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1 | 2205-2208 | 3)
and at different temperatures The cell constant, G*, is then given by
the equation:
G* = l
A = R k
(2 18)
Once the cell constant is determined, we can
use it for measuring the resistance or conductivity
of any solution The set up for the measurement
of the resistance is shown in Fig |
1 | 2206-2209 | The cell constant, G*, is then given by
the equation:
G* = l
A = R k
(2 18)
Once the cell constant is determined, we can
use it for measuring the resistance or conductivity
of any solution The set up for the measurement
of the resistance is shown in Fig 2 |
1 | 2207-2210 | 18)
Once the cell constant is determined, we can
use it for measuring the resistance or conductivity
of any solution The set up for the measurement
of the resistance is shown in Fig 2 5 |
1 | 2208-2211 | The set up for the measurement
of the resistance is shown in Fig 2 5 It consists of two resistances R3 and R4, a
variable resistance R1 and the conductivity cell
having the unknown resistance R2 |
1 | 2209-2212 | 2 5 It consists of two resistances R3 and R4, a
variable resistance R1 and the conductivity cell
having the unknown resistance R2 The
Wheatstone bridge is fed by an oscillator O (a
source of a |
1 | 2210-2213 | 5 It consists of two resistances R3 and R4, a
variable resistance R1 and the conductivity cell
having the unknown resistance R2 The
Wheatstone bridge is fed by an oscillator O (a
source of a c |
1 | 2211-2214 | It consists of two resistances R3 and R4, a
variable resistance R1 and the conductivity cell
having the unknown resistance R2 The
Wheatstone bridge is fed by an oscillator O (a
source of a c power in the audio frequency range
550 to 5000 cycles per second) |
1 | 2212-2215 | The
Wheatstone bridge is fed by an oscillator O (a
source of a c power in the audio frequency range
550 to 5000 cycles per second) P is a suitable
detector (a headphone or other electronic device)
and the bridge is balanced when no current passes
through the detector |
1 | 2213-2216 | c power in the audio frequency range
550 to 5000 cycles per second) P is a suitable
detector (a headphone or other electronic device)
and the bridge is balanced when no current passes
through the detector Under these conditions:
Unknown resistance R2 =
1
4
3
R R
R
(2 |
1 | 2214-2217 | power in the audio frequency range
550 to 5000 cycles per second) P is a suitable
detector (a headphone or other electronic device)
and the bridge is balanced when no current passes
through the detector Under these conditions:
Unknown resistance R2 =
1
4
3
R R
R
(2 19)
These days, inexpensive conductivity meters are
available which can directly read the conductance or resistance of the
solution in the conductivity cell |
1 | 2215-2218 | P is a suitable
detector (a headphone or other electronic device)
and the bridge is balanced when no current passes
through the detector Under these conditions:
Unknown resistance R2 =
1
4
3
R R
R
(2 19)
These days, inexpensive conductivity meters are
available which can directly read the conductance or resistance of the
solution in the conductivity cell Once the cell constant and the resistance
of the solution in the cell is determined, the conductivity of the solution
is given by the equation:
cell constant
G*
R
R
(2 |
1 | 2216-2219 | Under these conditions:
Unknown resistance R2 =
1
4
3
R R
R
(2 19)
These days, inexpensive conductivity meters are
available which can directly read the conductance or resistance of the
