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1 | 2305-2308 | 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 2 |
1 | 2306-2309 | All electrolytes of a particular type have the
same value for ‘A’ Fig 2 6: Molar conductivity versus c½ for acetic
acid (weak electrolyte) and potassium
chloride (strong electrolyte) in aqueous
solutions |
1 | 2307-2310 | Fig 2 6: Molar conductivity versus c½ for acetic
acid (weak electrolyte) and potassium
chloride (strong electrolyte) in aqueous
solutions Rationalised 2023-24
48
Chemistry
The molar conductivity of KCl solutions at different concentrations at
298 K are given below:
c/mol L–1
LLLLLm/S cm2 mol–1
0 |
1 | 2308-2311 | 2 6: Molar conductivity versus c½ for acetic
acid (weak electrolyte) and potassium
chloride (strong electrolyte) in aqueous
solutions Rationalised 2023-24
48
Chemistry
The molar conductivity of KCl solutions at different concentrations at
298 K are given below:
c/mol L–1
LLLLLm/S cm2 mol–1
0 000198
148 |
1 | 2309-2312 | 6: Molar conductivity versus c½ for acetic
acid (weak electrolyte) and potassium
chloride (strong electrolyte) in aqueous
solutions Rationalised 2023-24
48
Chemistry
The molar conductivity of KCl solutions at different concentrations at
298 K are given below:
c/mol L–1
LLLLLm/S cm2 mol–1
0 000198
148 61
0 |
1 | 2310-2313 | Rationalised 2023-24
48
Chemistry
The molar conductivity of KCl solutions at different concentrations at
298 K are given below:
c/mol L–1
LLLLLm/S cm2 mol–1
0 000198
148 61
0 000309
148 |
1 | 2311-2314 | 000198
148 61
0 000309
148 29
0 |
1 | 2312-2315 | 61
0 000309
148 29
0 000521
147 |
1 | 2313-2316 | 000309
148 29
0 000521
147 81
0 |
1 | 2314-2317 | 29
0 000521
147 81
0 000989
147 |
1 | 2315-2318 | 000521
147 81
0 000989
147 09
Show that a plot between Lm and c1/2 is a straight line |
1 | 2316-2319 | 81
0 000989
147 09
Show that a plot between Lm and c1/2 is a straight line Determine the
values of L°m
and A for KCl |
1 | 2317-2320 | 000989
147 09
Show that a plot between Lm and c1/2 is a straight line Determine the
values of L°m
and A for KCl Taking the square root of concentration we obtain:
c1/2/(mol L–1 )1/2
LLLLLm/S cm2mol–1
0 |
1 | 2318-2321 | 09
Show that a plot between Lm and c1/2 is a straight line Determine the
values of L°m
and A for KCl Taking the square root of concentration we obtain:
c1/2/(mol L–1 )1/2
LLLLLm/S cm2mol–1
0 01407
148 |
1 | 2319-2322 | Determine the
values of L°m
and A for KCl Taking the square root of concentration we obtain:
c1/2/(mol L–1 )1/2
LLLLLm/S cm2mol–1
0 01407
148 61
0 |
1 | 2320-2323 | Taking the square root of concentration we obtain:
c1/2/(mol L–1 )1/2
LLLLLm/S cm2mol–1
0 01407
148 61
0 01758
148 |
1 | 2321-2324 | 01407
148 61
0 01758
148 29
0 |
1 | 2322-2325 | 61
0 01758
148 29
0 02283
147 |
1 | 2323-2326 | 01758
148 29
0 02283
147 81
0 |
1 | 2324-2327 | 29
0 02283
147 81
0 03145
147 |
1 | 2325-2328 | 02283
147 81
0 03145
147 09
A plot of Lm ( y-axis) and c1/2 (x-axis) is shown in (Fig |
1 | 2326-2329 | 81
0 03145
147 09
A plot of Lm ( y-axis) and c1/2 (x-axis) is shown in (Fig 3 |
1 | 2327-2330 | 03145
147 09
A plot of Lm ( y-axis) and c1/2 (x-axis) is shown in (Fig 3 7) |
1 | 2328-2331 | 09
A plot of Lm ( y-axis) and c1/2 (x-axis) is shown in (Fig 3 7) It can be seen that it is nearly a straight line |
1 | 2329-2332 | 3 7) It can be seen that it is nearly a straight line From the intercept
(c1/2 = 0), we find that
L°m = 150 |
1 | 2330-2333 | 7) It can be seen that it is nearly a straight line From the intercept
