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PRACTICAL CHEMISTRY

37477 When $100 m l$ $N - N a O H$ solution and $10 m l$ of $10 N$ sulphuric acid solution are mixed together, the resulting solution will be

1 Alkali

2 Weakly acidic

3 Strongly acidic

4 Neutral

Explanation:

(d) If $N_{1} V_{1} = N_{2} V_{2}$ then the solution will be neutral
$∴$ $1 \times 100 = 10 \times 10$
$100 = 100$ $\Rightarrow$ solution in neutral

PRACTICAL CHEMISTRY

37478 Cosin used to detect end point of precipitation titration by adsorption is called

1 Absorption indicator

2 Adsorption indicator

3 Chemical indicator

4 Normal indicator

PRACTICAL CHEMISTRY

37479 $0.1914 g$ of an organic acid is dissolved in approx. $20 m l$ of water. $25 m l$ of $0.12 N$ $N a O H$ required for the complete neutralization of the acid solution. The equivalent weight of the acid is

1

65

2

64

3

63.8

4

62.5

Explanation:

(c) $V o l u m e = 25 m l = \frac{25}{1000} l i t r e$
$N o r m a l i t y = \frac{w t}{e q . w t \times V o l u m e}$ $\Rightarrow$ $0.12 = \frac{0.1914 \times 1000}{E \times 25}$
$e q . w t . = \frac{0.1914 \times 1000}{0.12 \times 25} = 63.8$

PRACTICAL CHEMISTRY

37480 Molecular weight of a tribasic acid is $W$ , its equivalent weight is

1

2 W

2

W / 3

3

3 W

4

W - 3

Explanation:

(b) $E q . w t = \frac{M o l e c u l a r W t .}{b a s i c i t y} = \frac{W}{3}$

PRACTICAL CHEMISTRY

37477 When $100 m l$ $N - N a O H$ solution and $10 m l$ of $10 N$ sulphuric acid solution are mixed together, the resulting solution will be

1 Alkali

2 Weakly acidic

3 Strongly acidic

4 Neutral

Explanation:

(d) If $N_{1} V_{1} = N_{2} V_{2}$ then the solution will be neutral
$∴$ $1 \times 100 = 10 \times 10$
$100 = 100$ $\Rightarrow$ solution in neutral

PRACTICAL CHEMISTRY

37478 Cosin used to detect end point of precipitation titration by adsorption is called

1 Absorption indicator

2 Adsorption indicator

3 Chemical indicator

4 Normal indicator

PRACTICAL CHEMISTRY

37479 $0.1914 g$ of an organic acid is dissolved in approx. $20 m l$ of water. $25 m l$ of $0.12 N$ $N a O H$ required for the complete neutralization of the acid solution. The equivalent weight of the acid is

1

65

2

64

3

63.8

4

62.5

Explanation:

(c) $V o l u m e = 25 m l = \frac{25}{1000} l i t r e$
$N o r m a l i t y = \frac{w t}{e q . w t \times V o l u m e}$ $\Rightarrow$ $0.12 = \frac{0.1914 \times 1000}{E \times 25}$
$e q . w t . = \frac{0.1914 \times 1000}{0.12 \times 25} = 63.8$

PRACTICAL CHEMISTRY

37480 Molecular weight of a tribasic acid is $W$ , its equivalent weight is

1

2 W

2

W / 3

3

3 W

4

W - 3

Explanation:

(b) $E q . w t = \frac{M o l e c u l a r W t .}{b a s i c i t y} = \frac{W}{3}$

PRACTICAL CHEMISTRY

37477 When $100 m l$ $N - N a O H$ solution and $10 m l$ of $10 N$ sulphuric acid solution are mixed together, the resulting solution will be

1 Alkali

2 Weakly acidic

3 Strongly acidic

4 Neutral

Explanation:

(d) If $N_{1} V_{1} = N_{2} V_{2}$ then the solution will be neutral
$∴$ $1 \times 100 = 10 \times 10$
$100 = 100$ $\Rightarrow$ solution in neutral

PRACTICAL CHEMISTRY

37478 Cosin used to detect end point of precipitation titration by adsorption is called

1 Absorption indicator

2 Adsorption indicator

3 Chemical indicator

4 Normal indicator

PRACTICAL CHEMISTRY

37479 $0.1914 g$ of an organic acid is dissolved in approx. $20 m l$ of water. $25 m l$ of $0.12 N$ $N a O H$ required for the complete neutralization of the acid solution. The equivalent weight of the acid is

1

65

2

64

3

63.8

4

62.5

Explanation:

(c) $V o l u m e = 25 m l = \frac{25}{1000} l i t r e$
$N o r m a l i t y = \frac{w t}{e q . w t \times V o l u m e}$ $\Rightarrow$ $0.12 = \frac{0.1914 \times 1000}{E \times 25}$
$e q . w t . = \frac{0.1914 \times 1000}{0.12 \times 25} = 63.8$

PRACTICAL CHEMISTRY

37480 Molecular weight of a tribasic acid is $W$ , its equivalent weight is

1

2 W

2

W / 3

3

3 W

4

W - 3

Explanation:

(b) $E q . w t = \frac{M o l e c u l a r W t .}{b a s i c i t y} = \frac{W}{3}$

PRACTICAL CHEMISTRY

37477 When $100 m l$ $N - N a O H$ solution and $10 m l$ of $10 N$ sulphuric acid solution are mixed together, the resulting solution will be

1 Alkali

2 Weakly acidic

3 Strongly acidic

4 Neutral

Explanation:

(d) If $N_{1} V_{1} = N_{2} V_{2}$ then the solution will be neutral
$∴$ $1 \times 100 = 10 \times 10$
$100 = 100$ $\Rightarrow$ solution in neutral

PRACTICAL CHEMISTRY

37478 Cosin used to detect end point of precipitation titration by adsorption is called

1 Absorption indicator

2 Adsorption indicator

3 Chemical indicator

4 Normal indicator

PRACTICAL CHEMISTRY

37479 $0.1914 g$ of an organic acid is dissolved in approx. $20 m l$ of water. $25 m l$ of $0.12 N$ $N a O H$ required for the complete neutralization of the acid solution. The equivalent weight of the acid is

1

65

2

64

3

63.8

4

62.5

Explanation:

(c) $V o l u m e = 25 m l = \frac{25}{1000} l i t r e$
$N o r m a l i t y = \frac{w t}{e q . w t \times V o l u m e}$ $\Rightarrow$ $0.12 = \frac{0.1914 \times 1000}{E \times 25}$
$e q . w t . = \frac{0.1914 \times 1000}{0.12 \times 25} = 63.8$

PRACTICAL CHEMISTRY

37480 Molecular weight of a tribasic acid is $W$ , its equivalent weight is

1

2 W

2

W / 3

3

3 W

4

W - 3

Explanation:

(b) $E q . w t = \frac{M o l e c u l a r W t .}{b a s i c i t y} = \frac{W}{3}$