solution in the conductivity cell Once the cell constant and the resistance
of the solution in the cell is determined, the conductivity of the solution
is given by the equation:
cell constant
G*
R
R
(2 20)
The conductivity of solutions of different electrolytes in the same
solvent and at a given temperature differs due to charge and size of the
Fig |
1 | 2217-2220 | 19)
These days, inexpensive conductivity meters are
available which can directly read the conductance or resistance of the
solution in the conductivity cell Once the cell constant and the resistance
of the solution in the cell is determined, the conductivity of the solution
is given by the equation:
cell constant
G*
R
R
(2 20)
The conductivity of solutions of different electrolytes in the same
solvent and at a given temperature differs due to charge and size of the
Fig 2 |
1 | 2218-2221 | Once the cell constant and the resistance
of the solution in the cell is determined, the conductivity of the solution
is given by the equation:
cell constant
G*
R
R
(2 20)
The conductivity of solutions of different electrolytes in the same
solvent and at a given temperature differs due to charge and size of the
Fig 2 5: Arrangement for measurement of
resistance of a solution of an
electrolyte |
1 | 2219-2222 | 20)
The conductivity of solutions of different electrolytes in the same
solvent and at a given temperature differs due to charge and size of the
Fig 2 5: Arrangement for measurement of
resistance of a solution of an
electrolyte Rationalised 2023-24
45
Electrochemistry
ions in which they dissociate, the concentration of ions or ease with
which the ions move under a potential gradient |
1 | 2220-2223 | 2 5: Arrangement for measurement of
resistance of a solution of an
electrolyte Rationalised 2023-24
45
Electrochemistry
ions in which they dissociate, the concentration of ions or ease with
which the ions move under a potential gradient It, therefore, becomes
necessary to define a physically more meaningful quantity called molar
conductivity denoted by the symbol Lm (Greek, lambda) |
1 | 2221-2224 | 5: Arrangement for measurement of
resistance of a solution of an
electrolyte Rationalised 2023-24
45
Electrochemistry
ions in which they dissociate, the concentration of ions or ease with
which the ions move under a potential gradient It, therefore, becomes
necessary to define a physically more meaningful quantity called molar
conductivity denoted by the symbol Lm (Greek, lambda) It is related
to the conductivity of the solution by the equation:
Molar conductivity = Lm = c
(2 |
1 | 2222-2225 | Rationalised 2023-24
45
Electrochemistry
ions in which they dissociate, the concentration of ions or ease with
which the ions move under a potential gradient It, therefore, becomes
necessary to define a physically more meaningful quantity called molar
conductivity denoted by the symbol Lm (Greek, lambda) It is related
to the conductivity of the solution by the equation:
Molar conductivity = Lm = c
(2 21)
In the above equation, if k is expressed in S m–1 and the concentration,
c in mol m–3 then the units of Lm are in S m2 mol–1 |
1 | 2223-2226 | It, therefore, becomes
necessary to define a physically more meaningful quantity called molar
conductivity denoted by the symbol Lm (Greek, lambda) It is related
to the conductivity of the solution by the equation:
Molar conductivity = Lm = c
(2 21)
In the above equation, if k is expressed in S m–1 and the concentration,
c in mol m–3 then the units of Lm are in S m2 mol–1 It may be noted that:
1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence
Lm(S cm2 mol–1) =
1
3
1
(S cm
)
1000 L m
× molarity (mol L )
If we use S cm–1 as the units for k and mol cm–3, the units of
concentration, then the units for Lm are S cm2 mol–1 |
1 | 2224-2227 | It is related
to the conductivity of the solution by the equation:
Molar conductivity = Lm = c
(2 21)
In the above equation, if k is expressed in S m–1 and the concentration,
c in mol m–3 then the units of Lm are in S m2 mol–1 It may be noted that:
1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence
Lm(S cm2 mol–1) =
1
3
1
(S cm
)
1000 L m
× molarity (mol L )
If we use S cm–1 as the units for k and mol cm–3, the units of
concentration, then the units for Lm are S cm2 mol–1 It can be calculated
by using the equation:
Lm (S cm2 mol–1) =
1
3
(S cm
) × 1000 (cm /L)
molarity (mol/L)
Both type of units are used in literature and are related to each
other by the equations:
1 S m2mol–1
= 104 S cm2mol–1 or
1 S cm2mol–1 = 10–4 S m2mol–1 |
1 | 2225-2228 | 21)
In the above equation, if k is expressed in S m–1 and the concentration,
c in mol m–3 then the units of Lm are in S m2 mol–1 It may be noted that:
1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence
Lm(S cm2 mol–1) =
1
3
1
(S cm
)
1000 L m
× molarity (mol L )
If we use S cm–1 as the units for k and mol cm–3, the units of
concentration, then the units for Lm are S cm2 mol–1 It can be calculated
by using the equation:
Lm (S cm2 mol–1) =
1
3
(S cm
) × 1000 (cm /L)
molarity (mol/L)
Both type of units are used in literature and are related to each
other by the equations:
1 S m2mol–1
= 104 S cm2mol–1 or
1 S cm2mol–1 = 10–4 S m2mol–1 Resistance of a conductivity cell filled with 0 |
1 | 2226-2229 | It may be noted that:
1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence
Lm(S cm2 mol–1) =
1
3
1
(S cm
)
1000 L m
× molarity (mol L )
If we use S cm–1 as the units for k and mol cm–3, the units of
concentration, then the units for Lm are S cm2 mol–1 It can be calculated
by using the equation:
Lm (S cm2 mol–1) =
1
3
(S cm
) × 1000 (cm /L)
molarity (mol/L)
Both type of units are used in literature and are related to each
other by the equations:
1 S m2mol–1
= 104 S cm2mol–1 or
1 S cm2mol–1 = 10–4 S m2mol–1 Resistance of a conductivity cell filled with 0 1 mol L–1 KCl solution is
100 W |
1 | 2227-2230 | It can be calculated
by using the equation:
Lm (S cm2 mol–1) =
1
3
(S cm
) × 1000 (cm /L)
molarity (mol/L)
Both type of units are used in literature and are related to each
other by the equations:
1 S m2mol–1
= 104 S cm2mol–1 or
1 S cm2mol–1 = 10–4 S m2mol–1 Resistance of a conductivity cell filled with 0 1 mol L–1 KCl solution is
100 W If the resistance of the same cell when filled with 0 |
1 | 2228-2231 | Resistance of a conductivity cell filled with 0 1 mol L–1 KCl solution is
100 W If the resistance of the same cell when filled with 0 02 mol L–1
KCl solution is 520 W , calculate the conductivity and molar conductivity
of 0 |
1 | 2229-2232 | 1 mol L–1 KCl solution is
100 W If the resistance of the same cell when filled with 0 02 mol L–1
KCl solution is 520 W , calculate the conductivity and molar conductivity
of 0 02 mol L–1 KCl solution |
1 | 2230-2233 | If the resistance of the same cell when filled with 0 02 mol L–1
KCl solution is 520 W , calculate the conductivity and molar conductivity
of 0 02 mol L–1 KCl solution The conductivity of 0 |
1 | 2231-2234 | 02 mol L–1
