QBManager

ELECTROCHEMISTRY

19837 If $96500$ coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15 a m p e r e s$ to deposite $20 m g$ of copper from a solution of copper sulphate is (Chemical equivalent of copper = $32$ )

1

5

min

20

sec

2

6

min

42

sec

3

4

min

40

sec

4

5

min

50

sec

Explanation:

(b) $m = Z i t$ $\Rightarrow$ $20 \times 10^{- 3} = (\frac{32}{96500}) \times 0.15 \times t$
$= 6.7 m i n = 6 m i n .42 s e c$

ELECTROCHEMISTRY

19838 How much current should be passed through acidified water for $100 s$ to liberate $0.224 l i t r e$ of $H_{2}$ ............. $A$

1

22.4

2

19.3

3

9.65

4

1

Explanation:

(b) $22.4$ litre $H_{2}$ = 1 mole $H_{2}$ = $N$ molecules of $H_{2} = 2 N$ atom of $H$
So charge required to liberate $22.4 l i t r e$ of $H_{2} = 2 N$ $e = 2 F$
Hence charge required to liberate $0.224 l i t r e$ of $H_{2}$ = $\frac{2 F}{22.4} \times 0.224$ $= \frac{2 F}{100} = 2 \times 965 C$
So current $i = \frac{Q}{t} = \frac{2 \times 965}{100} = 19.3 a m p$

ELECTROCHEMISTRY

19839 If a steady current of $4 a m p$ maintained for $40$ minutes, deposits $4.5 g m$ of zinc at the cathode and then the electro chemical equivalent will be

1

51 \times 10^{- 17} g m / C

2

28 \times 10^{- 6} g m / C

3

32 \times 10^{- 5} g m / C

4

47 \times 10^{- 5} g m / C

Explanation:

(d) $m = Z i t$ $\Rightarrow$ $Z = \frac{m}{i t} = \frac{4.5}{4 \times 40 \times 60} = 47 \times 10^{- 5} g / C$

ELECTROCHEMISTRY

19861 The electric conduction of a salt solution in water depends on the

1 Shape of its molecules

2 Size of its molecules

3 Size of solvent molecules

4 Extent of its ionization

ELECTROCHEMISTRY

19840 The current flowing in a copper voltameter is $3.2 A$ . The number of copper ions $(C u^{2 +})$ deposited at the cathode per minute is

1

0.5 \times 10^{20}

2

1.5 \times 10^{20}

3

3 \times 10^{20}

4

6 \times 10^{20}

Explanation:

(d) Charge supplied per minute = $3.2 \times 60$ = $192 C$
Charge $2 e$ liberates one $C u^{+ 2}$ ion
No of $C u^{+ 2}$ ion liberate by $192 C$
$= \frac{192}{2 e} = \frac{192}{2 \times 1.6 \times 10^{- 19}} = 6 \times 10^{20}$

ELECTROCHEMISTRY

19837 If $96500$ coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15 a m p e r e s$ to deposite $20 m g$ of copper from a solution of copper sulphate is (Chemical equivalent of copper = $32$ )

1

5

min

20

sec

2

6

min

42

sec

3

4

min

40

sec

4

5

min

50

sec

Explanation:

(b) $m = Z i t$ $\Rightarrow$ $20 \times 10^{- 3} = (\frac{32}{96500}) \times 0.15 \times t$
$= 6.7 m i n = 6 m i n .42 s e c$

ELECTROCHEMISTRY

19838 How much current should be passed through acidified water for $100 s$ to liberate $0.224 l i t r e$ of $H_{2}$ ............. $A$

1

22.4

2

19.3

3

9.65

4

1

Explanation:

(b) $22.4$ litre $H_{2}$ = 1 mole $H_{2}$ = $N$ molecules of $H_{2} = 2 N$ atom of $H$
So charge required to liberate $22.4 l i t r e$ of $H_{2} = 2 N$ $e = 2 F$
Hence charge required to liberate $0.224 l i t r e$ of $H_{2}$ = $\frac{2 F}{22.4} \times 0.224$ $= \frac{2 F}{100} = 2 \times 965 C$
So current $i = \frac{Q}{t} = \frac{2 \times 965}{100} = 19.3 a m p$

ELECTROCHEMISTRY

19839 If a steady current of $4 a m p$ maintained for $40$ minutes, deposits $4.5 g m$ of zinc at the cathode and then the electro chemical equivalent will be

1

51 \times 10^{- 17} g m / C

2

28 \times 10^{- 6} g m / C

3

32 \times 10^{- 5} g m / C

4

47 \times 10^{- 5} g m / C

Explanation:

(d) $m = Z i t$ $\Rightarrow$ $Z = \frac{m}{i t} = \frac{4.5}{4 \times 40 \times 60} = 47 \times 10^{- 5} g / C$