(c1/2 = 0), we find that
L°m = 150 0 S cm2 mol–1 and
A = – slope = 87 |
1 | 2331-2334 | It can be seen that it is nearly a straight line From the intercept
(c1/2 = 0), we find that
L°m = 150 0 S cm2 mol–1 and
A = – slope = 87 46 S cm2 mol–1/(mol/L–1)1/2 |
1 | 2332-2335 | From the intercept
(c1/2 = 0), we find that
L°m = 150 0 S cm2 mol–1 and
A = – slope = 87 46 S cm2 mol–1/(mol/L–1)1/2 Example 2 |
1 | 2333-2336 | 0 S cm2 mol–1 and
A = – slope = 87 46 S cm2 mol–1/(mol/L–1)1/2 Example 2 6
Example 2 |
1 | 2334-2337 | 46 S cm2 mol–1/(mol/L–1)1/2 Example 2 6
Example 2 6
Example 2 |
1 | 2335-2338 | Example 2 6
Example 2 6
Example 2 6
Example 2 |
1 | 2336-2339 | 6
Example 2 6
Example 2 6
Example 2 6
Example 2 |
1 | 2337-2340 | 6
Example 2 6
Example 2 6
Example 2 6
Solution
Solution
Solution
Solution
Solution
Fig |
1 | 2338-2341 | 6
Example 2 6
Example 2 6
Solution
Solution
Solution
Solution
Solution
Fig 2 |
1 | 2339-2342 | 6
Example 2 6
Solution
Solution
Solution
Solution
Solution
Fig 2 7: Variation of Lm against c½ |
1 | 2340-2343 | 6
Solution
Solution
Solution
Solution
Solution
Fig 2 7: Variation of Lm against c½ Rationalised 2023-24
49
Electrochemistry
Kohlrausch examined L°m values for a number of strong electrolytes
and observed certain regularities |
1 | 2341-2344 | 2 7: Variation of Lm against c½ Rationalised 2023-24
49
Electrochemistry
Kohlrausch examined L°m values for a number of strong electrolytes
and observed certain regularities He noted that the difference in L°m of
the electrolytes NaX and KX for any X is nearly constant |
1 | 2342-2345 | 7: Variation of Lm against c½ Rationalised 2023-24
49
Electrochemistry
Kohlrausch examined L°m values for a number of strong electrolytes
and observed certain regularities He noted that the difference in L°m of
the electrolytes NaX and KX for any X is nearly constant For example
at 298 K:
L°
m (KCl) – L°
m (NaCl) = L°
m (KBr) – L°
m (NaBr)
= L°
m (KI) – L°
m (NaI) ≃ 23 |
1 | 2343-2346 | Rationalised 2023-24
49
Electrochemistry
Kohlrausch examined L°m values for a number of strong electrolytes
and observed certain regularities He noted that the difference in L°m of
the electrolytes NaX and KX for any X is nearly constant For example
at 298 K:
L°
m (KCl) – L°
m (NaCl) = L°
m (KBr) – L°
m (NaBr)
= L°
m (KI) – L°
m (NaI) ≃ 23 4 S cm2 mol–1
and similarly it was found that
L°
m (NaBr)– L°
m (NaCl) = L°
m (KBr) – L°
m (KCl) ≃ 1 |
1 | 2344-2347 | He noted that the difference in L°m of
the electrolytes NaX and KX for any X is nearly constant For example
at 298 K:
L°
m (KCl) – L°
m (NaCl) = L°
m (KBr) – L°
m (NaBr)
= L°
m (KI) – L°
m (NaI) ≃ 23 4 S cm2 mol–1
and similarly it was found that
L°
m (NaBr)– L°
m (NaCl) = L°
m (KBr) – L°
m (KCl) ≃ 1 8 S cm2 mol–1
On the basis of the above observations he enunciated Kohlrausch
law of independent migration of ions |
1 | 2345-2348 | For example
at 298 K:
L°
m (KCl) – L°
m (NaCl) = L°
m (KBr) – L°
m (NaBr)
= L°
m (KI) – L°
m (NaI) ≃ 23 4 S cm2 mol–1
and similarly it was found that
L°
m (NaBr)– L°
m (NaCl) = L°
m (KBr) – L°
m (KCl) ≃ 1 8 S cm2 mol–1
On the basis of the above observations he enunciated Kohlrausch
law of independent migration of ions The law states that limiting
molar conductivity of an electrolyte can be represented as the sum of the
individual contributions of the anion and cation of the electrolyte |
1 | 2346-2349 | 4 S cm2 mol–1
and similarly it was found that
L°
m (NaBr)– L°
m (NaCl) = L°