KCl solution is 520 W , calculate the conductivity and molar conductivity
of 0 02 mol L–1 KCl solution The conductivity of 0 1 mol L–1 KCl
solution is 1 |
1 | 2232-2235 | 02 mol L–1 KCl solution The conductivity of 0 1 mol L–1 KCl
solution is 1 29 S/m |
1 | 2233-2236 | The conductivity of 0 1 mol L–1 KCl
solution is 1 29 S/m The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1 |
1 | 2234-2237 | 1 mol L–1 KCl
solution is 1 29 S/m The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1 29 S/m × 100 W = 129 m–1 = 1 |
1 | 2235-2238 | 29 S/m The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1 29 S/m × 100 W = 129 m–1 = 1 29 cm–1
Conductivity of 0 |
1 | 2236-2239 | The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1 29 S/m × 100 W = 129 m–1 = 1 29 cm–1
Conductivity of 0 02 mol L–1 KCl solution = cell constant / resistance
=
G*
R =
129 m–1
520
= 0 |
1 | 2237-2240 | 29 S/m × 100 W = 129 m–1 = 1 29 cm–1
Conductivity of 0 02 mol L–1 KCl solution = cell constant / resistance
=
G*
R =
129 m–1
520
= 0 248 S m–1
Concentration
= 0 |
1 | 2238-2241 | 29 cm–1
Conductivity of 0 02 mol L–1 KCl solution = cell constant / resistance
=
G*
R =
129 m–1
520
= 0 248 S m–1
Concentration
= 0 02 mol L–1
= 1000 × 0 |
1 | 2239-2242 | 02 mol L–1 KCl solution = cell constant / resistance
=
G*
R =
129 m–1
520
= 0 248 S m–1
Concentration
= 0 02 mol L–1
= 1000 × 0 02 mol m–3 = 20 mol m–3
Molar conductivity
=
m
c
=
–3
–1
–3
248 × 10
S m
20 mol m
= 124 × 10–4 S m2mol–1
Alternatively,
k =
1 |
1 | 2240-2243 | 248 S m–1
Concentration
= 0 02 mol L–1
= 1000 × 0 02 mol m–3 = 20 mol m–3
Molar conductivity
=
m
c
=
–3
–1
–3
248 × 10
S m
20 mol m
= 124 × 10–4 S m2mol–1
Alternatively,
k =
1 29 cm–1
520
= 0 |
1 | 2241-2244 | 02 mol L–1
= 1000 × 0 02 mol m–3 = 20 mol m–3
Molar conductivity
=
m
c
=
–3
–1
–3
248 × 10
S m
20 mol m
= 124 × 10–4 S m2mol–1
Alternatively,
k =
1 29 cm–1
520
= 0 248 × 10–2 S cm–1
Example 2 |
1 | 2242-2245 | 02 mol m–3 = 20 mol m–3
Molar conductivity
=
m
c
=
–3
–1
–3
248 × 10
S m
20 mol m
= 124 × 10–4 S m2mol–1
Alternatively,
k =
1 29 cm–1
520
= 0 248 × 10–2 S cm–1
Example 2 4
Example 2 |
1 | 2243-2246 | 29 cm–1
520
= 0 248 × 10–2 S cm–1
Example 2 4
Example 2 4
Example 2 |
1 | 2244-2247 | 248 × 10–2 S cm–1
Example 2 4
Example 2 4
Example 2 4
Example 2 |
1 | 2245-2248 | 4
Example 2 4
Example 2 4
Example 2 4
Example 2 |
1 | 2246-2249 | 4
Example 2 4
Example 2 4
Example 2 4
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
46
Chemistry
and
Lm = k × 1000 cm3 L–1 molarity–1
–2
–1
3
–1
–1
0 |
1 | 2247-2250 | 4
Example 2 4
Example 2 4
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
46
Chemistry
and
Lm = k × 1000 cm3 L–1 molarity–1
–2
–1
3
–1
–1
0 248×10
S cm
×1000 cm L
=
0 |
1 | 2248-2251 | 4
Example 2 4
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
46
Chemistry
and
Lm = k × 1000 cm3 L–1 molarity–1
–2
–1
3
–1
–1
0 248×10
S cm
×1000 cm L
=
0 02 mol L
= 124 S cm2 mol–1
The electrical resistance of a column of 0 |
1 | 2249-2252 | 4
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
46
Chemistry
and
Lm = k × 1000 cm3 L–1 molarity–1
–2
–1
3
–1
–1
0 248×10
S cm
×1000 cm L
=
0 02 mol L
= 124 S cm2 mol–1
The electrical resistance of a column of 0 05 mol L–1 NaOH solution of
diameter 1 cm and length 50 cm is 5 |
1 | 2250-2253 | 248×10
S cm
×1000 cm L
=
0 02 mol L
= 124 S cm2 mol–1
The electrical resistance of a column of 0 05 mol L–1 NaOH solution of
diameter 1 cm and length 50 cm is 5 55 × 103 ohm |
1 | 2251-2254 | 02 mol L
= 124 S cm2 mol–1
