QBManager

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306862 Equivalent wt. of $H_{3} P O_{4}$ in each of the reaction will be respectively [Molar mass of $H_{3} P O_{4}$ : 98 g/mol]
$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O$

1 98, 49, 32.67

2 49, 98, 32.67

3 98, 32.67, 49

4 32.67, 49, 98

Explanation:

$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O;$
$E W = \frac{M W}{1} = 98$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O;$
$E W = \frac{M W}{2} = 49$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O;$
$E W = \frac{M W}{3} = 32.67$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306863 One gram of a metal chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. The molecular formula of metal chloride is

1

M C l

2

M C l_{2}

3

M C l_{3}

4

M C l_{4}

Explanation:

Mass of metal chloride = 1g
Mass of chlorine = 0.835 g
Mass of metal = (1 – 0.835) = 0.165 g
Equivalent mass of metal $= \frac{0 .165 \times 35 .5}{0 .835} = 7 .01$
Valency of the metal
$= \frac{2 V . D}{E + 35 .5} = \frac{2 \times 85}{7 .01 + 35 .5} = 4$
Formula of the chloride $= M C l_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306864 In the reaction
$B r_{2} + N a_{2} C O_{3} \to N a B r + N a B r O_{3} + C O_{2}$
The equivalent weight of $N a B r O_{3}$ is

1

\frac{M o l . w t}{1}

2

\frac{M o l . w t}{10}

3

\frac{M o l . w t}{5}

4

\frac{M o l . w t}{4}

Explanation:

Valency factor for formation of $N a B r O_{3} = 5$
$∴$ Equivalent weight of $N a B r O_{3} = \frac{M o l . w t}{5}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306865 3 g of activated charcoal was added to 50 mL of acetic acid solution 0.06N in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

1 42 g

2 54 g

3 18 mg

4 36 mg

Explanation:

Let the weight of acetic acid initially be $w_{1}$ in 50 mL of 0.060 N solution.
Let the $N = \frac{w_{1} \times 1000}{M . w t . \times 50}$
$(N o r m a l i t y = 0 .06 N)$
$0 .06 = \frac{w_{1} \times 1000}{60 \times 50}$
$\Rightarrow w_{1} = \frac{0 .06 \times 60 \times 50}{1000} = 0 .18 g = 180 m g$
After an hour, the strength of acetic acid = 0.042 N
so, let the weight of acetic acid be $w_{2}$
$N = \frac{w_{2} \times 1000}{60 \times 50}$
$0 .042 = \frac{w_{2} \times 1000}{3000}$
$\Rightarrow w_{2} = 0 .126 g = 126 m g$
So amount of acetic acid adsorbed per 3 g = 180 - 126 mg = 54 mg
$∴$ Amount of acetic acid absorbed per
$g = \frac{54}{3} = 18 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306862 Equivalent wt. of $H_{3} P O_{4}$ in each of the reaction will be respectively [Molar mass of $H_{3} P O_{4}$ : 98 g/mol]
$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O$

1 98, 49, 32.67

2 49, 98, 32.67

3 98, 32.67, 49

4 32.67, 49, 98

Explanation:

$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O;$
$E W = \frac{M W}{1} = 98$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O;$
$E W = \frac{M W}{2} = 49$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O;$
$E W = \frac{M W}{3} = 32.67$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306863 One gram of a metal chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. The molecular formula of metal chloride is

1

M C l

2

M C l_{2}

3

M C l_{3}

4

M C l_{4}

Explanation:

Mass of metal chloride = 1g
Mass of chlorine = 0.835 g
Mass of metal = (1 – 0.835) = 0.165 g
Equivalent mass of metal $= \frac{0 .165 \times 35 .5}{0 .835} = 7 .01$
Valency of the metal
$= \frac{2 V . D}{E + 35 .5} = \frac{2 \times 85}{7 .01 + 35 .5} = 4$
Formula of the chloride $= M C l_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306864 In the reaction
$B r_{2} + N a_{2} C O_{3} \to N a B r + N a B r O_{3} + C O_{2}$
The equivalent weight of $N a B r O_{3}$ is

1

\frac{M o l . w t}{1}

2

\frac{M o l . w t}{10}

3

\frac{M o l . w t}{5}

4

\frac{M o l . w t}{4}

Explanation:

Valency factor for formation of $N a B r O_{3} = 5$
$∴$ Equivalent weight of $N a B r O_{3} = \frac{M o l . w t}{5}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306865 3 g of activated charcoal was added to 50 mL of acetic acid solution 0.06N in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

1 42 g

2 54 g

3 18 mg

4 36 mg

Explanation:

Let the weight of acetic acid initially be $w_{1}$ in 50 mL of 0.060 N solution.
Let the $N = \frac{w_{1} \times 1000}{M . w t . \times 50}$
$(N o r m a l i t y = 0 .06 N)$
$0 .06 = \frac{w_{1} \times 1000}{60 \times 50}$
$\Rightarrow w_{1} = \frac{0 .06 \times 60 \times 50}{1000} = 0 .18 g = 180 m g$
After an hour, the strength of acetic acid = 0.042 N
so, let the weight of acetic acid be $w_{2}$
$N = \frac{w_{2} \times 1000}{60 \times 50}$
$0 .042 = \frac{w_{2} \times 1000}{3000}$
$\Rightarrow w_{2} = 0 .126 g = 126 m g$
So amount of acetic acid adsorbed per 3 g = 180 - 126 mg = 54 mg
$∴$ Amount of acetic acid absorbed per
$g = \frac{54}{3} = 18 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306862 Equivalent wt. of $H_{3} P O_{4}$ in each of the reaction will be respectively [Molar mass of $H_{3} P O_{4}$ : 98 g/mol]
$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O$

