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PRACTICAL CHEMISTRY

37497 To what extent must a given solution of concentration of $40 m g$ silver nitrate per $m l$ be diluted to yield a solution of concentration of $16$ $m g$ $A g N O_{3}$ per $m l$

1 Each ml must be diluted to

2.5 m l

2 To each ml of solution

2.5 m l

of water should be added

3 To

2.5 m l

of solution

2 m l

of water should be added

4 To

1.5 m l

of solution

1.5 m l

of water should be added

Explanation:

(a) $40 \times 1 = 16 \times x$ $\Rightarrow$ $x = \frac{40}{16} = 2.5 m l$

PRACTICAL CHEMISTRY

37498 $10 m l$ of conc. $H_{2} S O_{4}$ ( $18$ molar) is diluted to $1$ litre. The approximate strength of dilute acid could be..... $N$

1

0.18

2

0.36

3

0.09

4

18

Explanation:

(b) $18 M$ of $H_{2} S O_{4} = 36 N o f H_{2} S O_{4}$
$N_{1} V_{1} = N_{2} V_{2}$
$36 \times 10$ = $N_{2} \times 1000$
$∴ N_{2} = 0.36 N$

PRACTICAL CHEMISTRY

37499 For preparing one litre $N / 10$ solution of $H_{2} S O_{4},$ we need $H_{2} S O_{4}$ ....... $g m s$

1

98

2

10

3

100

4

4.9

Explanation:

(d) $\frac{1}{10} = \frac{W \times 1000}{e q . w t \times v o l .} = \frac{W \times 1000}{49 \times 1000}$ $\Rightarrow$ $W = 4.9 g m s$

PRACTICAL CHEMISTRY

37500 What is the concentration of nitrate ions if equal volumes of $0.1 M$ $A g N O_{3}$ and $0.1 M$ $N a C l$ are mixed together

1

0.1 N

2

0.2 M

3

0.05 M

4

0.25 M

Explanation:

(c) $M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times V = M_{2} \times 2 V$ $\Rightarrow$ $M_{2} = \frac{1 \times V}{2 V} = 0.05 M$

PRACTICAL CHEMISTRY

37501 To change the strength of $25 m l$ of $0.15 M$ $H C l$ to $0.1 M$ the quantity of water that is to be added is....... $m l$

1

37.5

2

12.5

3

25

4

18.75

Explanation:

(b) $N_{1} V_{1} = N_{2} V_{2}$
$0.15 \times 25 = 0.1 V_{2} \Rightarrow V_{2} = 37.5 m l$
Water to be added = $37.5 - 25 = 12.5 m l$

PRACTICAL CHEMISTRY

37497 To what extent must a given solution of concentration of $40 m g$ silver nitrate per $m l$ be diluted to yield a solution of concentration of $16$ $m g$ $A g N O_{3}$ per $m l$

1 Each ml must be diluted to

2.5 m l

2 To each ml of solution

2.5 m l

of water should be added

3 To

2.5 m l

of solution

2 m l

of water should be added

4 To

1.5 m l

of solution

1.5 m l

of water should be added

Explanation:

(a) $40 \times 1 = 16 \times x$ $\Rightarrow$ $x = \frac{40}{16} = 2.5 m l$

PRACTICAL CHEMISTRY

37498 $10 m l$ of conc. $H_{2} S O_{4}$ ( $18$ molar) is diluted to $1$ litre. The approximate strength of dilute acid could be..... $N$

1

0.18

2

0.36

3

0.09

4

18

Explanation:

(b) $18 M$ of $H_{2} S O_{4} = 36 N o f H_{2} S O_{4}$
$N_{1} V_{1} = N_{2} V_{2}$
$36 \times 10$ = $N_{2} \times 1000$
$∴ N_{2} = 0.36 N$

PRACTICAL CHEMISTRY

37499 For preparing one litre $N / 10$ solution of $H_{2} S O_{4},$ we need $H_{2} S O_{4}$ ....... $g m s$

1

98

2

10

3

100

4

4.9

Explanation:

(d) $\frac{1}{10} = \frac{W \times 1000}{e q . w t \times v o l .} = \frac{W \times 1000}{49 \times 1000}$ $\Rightarrow$ $W = 4.9 g m s$

PRACTICAL CHEMISTRY

37500 What is the concentration of nitrate ions if equal volumes of $0.1 M$ $A g N O_{3}$ and $0.1 M$ $N a C l$ are mixed together