ELECTROCHEMISTRY

19861 The electric conduction of a salt solution in water depends on the

1 Shape of its molecules

2 Size of its molecules

3 Size of solvent molecules

4 Extent of its ionization

ELECTROCHEMISTRY

19840 The current flowing in a copper voltameter is $3.2 A$ . The number of copper ions $(C u^{2 +})$ deposited at the cathode per minute is

1

0.5 \times 10^{20}

2

1.5 \times 10^{20}

3

3 \times 10^{20}

4

6 \times 10^{20}

Explanation:

(d) Charge supplied per minute = $3.2 \times 60$ = $192 C$
Charge $2 e$ liberates one $C u^{+ 2}$ ion
No of $C u^{+ 2}$ ion liberate by $192 C$
$= \frac{192}{2 e} = \frac{192}{2 \times 1.6 \times 10^{- 19}} = 6 \times 10^{20}$

ELECTROCHEMISTRY

19837 If $96500$ coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15 a m p e r e s$ to deposite $20 m g$ of copper from a solution of copper sulphate is (Chemical equivalent of copper = $32$ )

1

5

min

20

sec

2

6

min

42

sec

3

4

min

40

sec

4

5

min

50

sec

Explanation:

(b) $m = Z i t$ $\Rightarrow$ $20 \times 10^{- 3} = (\frac{32}{96500}) \times 0.15 \times t$
$= 6.7 m i n = 6 m i n .42 s e c$

ELECTROCHEMISTRY

19838 How much current should be passed through acidified water for $100 s$ to liberate $0.224 l i t r e$ of $H_{2}$ ............. $A$

1

22.4

2

19.3

3

9.65

4

1

Explanation:

(b) $22.4$ litre $H_{2}$ = 1 mole $H_{2}$ = $N$ molecules of $H_{2} = 2 N$ atom of $H$
So charge required to liberate $22.4 l i t r e$ of $H_{2} = 2 N$ $e = 2 F$
Hence charge required to liberate $0.224 l i t r e$ of $H_{2}$ = $\frac{2 F}{22.4} \times 0.224$ $= \frac{2 F}{100} = 2 \times 965 C$
So current $i = \frac{Q}{t} = \frac{2 \times 965}{100} = 19.3 a m p$

ELECTROCHEMISTRY

19839 If a steady current of $4 a m p$ maintained for $40$ minutes, deposits $4.5 g m$ of zinc at the cathode and then the electro chemical equivalent will be

1

51 \times 10^{- 17} g m / C

2

28 \times 10^{- 6} g m / C

3

32 \times 10^{- 5} g m / C

4

47 \times 10^{- 5} g m / C

Explanation:

(d) $m = Z i t$ $\Rightarrow$ $Z = \frac{m}{i t} = \frac{4.5}{4 \times 40 \times 60} = 47 \times 10^{- 5} g / C$

ELECTROCHEMISTRY

19861 The electric conduction of a salt solution in water depends on the

1 Shape of its molecules

2 Size of its molecules

3 Size of solvent molecules

4 Extent of its ionization

ELECTROCHEMISTRY

19840 The current flowing in a copper voltameter is $3.2 A$ . The number of copper ions $(C u^{2 +})$ deposited at the cathode per minute is

1

0.5 \times 10^{20}

2

1.5 \times 10^{20}

3

3 \times 10^{20}

4

6 \times 10^{20}

Explanation:

(d) Charge supplied per minute = $3.2 \times 60$ = $192 C$
Charge $2 e$ liberates one $C u^{+ 2}$ ion
No of $C u^{+ 2}$ ion liberate by $192 C$
$= \frac{192}{2 e} = \frac{192}{2 \times 1.6 \times 10^{- 19}} = 6 \times 10^{20}$

ELECTROCHEMISTRY

19837 If $96500$ coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15 a m p e r e s$ to deposite $20 m g$ of copper from a solution of copper sulphate is (Chemical equivalent of copper = $32$ )

1

5

min

20

sec

2

6

min

42

sec

3

4

min

40

sec

4

5

min

50

sec

Explanation:

(b) $m = Z i t$ $\Rightarrow$ $20 \times 10^{- 3} = (\frac{32}{96500}) \times 0.15 \times t$
$= 6.7 m i n = 6 m i n .42 s e c$

ELECTROCHEMISTRY

19838 How much current should be passed through acidified water for $100 s$ to liberate $0.224 l i t r e$ of $H_{2}$ ............. $A$

1

22.4

2

19.3

3

9.65

4

1

Explanation:

(b) $22.4$ litre $H_{2}$ = 1 mole $H_{2}$ = $N$ molecules of $H_{2} = 2 N$ atom of $H$
So charge required to liberate $22.4 l i t r e$ of $H_{2} = 2 N$ $e = 2 F$
Hence charge required to liberate $0.224 l i t r e$ of $H_{2}$ = $\frac{2 F}{22.4} \times 0.224$ $= \frac{2 F}{100} = 2 \times 965 C$
So current $i = \frac{Q}{t} = \frac{2 \times 965}{100} = 19.3 a m p$

ELECTROCHEMISTRY

19839 If a steady current of $4 a m p$ maintained for $40$ minutes, deposits $4.5 g m$ of zinc at the cathode and then the electro chemical equivalent will be

1

51 \times 10^{- 17} g m / C

2

28 \times 10^{- 6} g m / C

3

32 \times 10^{- 5} g m / C

4

47 \times 10^{- 5} g m / C

Explanation:

(d) $m = Z i t$ $\Rightarrow$ $Z = \frac{m}{i t} = \frac{4.5}{4 \times 40 \times 60} = 47 \times 10^{- 5} g / C$

ELECTROCHEMISTRY

19861 The electric conduction of a salt solution in water depends on the

1 Shape of its molecules

2 Size of its molecules

3 Size of solvent molecules

4 Extent of its ionization

ELECTROCHEMISTRY

19840 The current flowing in a copper voltameter is $3.2 A$ . The number of copper ions $(C u^{2 +})$ deposited at the cathode per minute is

1

0.5 \times 10^{20}

2

1.5 \times 10^{20}

3

3 \times 10^{20}

4

6 \times 10^{20}

Explanation:

(d) Charge supplied per minute = $3.2 \times 60$ = $192 C$
Charge $2 e$ liberates one $C u^{+ 2}$ ion
No of $C u^{+ 2}$ ion liberate by $192 C$
$= \frac{192}{2 e} = \frac{192}{2 \times 1.6 \times 10^{- 19}} = 6 \times 10^{20}$

ELECTROCHEMISTRY

19837 If $96500$ coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of $0.15 a m p e r e s$ to deposite $20 m g$ of copper from a solution of copper sulphate is (Chemical equivalent of copper = $32$ )

1

5

min

20

sec

2

6

min

42

sec

3

4

min

40

sec

4

5

min

50

sec

Explanation:

(b) $m = Z i t$ $\Rightarrow$ $20 \times 10^{- 3} = (\frac{32}{96500}) \times 0.15 \times t$
$= 6.7 m i n = 6 m i n .42 s e c$

ELECTROCHEMISTRY

19838 How much current should be passed through acidified water for $100 s$ to liberate $0.224 l i t r e$ of $H_{2}$ ............. $A$

1

22.4

2

19.3

3

9.65

4

1

Explanation:

(b) $22.4$ litre $H_{2}$ = 1 mole $H_{2}$ = $N$ molecules of $H_{2} = 2 N$ atom of $H$
So charge required to liberate $22.4 l i t r e$ of $H_{2} = 2 N$ $e = 2 F$
Hence charge required to liberate $0.224 l i t r e$ of $H_{2}$ = $\frac{2 F}{22.4} \times 0.224$ $= \frac{2 F}{100} = 2 \times 965 C$
So current $i = \frac{Q}{t} = \frac{2 \times 965}{100} = 19.3 a m p$

ELECTROCHEMISTRY

19839 If a steady current of $4 a m p$ maintained for $40$ minutes, deposits $4.5 g m$ of zinc at the cathode and then the electro chemical equivalent will be

1

51 \times 10^{- 17} g m / C

2

28 \times 10^{- 6} g m / C

3

32 \times 10^{- 5} g m / C

4

47 \times 10^{- 5} g m / C

Explanation:

(d) $m = Z i t$ $\Rightarrow$ $Z = \frac{m}{i t} = \frac{4.5}{4 \times 40 \times 60} = 47 \times 10^{- 5} g / C$

ELECTROCHEMISTRY

19861 The electric conduction of a salt solution in water depends on the

1 Shape of its molecules

2 Size of its molecules

3 Size of solvent molecules

4 Extent of its ionization

ELECTROCHEMISTRY

19840 The current flowing in a copper voltameter is $3.2 A$ . The number of copper ions $(C u^{2 +})$ deposited at the cathode per minute is

1

0.5 \times 10^{20}

2

1.5 \times 10^{20}

3

3 \times 10^{20}

4

6 \times 10^{20}

Explanation:

(d) Charge supplied per minute = $3.2 \times 60$ = $192 C$
Charge $2 e$ liberates one $C u^{+ 2}$ ion
No of $C u^{+ 2}$ ion liberate by $192 C$
$= \frac{192}{2 e} = \frac{192}{2 \times 1.6 \times 10^{- 19}} = 6 \times 10^{20}$

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