1 98, 49, 32.67

2 49, 98, 32.67

3 98, 32.67, 49

4 32.67, 49, 98

Explanation:

$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O;$
$E W = \frac{M W}{1} = 98$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O;$
$E W = \frac{M W}{2} = 49$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O;$
$E W = \frac{M W}{3} = 32.67$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306863 One gram of a metal chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. The molecular formula of metal chloride is

1

M C l

2

M C l_{2}

3

M C l_{3}

4

M C l_{4}

Explanation:

Mass of metal chloride = 1g
Mass of chlorine = 0.835 g
Mass of metal = (1 – 0.835) = 0.165 g
Equivalent mass of metal $= \frac{0 .165 \times 35 .5}{0 .835} = 7 .01$
Valency of the metal
$= \frac{2 V . D}{E + 35 .5} = \frac{2 \times 85}{7 .01 + 35 .5} = 4$
Formula of the chloride $= M C l_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306864 In the reaction
$B r_{2} + N a_{2} C O_{3} \to N a B r + N a B r O_{3} + C O_{2}$
The equivalent weight of $N a B r O_{3}$ is

1

\frac{M o l . w t}{1}

2

\frac{M o l . w t}{10}

3

\frac{M o l . w t}{5}

4

\frac{M o l . w t}{4}

Explanation:

Valency factor for formation of $N a B r O_{3} = 5$
$∴$ Equivalent weight of $N a B r O_{3} = \frac{M o l . w t}{5}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306865 3 g of activated charcoal was added to 50 mL of acetic acid solution 0.06N in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

1 42 g

2 54 g

3 18 mg

4 36 mg

Explanation:

Let the weight of acetic acid initially be $w_{1}$ in 50 mL of 0.060 N solution.
Let the $N = \frac{w_{1} \times 1000}{M . w t . \times 50}$
$(N o r m a l i t y = 0 .06 N)$
$0 .06 = \frac{w_{1} \times 1000}{60 \times 50}$
$\Rightarrow w_{1} = \frac{0 .06 \times 60 \times 50}{1000} = 0 .18 g = 180 m g$
After an hour, the strength of acetic acid = 0.042 N
so, let the weight of acetic acid be $w_{2}$
$N = \frac{w_{2} \times 1000}{60 \times 50}$
$0 .042 = \frac{w_{2} \times 1000}{3000}$
$\Rightarrow w_{2} = 0 .126 g = 126 m g$
So amount of acetic acid adsorbed per 3 g = 180 - 126 mg = 54 mg
$∴$ Amount of acetic acid absorbed per
$g = \frac{54}{3} = 18 g$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306862 Equivalent wt. of $H_{3} P O_{4}$ in each of the reaction will be respectively [Molar mass of $H_{3} P O_{4}$ : 98 g/mol]
$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O$

1 98, 49, 32.67

2 49, 98, 32.67

3 98, 32.67, 49

4 32.67, 49, 98

Explanation:

$H_{3} P O_{4} + O H^{-} \to H_{2} {P O}_{4}^{-} + H_{2} O;$
$E W = \frac{M W}{1} = 98$
$H_{3} P O_{4} + 2 O H^{-} \to {H P O}_{4}^{2 -} + 2 H_{2} O;$
$E W = \frac{M W}{2} = 49$
$H_{3} P O_{4} + 3 O H^{-} \to {P O}_{4}^{3 -} + 3 H_{2} O;$
$E W = \frac{M W}{3} = 32.67$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306863 One gram of a metal chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. The molecular formula of metal chloride is

1

M C l

2

M C l_{2}

3

M C l_{3}

4

M C l_{4}

Explanation:

Mass of metal chloride = 1g
Mass of chlorine = 0.835 g
Mass of metal = (1 – 0.835) = 0.165 g
Equivalent mass of metal $= \frac{0 .165 \times 35 .5}{0 .835} = 7 .01$
Valency of the metal
$= \frac{2 V . D}{E + 35 .5} = \frac{2 \times 85}{7 .01 + 35 .5} = 4$
Formula of the chloride $= M C l_{4}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306864 In the reaction
$B r_{2} + N a_{2} C O_{3} \to N a B r + N a B r O_{3} + C O_{2}$
The equivalent weight of $N a B r O_{3}$ is

1

\frac{M o l . w t}{1}

2

\frac{M o l . w t}{10}

3

\frac{M o l . w t}{5}

4

\frac{M o l . w t}{4}

Explanation:

Valency factor for formation of $N a B r O_{3} = 5$
$∴$ Equivalent weight of $N a B r O_{3} = \frac{M o l . w t}{5}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306865 3 g of activated charcoal was added to 50 mL of acetic acid solution 0.06N in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

1 42 g

2 54 g

3 18 mg

4 36 mg

Explanation:

Let the weight of acetic acid initially be $w_{1}$ in 50 mL of 0.060 N solution.
Let the $N = \frac{w_{1} \times 1000}{M . w t . \times 50}$
$(N o r m a l i t y = 0 .06 N)$
$0 .06 = \frac{w_{1} \times 1000}{60 \times 50}$
$\Rightarrow w_{1} = \frac{0 .06 \times 60 \times 50}{1000} = 0 .18 g = 180 m g$
After an hour, the strength of acetic acid = 0.042 N
so, let the weight of acetic acid be $w_{2}$
$N = \frac{w_{2} \times 1000}{60 \times 50}$
$0 .042 = \frac{w_{2} \times 1000}{3000}$
$\Rightarrow w_{2} = 0 .126 g = 126 m g$
So amount of acetic acid adsorbed per 3 g = 180 - 126 mg = 54 mg
$∴$ Amount of acetic acid absorbed per
$g = \frac{54}{3} = 18 g$