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CHXII03:ELECTROCHEMISTRY

330374 The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of $0 .88 c m^{- 1}$ . The value of equivalent conductance of solution is

1

400 o h m^{- 1} c m^{2} g e q^{- 1}

2

295 o h m^{- 1} c m^{2} g e q^{- 1}

3

419 o h m^{- 1} c m^{2} g e q^{- 1}

4

425 o h m^{- 1} c m^{2} g e q^{- 1}

Explanation:

$Λ_{e q} = κ \times \frac{1000}{N} = \frac{1}{R} \times \frac{l}{A} \times \frac{1000}{N}$
$= \frac{1}{R} \times c e l l c o n s t a n t \times \frac{1000}{N} = \frac{1}{220} \times 0 .88 \times \frac{1000}{0 .01}$
$= 400 o h m^{- 1} c m^{2} g e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330375 The conductivity of a saturated solution of Sparingly soluble salt (x-factor = 1) is $1 .53 \times 1 0^{- 5} o h m^{- 1} c m^{- 1}$ and its conductance is $0 .765 o h m^{- 1} c m^{- 1} e q u i v^{- 1}$ . The molarity of sparingly soluble salt will be

1

1 0^{- 6}

2

25 \times 1 0^{- 9}

3

4 \times 1 0^{- 12}

4

2 \times 1 0^{- 2}

Explanation:

$0 .765 = \frac{1000 \times 1 .53 \times 1 0^{- 5}}{N o r m a l i t y}$
$N o r m a l i t y = 2 \times 1 0^{- 2} N$
$M o l a r i t y = \frac{2 \times 1 0^{- 2}}{1} = 2 \times 1 0^{- 2} M$

CHXII03:ELECTROCHEMISTRY

330376 What are the units of equivalent conductivity of a solution?

1 mho.

{cm}^{- 1}

2 ohm.

{cm}^{- 1} \cdot g

equiv

^{- 1}

3 mho.

{cm}^{- 2}

. g equiv

^{- 1}

4 mho.

{cm}^{2}

. equiv

^{- 1}

Explanation:

Unit of equivalent conductivity is $O h m^{- 1} c m^{2} e q^{- 1} o r m h o c m^{2} e q^{- 1} .$

CHXII03:ELECTROCHEMISTRY

330377 In a conductivity cell the two platinum electrodes, each of area 10 sq. cm is fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of $50 Ω$ find equivalent conductance of the salt solution?

1

125 S c m^{2} e q^{- 1}

2

120 S m^{2} e q^{- 1}

3

160 S c m^{2} e q^{- 1}

4

125 S c m^{2} e q^{- 1}

Explanation:

$50 = \frac{1}{Λ_{e q}} \times \frac{1 .5}{10 / 2} \times \frac{1000}{0 .05};$
$E q u i v a l e n t c o n d u c t a n c e = 125 S c m^{2} e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330374 The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of $0 .88 c m^{- 1}$ . The value of equivalent conductance of solution is

1

400 o h m^{- 1} c m^{2} g e q^{- 1}

2

295 o h m^{- 1} c m^{2} g e q^{- 1}

3

419 o h m^{- 1} c m^{2} g e q^{- 1}

4

425 o h m^{- 1} c m^{2} g e q^{- 1}

Explanation:

$Λ_{e q} = κ \times \frac{1000}{N} = \frac{1}{R} \times \frac{l}{A} \times \frac{1000}{N}$
$= \frac{1}{R} \times c e l l c o n s t a n t \times \frac{1000}{N} = \frac{1}{220} \times 0 .88 \times \frac{1000}{0 .01}$
$= 400 o h m^{- 1} c m^{2} g e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330375 The conductivity of a saturated solution of Sparingly soluble salt (x-factor = 1) is $1 .53 \times 1 0^{- 5} o h m^{- 1} c m^{- 1}$ and its conductance is $0 .765 o h m^{- 1} c m^{- 1} e q u i v^{- 1}$ . The molarity of sparingly soluble salt will be

1

1 0^{- 6}

2

25 \times 1 0^{- 9}

3

4 \times 1 0^{- 12}

4

2 \times 1 0^{- 2}

Explanation:

$0 .765 = \frac{1000 \times 1 .53 \times 1 0^{- 5}}{N o r m a l i t y}$
$N o r m a l i t y = 2 \times 1 0^{- 2} N$
$M o l a r i t y = \frac{2 \times 1 0^{- 2}}{1} = 2 \times 1 0^{- 2} M$

CHXII03:ELECTROCHEMISTRY

330376 What are the units of equivalent conductivity of a solution?

1 mho.

{cm}^{- 1}

2 ohm.

{cm}^{- 1} \cdot g

equiv

^{- 1}

3 mho.

{cm}^{- 2}

. g equiv

^{- 1}

4 mho.

{cm}^{2}

. equiv

^{- 1}

Explanation:

Unit of equivalent conductivity is $O h m^{- 1} c m^{2} e q^{- 1} o r m h o c m^{2} e q^{- 1} .$

CHXII03:ELECTROCHEMISTRY

330377 In a conductivity cell the two platinum electrodes, each of area 10 sq. cm is fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of $50 Ω$ find equivalent conductance of the salt solution?