m (KBr) – L°
m (KCl) ≃ 1 8 S cm2 mol–1
On the basis of the above observations he enunciated Kohlrausch
law of independent migration of ions The law states that limiting
molar conductivity of an electrolyte can be represented as the sum of the
individual contributions of the anion and cation of the electrolyte Thus,
if l°
Na+ and l°
Cl
– are limiting molar conductivity of the sodium and chloride
ions respectively, then the limiting molar conductivity for sodium chloride
is given by the equation:
L°m (NaCl) = l°
Na+ + l°
Cl
–
(2 |
1 | 2347-2350 | 8 S cm2 mol–1
On the basis of the above observations he enunciated Kohlrausch
law of independent migration of ions The law states that limiting
molar conductivity of an electrolyte can be represented as the sum of the
individual contributions of the anion and cation of the electrolyte Thus,
if l°
Na+ and l°
Cl
– are limiting molar conductivity of the sodium and chloride
ions respectively, then the limiting molar conductivity for sodium chloride
is given by the equation:
L°m (NaCl) = l°
Na+ + l°
Cl
–
(2 24)
In general, if an electrolyte on dissociation gives n+ cations and n–
anions then its limiting molar conductivity is given by:
L°m = n+ l°
+ + n– l°
–
(2 |
1 | 2348-2351 | The law states that limiting
molar conductivity of an electrolyte can be represented as the sum of the
individual contributions of the anion and cation of the electrolyte Thus,
if l°
Na+ and l°
Cl
– are limiting molar conductivity of the sodium and chloride
ions respectively, then the limiting molar conductivity for sodium chloride
is given by the equation:
L°m (NaCl) = l°
Na+ + l°
Cl
–
(2 24)
In general, if an electrolyte on dissociation gives n+ cations and n–
anions then its limiting molar conductivity is given by:
L°m = n+ l°
+ + n– l°
–
(2 25)
Here, l°
+ and l°
– are the limiting molar conductivities of the cation
and anion respectively |
1 | 2349-2352 | Thus,
if l°
Na+ and l°
Cl
– are limiting molar conductivity of the sodium and chloride
ions respectively, then the limiting molar conductivity for sodium chloride
is given by the equation:
L°m (NaCl) = l°
Na+ + l°
Cl
–
(2 24)
In general, if an electrolyte on dissociation gives n+ cations and n–
anions then its limiting molar conductivity is given by:
L°m = n+ l°
+ + n– l°
–
(2 25)
Here, l°
+ and l°
– are the limiting molar conductivities of the cation
and anion respectively The values of l° for some cations and anions at
298 K are given in Table 2 |
1 | 2350-2353 | 24)
In general, if an electrolyte on dissociation gives n+ cations and n–
anions then its limiting molar conductivity is given by:
L°m = n+ l°
+ + n– l°
–
(2 25)
Here, l°
+ and l°
– are the limiting molar conductivities of the cation
and anion respectively The values of l° for some cations and anions at
298 K are given in Table 2 4 |
1 | 2351-2354 | 25)
Here, l°
+ and l°
– are the limiting molar conductivities of the cation
and anion respectively The values of l° for some cations and anions at
298 K are given in Table 2 4 Table 2 |
1 | 2352-2355 | The values of l° for some cations and anions at
298 K are given in Table 2 4 Table 2 4:
Limiting Molar Conductivity for some
Ions in Water at 298 K
Weak Electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at
higher concentrations and hence for such electrolytes, the change in Lm
with dilution is due to increase in the degree of dissociation and
consequently the number of ions in total volume of solution that contains
1 mol of electrolyte |