The electrical resistance of a column of 0 05 mol L–1 NaOH solution of
diameter 1 cm and length 50 cm is 5 55 × 103 ohm Calculate its
resistivity, conductivity and molar conductivity |
1 | 2252-2255 | 05 mol L–1 NaOH solution of
diameter 1 cm and length 50 cm is 5 55 × 103 ohm Calculate its
resistivity, conductivity and molar conductivity A = p r2 = 3 |
1 | 2253-2256 | 55 × 103 ohm Calculate its
resistivity, conductivity and molar conductivity A = p r2 = 3 14 × 0 |
1 | 2254-2257 | Calculate its
resistivity, conductivity and molar conductivity A = p r2 = 3 14 × 0 52 cm2 = 0 |
1 | 2255-2258 | A = p r2 = 3 14 × 0 52 cm2 = 0 785 cm2 = 0 |
1 | 2256-2259 | 14 × 0 52 cm2 = 0 785 cm2 = 0 785 × 10–4 m2
l = 50 cm = 0 |
1 | 2257-2260 | 52 cm2 = 0 785 cm2 = 0 785 × 10–4 m2
l = 50 cm = 0 5 m
=
l
R
A
or
3
2
5 |
1 | 2258-2261 | 785 cm2 = 0 785 × 10–4 m2
l = 50 cm = 0 5 m
=
l
R
A
or
3
2
5 55
10
0 |
1 | 2259-2262 | 785 × 10–4 m2
l = 50 cm = 0 5 m
=
l
R
A
or
3
2
5 55
10
0 785cm
50cm
RA
l
= 87 |
1 | 2260-2263 | 5 m
=
l
R
A
or
3
2
5 55
10
0 785cm
50cm
RA
l
= 87 135 W cm
Conductivity =
= 1
=
1
87 |
1 | 2261-2264 | 55
10
0 785cm
50cm
RA
l
= 87 135 W cm
Conductivity =
= 1
=
1
87 135
S cm–1
= 0 |
1 | 2262-2265 | 785cm
50cm
RA
l
= 87 135 W cm
Conductivity =
= 1
=
1
87 135
S cm–1
= 0 01148 S cm–1
Molar conductivity,
m
=
× 1000
c
cm3 L–1
=
–1
3
–1
–1
0 |
1 | 2263-2266 | 135 W cm
Conductivity =
= 1
=
1
87 135
S cm–1
= 0 01148 S cm–1
Molar conductivity,
m
=
× 1000
c
cm3 L–1
=
–1
3
–1
–1
0 01148 S cm
×1000 cm L
0 |
1 | 2264-2267 | 135
S cm–1
= 0 01148 S cm–1
Molar conductivity,
m
=
× 1000
c
cm3 L–1
=
–1
3
–1
–1
0 01148 S cm
×1000 cm L
0 05 mol L
= 229 |
1 | 2265-2268 | 01148 S cm–1
Molar conductivity,
m
=
× 1000
c
cm3 L–1
=
–1
3
–1
–1
0 01148 S cm
×1000 cm L
0 05 mol L
= 229 6 S cm2 mol–1
If we want to calculate the values of different quantities in terms of ‘m’
instead of ‘cm’,
= RA
l
=
3
–4
2
5 |
1 | 2266-2269 | 01148 S cm
×1000 cm L
0 05 mol L
= 229 6 S cm2 mol–1
If we want to calculate the values of different quantities in terms of ‘m’
instead of ‘cm’,
= RA
l
=
3
–4
2
5 55 × 10 × 0 |
1 | 2267-2270 | 05 mol L
= 229 6 S cm2 mol–1
If we want to calculate the values of different quantities in terms of ‘m’
instead of ‘cm’,
= RA
l
=
3
–4
2
5 55 × 10 × 0 785×10
m
0 |
1 | 2268-2271 | 6 S cm2 mol–1
If we want to calculate the values of different quantities in terms of ‘m’
instead of ‘cm’,
= RA
l
=
3
–4
2
5 55 × 10 × 0 785×10
m
0 5 m
= 87 |
1 | 2269-2272 | 55 × 10 × 0 785×10
m
0 5 m
= 87 135 ×10–2 W m
= 1
=
100
87 |
1 | 2270-2273 | 785×10
m
0 5 m
= 87 135 ×10–2 W m
= 1
=
100
87 135 m
= 1 |
1 | 2271-2274 | 5 m
= 87 135 ×10–2 W m
= 1
=
100
87 135 m
= 1 148 S m–1
and
m =
c
=
–1
1 |
1 | 2272-2275 | 135 ×10–2 W m
= 1
=
100
87 135 m
= 1 148 S m–1
and
m =
c
=
–1
1 148 S m–3
50 mol m
= 229 |
1 | 2273-2276 | 135 m
= 1 148 S m–1
and
m =
c
=
–1
1 148 S m–3
50 mol m
= 229 6 × 10–4 S m2 mol–1 |
1 | 2274-2277 | 148 S m–1
and
m =
c
=
–1
1 148 S m–3
50 mol m
= 229 6 × 10–4 S m2 mol–1 Example 2 |
1 | 2275-2278 | 148 S m–3
50 mol m
= 229 6 × 10–4 S m2 mol–1 Example 2 5
Example 2 |
1 | 2276-2279 | 6 × 10–4 S m2 mol–1 Example 2 5
Example 2 5
Example 2 |
1 | 2277-2280 | Example 2 5
Example 2 5
Example 2 5
Example 2 |
1 | 2278-2281 | 5
Example 2 5
Example 2 5
Example 2 5
Example 2 |
1 | 2279-2282 | 5
Example 2 5
Example 2 5
Example 2 5
Solution
Solution
Solution
Solution
Solution
Both conductivity and molar conductivity change with the
concentration of the electrolyte |
1 | 2280-2283 | 5
Example 2 5
Example 2 5
Solution
