Equivalent, Molar Conductivity and Cell Constant
Trending Questions
Q. The resistance of 0.01 N NaCl solution at 25∘C is 200 Ω. Cell constant of conductivity cell is 1 cm−1. The equivalent conductance is :
- 5×102 Ω−1 cm2 (g.eq)−1
- 6×103 Ω−1 cm2 (g.eq)−1
- 7×104 Ω−1 cm2 (g.eq)−1
- 8×105 Ω−1 cm2 (g.eq)−1
Q. The molar ionic conductances at infinite dilution of Mg2+ and Cl− are 106.1 and 76.3 ohm−1 ohm−1 cm2 mol−1 respectively. The Molar conductance of solution of MgCl2 at
infinite dilution is
infinite dilution is
- 29.8ohm−1cm2mol−1
- 183.4ohm−1cm2mol−1
- 285.7ohm−1cm2mol−1
- 258.7ohm−1cm2mol−1
Q.
100 mL of 1.0 M solution of a monobasic acid (pKa=5) is titrated against Ca(OH)2 solution. At equivalence point the pH of solution is ?
6
4.5
9.5
7
Q.
Conductivity (unit Siemen's) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel then the unit of the constant of proportionality is
[AIEEE 2002]
Q. At 25oC for the given cell,
Ag(s)|Ag+(aq, x M)||Ag+(aq, 1.0 M)|Ag(s)
Ecell=0.26 V
What will be the value of log10(x) ?
Here x is the concentration of electrolyte
Ag(s)|Ag+(aq, x M)||Ag+(aq, 1.0 M)|Ag(s)
Ecell=0.26 V
What will be the value of log10(x) ?
Here x is the concentration of electrolyte
- −2.2
- +2.2
- +4.4
- −4.4
Q. The factor(s) which affect conductivity of a solution is/are:
- All of the above.
- Charge and size of ions
- Concentration of ions
- Ease with which the ions move under a potential gradient
Q.
Equivalent conductance of 0.1 M HA(weak acid) solution is 1Scm2equivalent−1 and that at infinite dilution is 1Scm2equivalent−1 Hence pH of HA solution is
1.3
1.7
2.3
3
Q.
Specific conductance of 0.1 MCH3COOH at 25∘ is 3.9×10−4ohm−1cm−1. If λ∞(H3O+) and λ∞(VH3COO−) at 25∘C are 349.0 and 41.0 ohm−1cm2mol−1 respectively, degree of ionization of CH3COOH at the given concentration is :
2.0%
1.0%
4.0%
5.0%
Q. The molar conductivity of a 0.5mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76×10−3S cm−1 at 298 K is
- 0.086 S cm2/mol
- 28.8 S cm2/mol
- 2.88 S cm2/mol
- 11.52 S cm2/mol
Q. The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55×103 ohm. Calculate its resistivity, conductivity, and molar conductivity.
- 87.135×10−2ohm m, 1.148 Sm−1, 229.6×10−4S m2 mol−1
- 67.135×10−2ohm m, 1.489 Sm−1, 229.6×10−4S m2 mol−1
- 87.135×10−2ohm m, 1.296 Sm−1, 229.6×10−4S m2 mol−1
- 77.135×10−2ohm m, 1.148 Sm−1, 229.6×10−4S m2 mol−1
Q.
At 298 K, the conductivity of a saturated solution of AgCl in water is 2.6×10−6S cm−1. Its solubility product at 298 K (given : λ∞(Ag+)=63.0S cm2mol−1, λ∞(Cl−)=67.0 S cm2mol−1)
4.0×1010M2
4.0×10−16M2
2×10−8M2
2.0×10−5M2
Q. The specific conductance of 0.01 M solution of acetic was found to be 0.0163 S m−1 at 250C. Calculate the degree of dissociation of the acid.
Molar conductance of acetic acid at infinite dilution is 390.7×10−4 S m2mol−1 at 25oC.
Molar conductance of acetic acid at infinite dilution is 390.7×10−4 S m2mol−1 at 25oC.
Q. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm2 . The conductance of this solution was found to be 5×10−7S. The pH of the solution is 4. The value of limiting molar conductivity (∧0m) of this weak monobasic acid in aqueous solution is Z×102 S cm−1mol−1 . The value of Z is?
Q. The resistance of decinormal solution of a salt occupying a volume between two platinum electrodes 1.80 cm apart and 5.4 cm2 area was found to be 32 ohm. Calculate ∧eq.
