Equivalent, Molar Conductivity and Cell Constant
Trending Questions
Q.
Gram equivalent volume ofO2 at stp is 1-11.2 , 1-5.6l, 3-22.4l, 4-2.8
Q. how to find the number of electrons in 200g of CaCO3
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. Specific conductivity of 0.01N H2SO4 solution is 6×10^-3 S cm^-1. Its molar conductivity is
Q. 2. Give solubility order for(i) alkali metal hydroxide(iii) alkali metal per chlorates(v) alkaline earth metal carbonates(vii) alkali metal bicarbonates(ix) lead (II) halides(ii) Sodium halides(iv) alkali metal oxides(vi) alkali metal carbonates(viii) silver halides(x) mercury (I) halides
Q. Total number of electrons present in 1.8 mL of water is (density of water is 1 g mL–1)
1.8 mL जल (जल का घनत्व 1 g mL–1 है) में उपस्थित इलेक्ट्रॉनों की कुल संख्या है
1.8 mL जल (जल का घनत्व 1 g mL–1 है) में उपस्थित इलेक्ट्रॉनों की कुल संख्या है
- NA
- 0.1 NA
- 10 NA
- 0.2 NA
Q.
What is the molar conductivity of acetic acid?
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. 42 One gram equivalent of a substance is present in: 1.0.25mol of 02 2.0.5mol of 02 3.1mol of 02 4.8mol of 02
Q. Electrolytes which have low value of equivalent conductance at high concentrations are termed as strong electrolytes.
- True
- False
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. 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. Permutit is:
- hydrated sodium aluminium silicate
- sodium hexametaphosphate
- sodium silicate
- sodium meta-aluminate
Q.
Specific conductance of 0.1 M HA is 3.75×10−4ohm−1cm−1. If λ∞(HA)=250 ohm−1cm2mol−1, the dissociation constant Ka of HA is :
1.0×10−5
2.25×10−4
2.25×10−5
2.25×10−13
Q. Equivalent conductivity ∧eq
- k×1000mohm−1cm2Eq−1, m−molality
- k×1000Mohm−1cm2Eq−1, M−molarity
- k×1000Nohm−1cm2Eq−1, N−normality
- None of the above
Q.
The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductiviy of 5.76×10−3 S cm−1 at 298 K is
2.88 cm2/mol
11.52 S cm2/mol
0.086 S cm2/mol
28.8 cm2/mol
Q. Why Conductivity decreases with concentration ?
Q. At a particular temperature, the ratio of equivalent conductance to specific conductance of a 0.01 N NaCl solution is:
105cm3 equiv−1
102cm3 equiv−1
103cm3 equiv−1
10 cm3 equiv−1
Q.
In infinite dilutions, the molar conductances of Ba2+ and Cl- are 127 and
76 . The equivalent conductivity of at infinite
dilution is
[CBSE 2000]
101.5
139.5
203.5
279.5
Q. Maximum no. Of electrons in cu havinh m less than or equal to 1
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. The conductivity of a saturated solution of BaSO4 is 3.06×10−6ohm−1cm−1 and its molar conductance is 1.53 ohm−1 cm2 mol−1. What will be the the Ksp of BaSO4 ?
4×10−12
4×10−6
2.5×10−13
2.5×10−9
Q. The ionic conductivity of Ba2+ and Cl− at infinite dilution are 127 and 76 ohm−1 cm2 mol−1 respectively .The equivalent conductivity of BaCl2 at infinity dilution (in ohm−1 cm2 eq−1) would be :
- 279
- 203
- 139.5
- 101.5