Solution
Solution
Solution
Solution
Both conductivity and molar conductivity change with the
concentration of the electrolyte Conductivity always decreases with
decrease in concentration both, for weak and strong electrolytes |
1 | 2281-2284 | 5
Example 2 5
Solution
Solution
Solution
Solution
Solution
Both conductivity and molar conductivity change with the
concentration of the electrolyte Conductivity always decreases with
decrease in concentration both, for weak and strong electrolytes This can be explained by the fact that the number of ions per unit
volume that carry the current in a solution decreases on dilution |
1 | 2282-2285 | 5
Solution
Solution
Solution
Solution
Solution
Both conductivity and molar conductivity change with the
concentration of the electrolyte Conductivity always decreases with
decrease in concentration both, for weak and strong electrolytes This can be explained by the fact that the number of ions per unit
volume that carry the current in a solution decreases on dilution The conductivity of a solution at any given concentration is the
conductance of one unit volume of solution kept between two
2 |
1 | 2283-2286 | Conductivity always decreases with
decrease in concentration both, for weak and strong electrolytes This can be explained by the fact that the number of ions per unit
volume that carry the current in a solution decreases on dilution The conductivity of a solution at any given concentration is the
conductance of one unit volume of solution kept between two
2 4 |
1 | 2284-2287 | This can be explained by the fact that the number of ions per unit
volume that carry the current in a solution decreases on dilution The conductivity of a solution at any given concentration is the
conductance of one unit volume of solution kept between two
2 4 2 Variation of
Conductivity
and Molar
Conductivity
with
Concentration
Rationalised 2023-24
47
Electrochemistry
platinum electrodes with unit area of cross section and at a distance
of unit length |
1 | 2285-2288 | The conductivity of a solution at any given concentration is the
conductance of one unit volume of solution kept between two
2 4 2 Variation of
Conductivity
and Molar
Conductivity
with
Concentration
Rationalised 2023-24
47
Electrochemistry
platinum electrodes with unit area of cross section and at a distance
of unit length This is clear from the equation:
=
A =
G
l
(both A and l are unity in their appropriate units in
m or cm)
Molar conductivity of a solution at a given concentration is the
conductance of the volume V of solution containing one mole of
electrolyte kept between two electrodes with area of cross section A and
distance of unit length |
1 | 2286-2289 | 4 2 Variation of
Conductivity
and Molar
Conductivity
with
Concentration
Rationalised 2023-24
47
Electrochemistry
platinum electrodes with unit area of cross section and at a distance
of unit length This is clear from the equation:
=
A =
G
l
(both A and l are unity in their appropriate units in
m or cm)
Molar conductivity of a solution at a given concentration is the
conductance of the volume V of solution containing one mole of
electrolyte kept between two electrodes with area of cross section A and
distance of unit length Therefore,
κ
Λ
κ
=
=
m
A
l
Since l = 1 and A = V ( volume containing 1 mole of electrolyte)
Lm = k V
(2 |
1 | 2287-2290 | 2 Variation of
Conductivity
and Molar
Conductivity
with
Concentration
Rationalised 2023-24
47
Electrochemistry
platinum electrodes with unit area of cross section and at a distance