1

0.1 N

2

0.2 M

3

0.05 M

4

0.25 M

Explanation:

(c) $M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times V = M_{2} \times 2 V$ $\Rightarrow$ $M_{2} = \frac{1 \times V}{2 V} = 0.05 M$

PRACTICAL CHEMISTRY

37501 To change the strength of $25 m l$ of $0.15 M$ $H C l$ to $0.1 M$ the quantity of water that is to be added is....... $m l$

1

37.5

2

12.5

3

25

4

18.75

Explanation:

(b) $N_{1} V_{1} = N_{2} V_{2}$
$0.15 \times 25 = 0.1 V_{2} \Rightarrow V_{2} = 37.5 m l$
Water to be added = $37.5 - 25 = 12.5 m l$

PRACTICAL CHEMISTRY

37497 To what extent must a given solution of concentration of $40 m g$ silver nitrate per $m l$ be diluted to yield a solution of concentration of $16$ $m g$ $A g N O_{3}$ per $m l$

1 Each ml must be diluted to

2.5 m l

2 To each ml of solution

2.5 m l

of water should be added

3 To

2.5 m l

of solution

2 m l

of water should be added

4 To

1.5 m l

of solution

1.5 m l

of water should be added

Explanation:

(a) $40 \times 1 = 16 \times x$ $\Rightarrow$ $x = \frac{40}{16} = 2.5 m l$

PRACTICAL CHEMISTRY

37498 $10 m l$ of conc. $H_{2} S O_{4}$ ( $18$ molar) is diluted to $1$ litre. The approximate strength of dilute acid could be..... $N$

1

0.18

2

0.36

3

0.09

4

18

Explanation:

(b) $18 M$ of $H_{2} S O_{4} = 36 N o f H_{2} S O_{4}$
$N_{1} V_{1} = N_{2} V_{2}$
$36 \times 10$ = $N_{2} \times 1000$
$∴ N_{2} = 0.36 N$

PRACTICAL CHEMISTRY

37499 For preparing one litre $N / 10$ solution of $H_{2} S O_{4},$ we need $H_{2} S O_{4}$ ....... $g m s$

1

98

2

10

3

100

4

4.9

Explanation:

(d) $\frac{1}{10} = \frac{W \times 1000}{e q . w t \times v o l .} = \frac{W \times 1000}{49 \times 1000}$ $\Rightarrow$ $W = 4.9 g m s$

PRACTICAL CHEMISTRY

37500 What is the concentration of nitrate ions if equal volumes of $0.1 M$ $A g N O_{3}$ and $0.1 M$ $N a C l$ are mixed together

1

0.1 N

2

0.2 M

3

0.05 M

4

0.25 M

Explanation:

(c) $M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times V = M_{2} \times 2 V$ $\Rightarrow$ $M_{2} = \frac{1 \times V}{2 V} = 0.05 M$

PRACTICAL CHEMISTRY

37501 To change the strength of $25 m l$ of $0.15 M$ $H C l$ to $0.1 M$ the quantity of water that is to be added is....... $m l$

1

37.5

2

12.5

3

25

4

18.75

Explanation:

(b) $N_{1} V_{1} = N_{2} V_{2}$
$0.15 \times 25 = 0.1 V_{2} \Rightarrow V_{2} = 37.5 m l$
Water to be added = $37.5 - 25 = 12.5 m l$

PRACTICAL CHEMISTRY

37497 To what extent must a given solution of concentration of $40 m g$ silver nitrate per $m l$ be diluted to yield a solution of concentration of $16$ $m g$ $A g N O_{3}$ per $m l$

1 Each ml must be diluted to

2.5 m l

2 To each ml of solution

2.5 m l

of water should be added

3 To

2.5 m l

of solution

2 m l

of water should be added

4 To

1.5 m l

of solution

1.5 m l

of water should be added

Explanation:

(a) $40 \times 1 = 16 \times x$ $\Rightarrow$ $x = \frac{40}{16} = 2.5 m l$

PRACTICAL CHEMISTRY

37498 $10 m l$ of conc. $H_{2} S O_{4}$ ( $18$ molar) is diluted to $1$ litre. The approximate strength of dilute acid could be..... $N$

1

0.18

2

0.36

3

0.09

4

18

Explanation:

(b) $18 M$ of $H_{2} S O_{4} = 36 N o f H_{2} S O_{4}$
$N_{1} V_{1} = N_{2} V_{2}$
$36 \times 10$ = $N_{2} \times 1000$
$∴ N_{2} = 0.36 N$

PRACTICAL CHEMISTRY

37499 For preparing one litre $N / 10$ solution of $H_{2} S O_{4},$ we need $H_{2} S O_{4}$ ....... $g m s$

1

98

2

10

3

100

4

4.9

Explanation:

(d) $\frac{1}{10} = \frac{W \times 1000}{e q . w t \times v o l .} = \frac{W \times 1000}{49 \times 1000}$ $\Rightarrow$ $W = 4.9 g m s$

PRACTICAL CHEMISTRY

37500 What is the concentration of nitrate ions if equal volumes of $0.1 M$ $A g N O_{3}$ and $0.1 M$ $N a C l$ are mixed together

1

0.1 N

2

0.2 M

3

0.05 M

4

0.25 M

Explanation:

(c) $M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times V = M_{2} \times 2 V$ $\Rightarrow$ $M_{2} = \frac{1 \times V}{2 V} = 0.05 M$

PRACTICAL CHEMISTRY

37501 To change the strength of $25 m l$ of $0.15 M$ $H C l$ to $0.1 M$ the quantity of water that is to be added is....... $m l$

1

37.5

2

12.5

3

25

4

18.75

Explanation:

(b) $N_{1} V_{1} = N_{2} V_{2}$
$0.15 \times 25 = 0.1 V_{2} \Rightarrow V_{2} = 37.5 m l$
Water to be added = $37.5 - 25 = 12.5 m l$

PRACTICAL CHEMISTRY

37497 To what extent must a given solution of concentration of $40 m g$ silver nitrate per $m l$ be diluted to yield a solution of concentration of $16$ $m g$ $A g N O_{3}$ per $m l$

1 Each ml must be diluted to

2.5 m l

2 To each ml of solution

2.5 m l

of water should be added

3 To

2.5 m l

of solution

2 m l

of water should be added

4 To

1.5 m l

of solution

1.5 m l

of water should be added

Explanation:

(a) $40 \times 1 = 16 \times x$ $\Rightarrow$ $x = \frac{40}{16} = 2.5 m l$

PRACTICAL CHEMISTRY

37498 $10 m l$ of conc. $H_{2} S O_{4}$ ( $18$ molar) is diluted to $1$ litre. The approximate strength of dilute acid could be..... $N$

1

0.18

2

0.36

3

0.09

4

18

Explanation:

(b) $18 M$ of $H_{2} S O_{4} = 36 N o f H_{2} S O_{4}$
$N_{1} V_{1} = N_{2} V_{2}$
$36 \times 10$ = $N_{2} \times 1000$
$∴ N_{2} = 0.36 N$

PRACTICAL CHEMISTRY

37499 For preparing one litre $N / 10$ solution of $H_{2} S O_{4},$ we need $H_{2} S O_{4}$ ....... $g m s$

1

98

2

10

3

100

4

4.9

Explanation:

(d) $\frac{1}{10} = \frac{W \times 1000}{e q . w t \times v o l .} = \frac{W \times 1000}{49 \times 1000}$ $\Rightarrow$ $W = 4.9 g m s$

PRACTICAL CHEMISTRY

37500 What is the concentration of nitrate ions if equal volumes of $0.1 M$ $A g N O_{3}$ and $0.1 M$ $N a C l$ are mixed together

1

0.1 N

2

0.2 M

3

0.05 M

4

0.25 M

Explanation:

(c) $M_{1} V_{1} = M_{2} V_{2}$
$0.1 \times V = M_{2} \times 2 V$ $\Rightarrow$ $M_{2} = \frac{1 \times V}{2 V} = 0.05 M$

PRACTICAL CHEMISTRY

37501 To change the strength of $25 m l$ of $0.15 M$ $H C l$ to $0.1 M$ the quantity of water that is to be added is....... $m l$

1

37.5

2

12.5

3

25

4

18.75

Explanation:

(b) $N_{1} V_{1} = N_{2} V_{2}$
$0.15 \times 25 = 0.1 V_{2} \Rightarrow V_{2} = 37.5 m l$
Water to be added = $37.5 - 25 = 12.5 m l$

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