1 | 2353-2356 | 4 Table 2 4:
Limiting Molar Conductivity for some
Ions in Water at 298 K
Weak Electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at
higher concentrations and hence for such electrolytes, the change in Lm
with dilution is due to increase in the degree of dissociation and
consequently the number of ions in total volume of solution that contains
1 mol of electrolyte In such cases Lm increases steeply (Fig |
1 | 2354-2357 | Table 2 4:
Limiting Molar Conductivity for some
Ions in Water at 298 K
Weak Electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at
higher concentrations and hence for such electrolytes, the change in Lm
with dilution is due to increase in the degree of dissociation and
consequently the number of ions in total volume of solution that contains
1 mol of electrolyte In such cases Lm increases steeply (Fig 2 |
1 | 2355-2358 | 4:
Limiting Molar Conductivity for some
Ions in Water at 298 K
Weak Electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at
higher concentrations and hence for such electrolytes, the change in Lm
with dilution is due to increase in the degree of dissociation and
consequently the number of ions in total volume of solution that contains
1 mol of electrolyte In such cases Lm increases steeply (Fig 2 6) on
dilution, especially near lower concentrations |
1 | 2356-2359 | In such cases Lm increases steeply (Fig 2 6) on
dilution, especially near lower concentrations Therefore, L°m cannot be
obtained by extrapolation of Lm to zero concentration |
1 | 2357-2360 | 2 6) on
dilution, especially near lower concentrations Therefore, L°m cannot be
obtained by extrapolation of Lm to zero concentration At infinite dilution
(i |
1 | 2358-2361 | 6) on
dilution, especially near lower concentrations Therefore, L°m cannot be
obtained by extrapolation of Lm to zero concentration At infinite dilution
(i e |
1 | 2359-2362 | Therefore, L°m cannot be
obtained by extrapolation of Lm to zero concentration At infinite dilution
(i e , concentration c ® zero) electrolyte dissociates completely (a =1),
but at such low concentration the conductivity of the solution is so low
that it cannot be measured accurately |
1 | 2360-2363 | At infinite dilution
(i e , concentration c ® zero) electrolyte dissociates completely (a =1),
but at such low concentration the conductivity of the solution is so low
that it cannot be measured accurately Therefore, L°m for weak electrolytes
is obtained by using Kohlrausch law of independent migration of ions
(Example 2 |
1 | 2361-2364 | e , concentration c ® zero) electrolyte dissociates completely (a =1),
but at such low concentration the conductivity of the solution is so low
that it cannot be measured accurately Therefore, L°m for weak electrolytes
is obtained by using Kohlrausch law of independent migration of ions
(Example 2 8) |
1 | 2362-2365 | , concentration c ® zero) electrolyte dissociates completely (a =1),
but at such low concentration the conductivity of the solution is so low
that it cannot be measured accurately Therefore, L°m for weak electrolytes
is obtained by using Kohlrausch law of independent migration of ions
(Example 2 8) At any concentration c, if a is the degree of dissociation
Ion
lllll0/(S cm2mol–1)
Ion
lllll 0/(S cm2 mol–1)
H+
349 |
1 | 2363-2366 | Therefore, L°m for weak electrolytes
is obtained by using Kohlrausch law of independent migration of ions
(Example 2 8) At any concentration c, if a is the degree of dissociation