of unit length This is clear from the equation:
=
A =
G
l
(both A and l are unity in their appropriate units in
m or cm)
Molar conductivity of a solution at a given concentration is the
conductance of the volume V of solution containing one mole of
electrolyte kept between two electrodes with area of cross section A and
distance of unit length Therefore,
κ
Λ
κ
=
=
m
A
l
Since l = 1 and A = V ( volume containing 1 mole of electrolyte)
Lm = k V
(2 22)
Molar conductivity increases with
decrease in concentration |
1 | 2288-2291 | This is clear from the equation:
=
A =
G
l
(both A and l are unity in their appropriate units in
m or cm)
Molar conductivity of a solution at a given concentration is the
conductance of the volume V of solution containing one mole of
electrolyte kept between two electrodes with area of cross section A and
distance of unit length Therefore,
κ
Λ
κ
=
=
m
A
l
Since l = 1 and A = V ( volume containing 1 mole of electrolyte)
Lm = k V
(2 22)
Molar conductivity increases with
decrease in concentration This is
because the total volume, V, of solution
containing one mole of electrolyte also
increases |
1 | 2289-2292 | Therefore,
κ
Λ
κ
=
=
m
A
l
Since l = 1 and A = V ( volume containing 1 mole of electrolyte)
Lm = k V
(2 22)
Molar conductivity increases with
decrease in concentration This is
because the total volume, V, of solution
containing one mole of electrolyte also
increases It has been found that decrease
in k on dilution of a solution is more
than compensated by increase in its
volume |
1 | 2290-2293 | 22)
Molar conductivity increases with
decrease in concentration This is
because the total volume, V, of solution
containing one mole of electrolyte also
increases It has been found that decrease
in k on dilution of a solution is more
than compensated by increase in its
volume Physically, it means that at a
given concentration, Lm can be defined
as the conductance of the electrolytic
solution kept between the electrodes of a
conductivity cell at unit distance but
having area of cross section large enough
to accommodate sufficient volume of
solution that contains one mole of the
electrolyte |
1 | 2291-2294 | This is
because the total volume, V, of solution
containing one mole of electrolyte also
increases It has been found that decrease
in k on dilution of a solution is more
than compensated by increase in its
volume Physically, it means that at a
given concentration, Lm can be defined
as the conductance of the electrolytic
solution kept between the electrodes of a
conductivity cell at unit distance but
having area of cross section large enough
to accommodate sufficient volume of
solution that contains one mole of the
electrolyte When
concentration
approaches zero, the molar conductivity
is
known
as
limiting
molar
conductivity and is represented by the
symbol L°m |
1 | 2292-2295 | It has been found that decrease
in k on dilution of a solution is more
than compensated by increase in its
volume Physically, it means that at a
given concentration, Lm can be defined
as the conductance of the electrolytic
solution kept between the electrodes of a
conductivity cell at unit distance but
having area of cross section large enough
to accommodate sufficient volume of
solution that contains one mole of the
electrolyte When
concentration
approaches zero, the molar conductivity
is
known
as
limiting
molar
conductivity and is represented by the
symbol L°m The variation in Lm with
concentration is different (Fig |
