of AgCl, the equivalent mass of the metal is

1 20.04

2

43 .80

3 40.08

4

21 .90

Explanation:

$\frac{E q u i v a l e n t m a s s o f m e t a l + E q u i v a l e n t m a s s o f C l^{-}}{E q u i v a l e n t m a s s o f A g + E q u i v a l e n t m a s s o f C l^{-}}$
$= \frac{0 .2}{0 .5} \Rightarrow \frac{E + 35 .5}{108 + 35 .5} = \frac{0 .2}{0 .5} \Rightarrow E = 21 .90$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306848 $50 c m^{3} o f 0 .04 M K_{2} C r_{2} O_{7}$ in acidic medium oxidizes a sample of $H_{2} S$ gas to sulphur. Volume of $0 .03 M K M n O_{4}$ required to oxidize the same amount of $H_{2} S$ gas to sulphur, in acidic medium is

1

60 c m^{3}

2

80 c m^{3}

3

90 c m^{3}

4

120 c m^{3}

Explanation:

$K_{2} C r_{2} O_{7} a n d K M n O_{4}$
both are good oxidizing agents in acidic medium. The reaction is as follows:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Normality of $C r_{2} O_{7}^{2 -} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .04 \times 6 = 0 .24 N$
Normality of ${M n O}_{4}^{-} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .03 \times 5 = 0 .15 N$
$N_{1} V_{1} = N_{2} V_{2}$
$(K_{2} C r_{2} O_{7}) (K M n O_{4})$
$\Rightarrow 0 .24 \times 50 = 0 .15 \times V_{2}$
$V_{2} = \frac{0 .24 \times 50}{0 .15} = 80 c m^{3}$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306849 Assertion :
Equivalent weight of $Cu$ in $CuO$ is 31.8 and in ${Cu}_{2} O$ is 63.6.
Reason :
Equivalent weight of an element
$= \frac{Atomic weight}{x - factor}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The equivalent weight (eq. wt.) of copper $(Cu)$ in copper oxide $(CuO)$ is calculated as follows:The atomic weight (At.wt) of copper $(Cu)$ is $63.6 g / mol$ , and its valency is 2 .
Equivalent weight of $Cu$ in $CuO = \frac{63.6}{2} = 31.8$
Equivalent weight of $Cu$ in ${Cu}_{2} O = \frac{63.6}{1} = 63.6$
So, the option (1) is correct

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306846 The equivalent weight of $H_{2} O_{2}$ in
$2 {MnO}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 {Mn}^{2 +} + 5 O_{2} + 8 H_{2} O$ is

1 17

2 34

3 68

4 85

Explanation:

${2 M n O}_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} \to 2 M n^{2 +} + 5 O_{2} + 8 H_{2} O$
$H_{2} \overset{- 1}{O_{2}} \to 2 H_{2} \overset{- 2}{O_{2}} + 2 H^{+}$
Total change in oxidation number
$= - 4 - (- 2) = - 2$
The equivalent of $H_{2} O_{2}$
$= \frac{M . w t o f H_{2} O_{2}}{C h a n g e i n o x i d a t i o n n u m b e r} = \frac{34}{2} = 17$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306847 If 0.20 g chloride of a certain metal, when dissolved in water and treated with excess of $A g N O_{3}$ , yields 0.50 g of AgCl, the equivalent mass of the metal is

1 20.04

2

43 .80

3 40.08

4

21 .90

Explanation:

$\frac{E q u i v a l e n t m a s s o f m e t a l + E q u i v a l e n t m a s s o f C l^{-}}{E q u i v a l e n t m a s s o f A g + E q u i v a l e n t m a s s o f C l^{-}}$
$= \frac{0 .2}{0 .5} \Rightarrow \frac{E + 35 .5}{108 + 35 .5} = \frac{0 .2}{0 .5} \Rightarrow E = 21 .90$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306848 $50 c m^{3} o f 0 .04 M K_{2} C r_{2} O_{7}$ in acidic medium oxidizes a sample of $H_{2} S$ gas to sulphur. Volume of $0 .03 M K M n O_{4}$ required to oxidize the same amount of $H_{2} S$ gas to sulphur, in acidic medium is

1

60 c m^{3}

2

80 c m^{3}

3

90 c m^{3}

4

120 c m^{3}

Explanation:

$K_{2} C r_{2} O_{7} a n d K M n O_{4}$
both are good oxidizing agents in acidic medium. The reaction is as follows:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Normality of $C r_{2} O_{7}^{2 -} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .04 \times 6 = 0 .24 N$
Normality of ${M n O}_{4}^{-} = M o l a r i t y \times n o . o f e l e c t r o n s i n v o l v e d$
$= 0 .03 \times 5 = 0 .15 N$
$N_{1} V_{1} = N_{2} V_{2}$
$(K_{2} C r_{2} O_{7}) (K M n O_{4})$
$\Rightarrow 0 .24 \times 50 = 0 .15 \times V_{2}$
$V_{2} = \frac{0 .24 \times 50}{0 .15} = 80 c m^{3}$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306849 Assertion :
Equivalent weight of $Cu$ in $CuO$ is 31.8 and in ${Cu}_{2} O$ is 63.6.
Reason :
Equivalent weight of an element
$= \frac{Atomic weight}{x - factor}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

The equivalent weight (eq. wt.) of copper $(Cu)$ in copper oxide $(CuO)$ is calculated as follows:The atomic weight (At.wt) of copper $(Cu)$ is $63.6 g / mol$ , and its valency is 2 .
Equivalent weight of $Cu$ in $CuO = \frac{63.6}{2} = 31.8$
Equivalent weight of $Cu$ in ${Cu}_{2} O = \frac{63.6}{1} = 63.6$
So, the option (1) is correct