Ionic Conductance
Trending Questions
If Λom(HA)=190 S cm2mol−1, the ionization constant (Ka) of HA is equal to ____ ×10−6.
(Round off to the Nearest Integer)
- 1.166 cm−1
- 2.173 cm−1
- 3.324 cm−1
- 4.616 cm−1
Calculate the equivalent conductance (in ohm−1 cm2 eq−1) of acetic acid at infinite dilution.
- 1.166 cm-1
- 2.173 cm-1
- 3.324 cm-1
- 4.616 cm-1
- 0.912 cm−1
- 0.512 cm−1
- 0.728 cm−1
- 0.632 cm−1
- 0.15
- 1.5
- 0.015
- 150
- formation of micelles
- emulsification of oil or grease
- lowering of surface tension of water
- all of the above
- 50 m−1
- 25 m−1
- 56 m−1
- 28 m−1
Limiting molar conductivities at 25oC:
HCl=426 Ω−1cm2mol−1
NaCl=126 Ω−1cm2mol−1
NaC(sodium crotonate)=83 Ω−1cm2mol−1.
(Crotonate ion is represented by C−)
- 10−3
1.11×10−5
1.11×10−4
0.01
- 0.25 S cm−1
- 0.025 S cm−1
- 0.0025 S cm−1
- None of the above
- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Assertion is incorrect but Reason is correct
- minimium molar conductivity
- limiting molar conductivity
- maximum molar conductivity
- None of these
The emf of a concentration cell consisting of two zinc electrode, one dipping into M/4 solution of zinc sulphate and the other into M/16 solution of the same salt at 250C :
- 0.0125 V
- 0.0250 V
- 0.0178 V
- 0.0356 V
- 2
- 3
- 5
- 6
- R = 66.67 ohm.
- R = 86.67 ohm.
- R = 96.67 ohm.
- R = 76.67 ohm.
3Cl2+6NaOH→5NaCl+NaClO3+3H2O
- 59.1
- 42.6
- 71
- 35.5
- I : KCl; II : CH3OH;
III : CH3(CH2)11OSO−3Na+ - I : CH3(CH2)11OSO−3Na+;
II : CH3OH; III : KCl - I : KCl;
II : CH3(CH2)11OSO−3Na+;
III : CH3OH - I : CH3OH; II : KCl;
III : CH3(CH2)11OSO−3Na+
- 217.3 Ω−1cm2
- 272.1 Ω−1cm2
- 191.05 Ω−1cm2
- 382.1 Ω−1cm2
- 204 S cm2 mol−1
- 2.04 S cm2 mol−1
- 20.4 S cm2 mol−1
- 0.204 S cm2 mol−1
For a solution of an electrolyte at a particular concentration and temperature, which of the following is not the function of cell constant?
1) specific conductance
2) molar conductance
3) equivalent conductance
4) conductance
- 1, 2
- 1, 2, 3
- 2, 3, 4
- 1, 4
- λc=λ∞+(B)C
- λc=λ∞−(B)C
- λc=λ∞−B√C
- λc=λ∞+B√C
- 0.2
- 0.02
- 0.002
- None of these
- R = 96.67 ohm.
- R = 76.67 ohm.
- R = 86.67 ohm.
- R = 66.67 ohm.
- 1.166 cm−1
- 2.173 cm−1
- 3.324 cm−1
- 4.616 cm−1
The molar conductivities of ∧0NaOAC and ∧0HCl at infinite dilution in water at 250C and 91.0 and 426.2Scm2/mol respectively. To calculate ∧0HOAc, the additional value required is:
- ∧0NaCl
- ∧0H2O
- ∧0NaOH
- ∧0KCl
|
List I |
List II |
|
P. |
(C2H5)3NX+CH3COOHY |
1. |
Conductivity decreases and then increases |
Q. |
KI (0.1 M)X+AgNO3(0.01 M)Y |
2. |
Conductivity decreases and then does notchange much |
R. |
CH3COOHX+KOHY |
3. |
Conductivity increases and then does notchange much |
S. |
NaOHX+HIY |
4. |
Conductivity does not change much andthen increases |
P |
Q |
R |
S |
|
(A) |
3 |
4 |
2 |
1 |
(B) |
4 |
3 |
2 |
1 |
(C) |
2 |
3 |
4 |
1 |
(D) |
1 |
4 |
3 |
2 |
- (B)
- (D)
- (C)
- (A)
- λc=λ∞−b√c
- λc=λ∞+b√c
- λc=λ∞+bc
- λc=λ∞−bc
- λC=λ∞+(B)C
- λC=λ∞−(B)C
- λC=λ∞−(B)√C
- λC=λ∞+(B)√C
- I : KCl; II : CH3OH;
III : CH3(CH2)11OSO−3Na+ - I : CH3(CH2)11OSO−3Na+;
II : CH3OH; III : KCl - I : KCl;
II : CH3(CH2)11OSO−3Na+;
III : CH3OH - I : CH3OH; II : KCl;
III : CH3(CH2)11OSO−3Na+