1

125 S c m^{2} e q^{- 1}

2

120 S m^{2} e q^{- 1}

3

160 S c m^{2} e q^{- 1}

4

125 S c m^{2} e q^{- 1}

Explanation:

$50 = \frac{1}{Λ_{e q}} \times \frac{1 .5}{10 / 2} \times \frac{1000}{0 .05};$
$E q u i v a l e n t c o n d u c t a n c e = 125 S c m^{2} e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330374 The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of $0 .88 c m^{- 1}$ . The value of equivalent conductance of solution is

1

400 o h m^{- 1} c m^{2} g e q^{- 1}

2

295 o h m^{- 1} c m^{2} g e q^{- 1}

3

419 o h m^{- 1} c m^{2} g e q^{- 1}

4

425 o h m^{- 1} c m^{2} g e q^{- 1}

Explanation:

$Λ_{e q} = κ \times \frac{1000}{N} = \frac{1}{R} \times \frac{l}{A} \times \frac{1000}{N}$
$= \frac{1}{R} \times c e l l c o n s t a n t \times \frac{1000}{N} = \frac{1}{220} \times 0 .88 \times \frac{1000}{0 .01}$
$= 400 o h m^{- 1} c m^{2} g e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330375 The conductivity of a saturated solution of Sparingly soluble salt (x-factor = 1) is $1 .53 \times 1 0^{- 5} o h m^{- 1} c m^{- 1}$ and its conductance is $0 .765 o h m^{- 1} c m^{- 1} e q u i v^{- 1}$ . The molarity of sparingly soluble salt will be

1

1 0^{- 6}

2

25 \times 1 0^{- 9}

3

4 \times 1 0^{- 12}

4

2 \times 1 0^{- 2}

Explanation:

$0 .765 = \frac{1000 \times 1 .53 \times 1 0^{- 5}}{N o r m a l i t y}$
$N o r m a l i t y = 2 \times 1 0^{- 2} N$
$M o l a r i t y = \frac{2 \times 1 0^{- 2}}{1} = 2 \times 1 0^{- 2} M$

CHXII03:ELECTROCHEMISTRY

330376 What are the units of equivalent conductivity of a solution?

1 mho.

{cm}^{- 1}

2 ohm.

{cm}^{- 1} \cdot g

equiv

^{- 1}

3 mho.

{cm}^{- 2}

. g equiv

^{- 1}

4 mho.

{cm}^{2}

. equiv

^{- 1}

Explanation:

Unit of equivalent conductivity is $O h m^{- 1} c m^{2} e q^{- 1} o r m h o c m^{2} e q^{- 1} .$

CHXII03:ELECTROCHEMISTRY

330377 In a conductivity cell the two platinum electrodes, each of area 10 sq. cm is fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of $50 Ω$ find equivalent conductance of the salt solution?

1

125 S c m^{2} e q^{- 1}

2

120 S m^{2} e q^{- 1}

3

160 S c m^{2} e q^{- 1}

4

125 S c m^{2} e q^{- 1}

Explanation:

$50 = \frac{1}{Λ_{e q}} \times \frac{1 .5}{10 / 2} \times \frac{1000}{0 .05};$
$E q u i v a l e n t c o n d u c t a n c e = 125 S c m^{2} e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330374 The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of $0 .88 c m^{- 1}$ . The value of equivalent conductance of solution is

1

400 o h m^{- 1} c m^{2} g e q^{- 1}

2

295 o h m^{- 1} c m^{2} g e q^{- 1}

3

419 o h m^{- 1} c m^{2} g e q^{- 1}

4

425 o h m^{- 1} c m^{2} g e q^{- 1}

Explanation:

$Λ_{e q} = κ \times \frac{1000}{N} = \frac{1}{R} \times \frac{l}{A} \times \frac{1000}{N}$
$= \frac{1}{R} \times c e l l c o n s t a n t \times \frac{1000}{N} = \frac{1}{220} \times 0 .88 \times \frac{1000}{0 .01}$
$= 400 o h m^{- 1} c m^{2} g e q^{- 1}$

CHXII03:ELECTROCHEMISTRY

330375 The conductivity of a saturated solution of Sparingly soluble salt (x-factor = 1) is $1 .53 \times 1 0^{- 5} o h m^{- 1} c m^{- 1}$ and its conductance is $0 .765 o h m^{- 1} c m^{- 1} e q u i v^{- 1}$ . The molarity of sparingly soluble salt will be

1

1 0^{- 6}

2

25 \times 1 0^{- 9}

3

4 \times 1 0^{- 12}

4

2 \times 1 0^{- 2}

Explanation:

$0 .765 = \frac{1000 \times 1 .53 \times 1 0^{- 5}}{N o r m a l i t y}$
$N o r m a l i t y = 2 \times 1 0^{- 2} N$
$M o l a r i t y = \frac{2 \times 1 0^{- 2}}{1} = 2 \times 1 0^{- 2} M$

CHXII03:ELECTROCHEMISTRY

330376 What are the units of equivalent conductivity of a solution?

1 mho.

{cm}^{- 1}

2 ohm.

{cm}^{- 1} \cdot g

equiv

^{- 1}

3 mho.

{cm}^{- 2}

. g equiv

^{- 1}

4 mho.

{cm}^{2}

. equiv

^{- 1}

Explanation:

Unit of equivalent conductivity is $O h m^{- 1} c m^{2} e q^{- 1} o r m h o c m^{2} e q^{- 1} .$

CHXII03:ELECTROCHEMISTRY

330377 In a conductivity cell the two platinum electrodes, each of area 10 sq. cm is fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of $50 Ω$ find equivalent conductance of the salt solution?

1

125 S c m^{2} e q^{- 1}

2

120 S m^{2} e q^{- 1}

3

160 S c m^{2} e q^{- 1}

4

125 S c m^{2} e q^{- 1}

Explanation:

$50 = \frac{1}{Λ_{e q}} \times \frac{1 .5}{10 / 2} \times \frac{1000}{0 .05};$
$E q u i v a l e n t c o n d u c t a n c e = 125 S c m^{2} e q^{- 1}$

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