1 | 2293-2296 | Physically, it means that at a
given concentration, Lm can be defined
as the conductance of the electrolytic
solution kept between the electrodes of a
conductivity cell at unit distance but
having area of cross section large enough
to accommodate sufficient volume of
solution that contains one mole of the
electrolyte When
concentration
approaches zero, the molar conductivity
is
known
as
limiting
molar
conductivity and is represented by the
symbol L°m The variation in Lm with
concentration is different (Fig 2 |
1 | 2294-2297 | When
concentration
approaches zero, the molar conductivity
is
known
as
limiting
molar
conductivity and is represented by the
symbol L°m The variation in Lm with
concentration is different (Fig 2 6) for
strong and weak electrolytes |
1 | 2295-2298 | The variation in Lm with
concentration is different (Fig 2 6) for
strong and weak electrolytes Strong Electrolytes
For strong electrolytes, Lm increases slowly with dilution and can be
represented by the equation:
Lm = L°m – A c ½
(2 |
1 | 2296-2299 | 2 6) for
strong and weak electrolytes Strong Electrolytes
For strong electrolytes, Lm increases slowly with dilution and can be
represented by the equation:
Lm = L°m – A c ½
(2 23)
It can be seen that if we plot (Fig |
1 | 2297-2300 | 6) for
strong and weak electrolytes Strong Electrolytes
For strong electrolytes, Lm increases slowly with dilution and can be
represented by the equation:
Lm = L°m – A c ½
(2 23)
It can be seen that if we plot (Fig 2 |
1 | 2298-2301 | Strong Electrolytes
For strong electrolytes, Lm increases slowly with dilution and can be
represented by the equation:
Lm = L°m – A c ½
(2 23)
It can be seen that if we plot (Fig 2 6) Lm against
c1/2, we obtain a straight line with intercept equal to L°m and slope
equal to ‘–A’ |
1 | 2299-2302 | 23)
It can be seen that if we plot (Fig 2 6) Lm against
c1/2, we obtain a straight line with intercept equal to L°m and slope
equal to ‘–A’ The value of the constant ‘A’ for a given solvent and
temperature depends on the type of electrolyte i |
1 | 2300-2303 | 2 6) Lm against
c1/2, we obtain a straight line with intercept equal to L°m and slope
equal to ‘–A’ The value of the constant ‘A’ for a given solvent and
temperature depends on the type of electrolyte i e |
1 | 2301-2304 | 6) Lm against
c1/2, we obtain a straight line with intercept equal to L°m and slope
equal to ‘–A’ The value of the constant ‘A’ for a given solvent and
temperature depends on the type of electrolyte i e , the charges on the
cation and anion produced on the dissociation of the electrolyte in the
solution |
1 | 2302-2305 | The value of the constant ‘A’ for a given solvent and
temperature depends on the type of electrolyte i e , the charges on the
cation and anion produced on the dissociation of the electrolyte in the
solution Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2
electrolytes respectively |
1 | 2303-2306 | e , the charges on the
cation and anion produced on the dissociation of the electrolyte in the
solution Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2
electrolytes respectively All electrolytes of a particular type have the
same value for ‘A’ |
1 | 2304-2307 | , the charges on the
cation and anion produced on the dissociation of the electrolyte in the
solution Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2
electrolytes respectively All electrolytes of a particular type have the
same value for ‘A’ Fig |
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