- 114.1Scm2eq−1
- 124.1Scm2eq−1
- 104.1Scm2eq−1
- 134.1Scm2eq−1
Q. Resistance of 0.2 M solution of an electrolyte is 50Ω.The specific conductance of the solution is 1.4 Sm−1. The resistance of 0.5 M solution of the same electrolyte is 280Ω. The molar conductivity of 0.5 M solution of the electrolyte in Sm2mol−1 is
- 5×10−4
- 5×10−3
- 5×103
- 5×102
Q. The specific conductance of a 0.20 molL−1
solution of an electrolyte at 200C is 2.48×10−4ohm−1cm−1. The molar conductivity of the solution is :
solution of an electrolyte at 200C is 2.48×10−4ohm−1cm−1. The molar conductivity of the solution is :
- 1.24 ohm−1 cm2 mol−1
- 4.96 ohm−1 cm2 mol−1
- 0.24 ohm−1 cm2 mol−1
- 0.49 ohm−1 cm2mol−1
Q. The resistance of 0.01 N solution of an electrolyte is 210 Ω at 298 K with a cell constant of 0.88 cm−1. Calculate the conductivity and equivalent conductivity of the solution.
- 4.19×10−3ohm−1 cm−1, 419.05 S cm2 eq−1
- 5.19×10−3ohm−1 cm−1, 519.05 S cm2 eq−1
- 6.19×10−3ohm−1 cm−1, 619.05 S cm2 eq−1
- 7.19×10−3ohm−1 cm−1, 719.05 S cm2 eq−1
Q. Which of the following graph is correct?
Λm - Molar conductivity
C - concentration of electrolyte
Λm - Molar conductivity
C - concentration of electrolyte
Q.
Specific conductance of 0.1 MCH3COOH at 25∘ is 3.9×10−4ohm−1cm−1. If λ∞(H3O+) and λ∞(VH3COO−) at 25∘C are 349.0 and 41.0 ohm−1cm2mol−1 respectively, degree of ionization of CH3COOH at the given concentration is :
2.0%
1.0%
4.0%
5.0%
Q. The molar ionic conductances at infinite dilution of Mg2+ and Cl− are 106.1 and 76.3 ohm−1 ohm−1 cm2 mol−1 respectively. The Molar conductance of solution of MgCl2 at
infinite dilution is
infinite dilution is
- 29.8ohm−1cm2mol−1
- 183.4ohm−1cm2mol−1
- 285.7ohm−1cm2mol−1
- 258.7ohm−1cm2mol−1
Q. Specific conductance of 0.2 M aqueous solution is 16.4×10−2 ohm−1 cm−1. The molar conductance of solution is
- 820 ohm−1 cm2 mol−1
- 1201 ohm−1 cm2 mol−1
- 315 ohm−1 cm2 mol−1
- 519 ohm−1 cm2 mol−1
Q. The specific conductance of a 0.20 molL−1
solution of an electrolyte at 200C is 2.48×10−4ohm−1cm−1. The molar conductivity of the solution is :
solution of an electrolyte at 200C is 2.48×10−4ohm−1cm−1. The molar conductivity of the solution is :
- 1.24 ohm−1 cm2 mol−1
- 4.96 ohm−1 cm2 mol−1
- 0.24 ohm−1 cm2 mol−1
- 0.49 ohm−1 cm2mol−1
Q. Resistance of a 0.1 N KCN solution is 245 Ω. If the electrodes in the cell are 5 cm apart and area having 100 cm2 each, the molar conductance of the solution will be:
Q. Conductance of 0.1 M KCl (conductivity = x ohm−1cm−1)
filled in a conductivity cell is y ohm−1. If the conductance of 0.1 M NaOH filled in the same cell is z ohm−1, molar conductance of NaOH will be :
filled in a conductivity cell is y ohm−1. If the conductance of 0.1 M NaOH filled in the same cell is z ohm−1, molar conductance of NaOH will be :
- 0.1xzy
- 10xzy
- 103xzy
- 104xzy
Q. The resistance and conductivity of 0.02 M KCl solution are 82.4 ohm and 0.002768Scm−1, respectively. When filled with 0.005NK2SO4, the solution had a resistance of 324 ohm. Calculate equivalent conductivity of K2SO4 solution.
- 381.6Scm2mol−1
- 581.6Scm2mol−1
- 681.6Scm2mol−1
- 281.6Scm2mol−1