Ion
lllll0/(S cm2mol–1)
Ion
lllll 0/(S cm2 mol–1)
H+
349 6
OH–
199 |
1 | 2364-2367 | 8) At any concentration c, if a is the degree of dissociation
Ion
lllll0/(S cm2mol–1)
Ion
lllll 0/(S cm2 mol–1)
H+
349 6
OH–
199 1
Na+
50 |
1 | 2365-2368 | At any concentration c, if a is the degree of dissociation
Ion
lllll0/(S cm2mol–1)
Ion
lllll 0/(S cm2 mol–1)
H+
349 6
OH–
199 1
Na+
50 1
Cl–
76 |
1 | 2366-2369 | 6
OH–
199 1
Na+
50 1
Cl–
76 3
K+
73 |
1 | 2367-2370 | 1
Na+
50 1
Cl–
76 3
K+
73 5
Br–
78 |
1 | 2368-2371 | 1
Cl–
76 3
K+
73 5
Br–
78 1
Ca2+
119 |
1 | 2369-2372 | 3
K+
73 5
Br–
78 1
Ca2+
119 0
CH3COO–
40 |
1 | 2370-2373 | 5
Br–
78 1
Ca2+
119 0
CH3COO–
40 9
Mg2+
106 |
1 | 2371-2374 | 1
Ca2+
119 0
CH3COO–
40 9
Mg2+
106 0
SO4
2
160 |
1 | 2372-2375 | 0
CH3COO–
40 9
Mg2+
106 0
SO4
2
160 0
Rationalised 2023-24
50
Chemistry
then it can be approximated to the ratio of molar conductivity Lm at the
concentration c to limiting molar conductivity, L
0
m |
1 | 2373-2376 | 9
Mg2+
106 0
SO4
2
160 0
Rationalised 2023-24
50
Chemistry
then it can be approximated to the ratio of molar conductivity Lm at the
concentration c to limiting molar conductivity, L
0
m Thus we have:
°
=
m
m
(2 |
1 | 2374-2377 | 0
SO4
2
160 0
Rationalised 2023-24
50
Chemistry
then it can be approximated to the ratio of molar conductivity Lm at the
concentration c to limiting molar conductivity, L
0
m Thus we have:
°
=
m
m
(2 26)
But we know that for a weak electrolyte like acetic acid (Class XI,
Unit 7),
2
2
2
2
=
=
=
a
1
1
m
m
m
m
m
m
m
m
c
c
c
K
(2 |
1 | 2375-2378 | 0
Rationalised 2023-24
50
Chemistry
then it can be approximated to the ratio of molar conductivity Lm at the
concentration c to limiting molar conductivity, L
0
m Thus we have:
°
=
m
m
(2 26)
But we know that for a weak electrolyte like acetic acid (Class XI,
Unit 7),
2
2
2
2
=
=
=
a
1
1
m
m
m
m
m
m
m
m
c
c
c
K
(2 27)
Applications of Kohlrausch law
Using Kohlrausch law of independent migration of ions, it is possible to
calculate L
0
m for any electrolyte from the lo of individual ions |
1 | 2376-2379 | Thus we have:
°
=
m
m
(2 26)
But we know that for a weak electrolyte like acetic acid (Class XI,
Unit 7),
2
2
2
2
=
=
=
a
1
1
m
m
m
m
m
m
m
m
c
c
c
K
(2 27)
Applications of Kohlrausch law
Using Kohlrausch law of independent migration of ions, it is possible to
calculate L
0
m for any electrolyte from the lo of individual ions Moreover,
for weak electrolytes like acetic acid it is possible to determine the value
of its dissociation constant once we know the L
0
m and Lm at a given
concentration c |
1 | 2377-2380 | 26)
But we know that for a weak electrolyte like acetic acid (Class XI,
Unit 7),
2
2
2
2
=
=
=
a
1
1
m
m
m
m
m
m
m
m
c
c
c
K
(2 27)
Applications of Kohlrausch law
Using Kohlrausch law of independent migration of ions, it is possible to
calculate L
0
m for any electrolyte from the lo of individual ions Moreover,
for weak electrolytes like acetic acid it is possible to determine the value
of its dissociation constant once we know the L
0
m and Lm at a given
concentration c Calculate L
0
m for CaCl2 and MgSO4 from the data given in Table 3 |
1 | 2378-2381 | 27)
Applications of Kohlrausch law
Using Kohlrausch law of independent migration of ions, it is possible to
calculate L
0
m for any electrolyte from the lo of individual ions Moreover,
for weak electrolytes like acetic acid it is possible to determine the value
of its dissociation constant once we know the L
0
m and Lm at a given
concentration c Calculate L
0
m for CaCl2 and MgSO4 from the data given in Table 3 4 |
1 | 2379-2382 | Moreover,
for weak electrolytes like acetic acid it is possible to determine the value
of its dissociation constant once we know the L
0
m and Lm at a given
concentration c Calculate L
0
m for CaCl2 and MgSO4 from the data given in Table 3 4 We know from Kohlrausch law that
mCaCl2
=
2+
–
Ca
2Cl
= 119 |
1 | 2380-2383 | Calculate L
0
m for CaCl2 and MgSO4 from the data given in Table 3 4 We know from Kohlrausch law that
mCaCl2
=
2+
–
Ca
2Cl
= 119 0 S cm2 mol–1 + 2(76 |
1 | 2381-2384 | 4 We know from Kohlrausch law that
mCaCl2
=
2+
–
Ca
2Cl
= 119 0 S cm2 mol–1 + 2(76 3) S cm2 mol–1
= (119 |
1 | 2382-2385 | We know from Kohlrausch law that
mCaCl2
=
2+
–
Ca
2Cl
= 119 0 S cm2 mol–1 + 2(76 3) S cm2 mol–1
= (119 0 + 152 |
1 | 2383-2386 | 0 S cm2 mol–1 + 2(76 3) S cm2 mol–1
= (119 0 + 152 6) S cm2 mol–1
= 271 |
1 | 2384-2387 | 3) S cm2 mol–1
= (119 0 + 152 6) S cm2 mol–1
= 271 6 S cm2 mol–1
mMgSO4
=
2–
2+
4
Mg
SO
= 106 |
1 | 2385-2388 | 0 + 152 6) S cm2 mol–1
= 271 6 S cm2 mol–1
mMgSO4
=
2–
2+
4
Mg
SO
= 106 0 S cm2 mol–1 + 160 |
1 | 2386-2389 | 6) S cm2 mol–1
= 271 6 S cm2 mol–1
mMgSO4
=
2–
2+
4
Mg
SO
= 106 0 S cm2 mol–1 + 160 0 S cm2 mol–1
= 266 S cm2 mol–1 |
1 | 2387-2390 | 6 S cm2 mol–1
mMgSO4
=
2–
2+
4
Mg
SO
= 106 0 S cm2 mol–1 + 160 0 S cm2 mol–1
= 266 S cm2 mol–1 L
0
m for NaCl, HCl and NaAc are 126 |
1 | 2388-2391 | 0 S cm2 mol–1 + 160 0 S cm2 mol–1
= 266 S cm2 mol–1 L
0
m for NaCl, HCl and NaAc are 126 4, 425 |
1 | 2389-2392 | 0 S cm2 mol–1
= 266 S cm2 mol–1 L
0
m for NaCl, HCl and NaAc are 126 4, 425 9 and 91 |
1 | 2390-2393 | L
0
m for NaCl, HCl and NaAc are 126 4, 425 9 and 91 0 S cm2 mol–1
respectively |
1 | 2391-2394 | 4, 425 9 and 91 0 S cm2 mol–1
respectively Calculate L
0 for HAc |
1 | 2392-2395 | 9 and 91 0 S cm2 mol–1
respectively Calculate L
0 for HAc
mHAc
=
+
–
H
Ac
+
–
–
+
–
+
H
Cl
Ac
Na
Cl
Na
=
HCl
NaAc
NaCl
m
m
m
= (425 |
1 | 2393-2396 | 0 S cm2 mol–1
respectively Calculate L
0 for HAc
mHAc
=
+
–
H
Ac
+
–
–
+
–
+
H
Cl
Ac
Na
Cl
Na
=
HCl
NaAc
NaCl
m
m
m
= (425 9 + 91 |
1 | 2394-2397 | Calculate L
0 for HAc
mHAc
=
+
–
H
Ac
+
–
–
+
–
+
H
Cl
Ac
Na
Cl
Na
=
HCl
NaAc
NaCl
m
m
m
= (425 9 + 91 0 – 126 |
1 | 2395-2398 |
mHAc
=
+
–
H
Ac
+
–
–
+
–
+
H
Cl
Ac
Na
Cl
Na
=
HCl
NaAc
NaCl
m
m
m
= (425 9 + 91 0 – 126 4 ) S cm2 mol –1
= 390 |
1 | 2396-2399 | 9 + 91 0 – 126 4 ) S cm2 mol –1
= 390 5 S cm2 mol–1 |
1 | 2397-2400 | 0 – 126 4 ) S cm2 mol –1
= 390 5 S cm2 mol–1 The conductivity of 0 |
1 | 2398-2401 | 4 ) S cm2 mol –1
= 390 5 S cm2 mol–1 The conductivity of 0 001028 mol L–1 acetic acid is 4 |
1 | 2399-2402 | 5 S cm2 mol–1 The conductivity of 0 001028 mol L–1 acetic acid is 4 95 × 10–5 S cm–1 |
1 | 2400-2403 | The conductivity of 0 001028 mol L–1 acetic acid is 4 95 × 10–5 S cm–1 Calculate its dissociation constant if L
0
m for acetic acid is
390 |
1 | 2401-2404 | 001028 mol L–1 acetic acid is 4 95 × 10–5 S cm–1 Calculate its dissociation constant if L
0
m for acetic acid is
390 5 S cm2 mol–1 |
1 | 2402-2405 | 95 × 10–5 S cm–1 Calculate its dissociation constant if L
0
m for acetic acid is
390 5 S cm2 mol–1 m
= |
1 | 2403-2406 | Calculate its dissociation constant if L
0
m for acetic acid is
390 5 S cm2 mol–1 m
= 5
1
3
1
4 95 10
Scm
1000cm
0 001028 mol L
L
c
= 48 |
1 | 2404-2407 | 5 S cm2 mol–1 m
= 5
1
3
1
4 95 10
Scm
1000cm
0 001028 mol L
L
c
= 48 15 S cm3 mol–1
a
=
2
1
2
1
48 |
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