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प्रश्न
The molar conductivity of a 0.5 mol dm−3 solution of AgNO3 with electrolytic conductivity of 5.76 × 10−3 S cm−1 at 298 K is ____________.
पर्याय
2.88 S cm2 mol−1
11.52 S cm2 mol−1
0.086 S cm2 mol−1
28.8 S cm2 mol−1
उत्तर
The molar conductivity of a 0.5 mol dm−3 solution of AgNO3 with electrolytic conductivity of 5.76 × 10−3 S cm−1 at 298 K is 11.52 S cm2 mol−1.
Explanation:
Λ = `"κ"/"M" xx 10^-3 "mol"^-1 "m"^3`
= `(5.76 xx 10^-3 xx "S cm"^-1 xx 10^-3)/0.5` mol−1 m3
= `(5.76 xx 10^-3 xx 10^-3 xx 10^6)/0.5` S cm−1 mol−1 cm3
= 11.52 S cm2 mol−1
APPEARS IN
संबंधित प्रश्न
| Electrolyte | KCl | KNO3 | HCl | NaOAC | NaCl |
| Λ_ (S cm mol−1) |
149.9 | 145.0 | 426.2 | 91.0 | 126.5 |
Calculate \[\ce{Λ^∘_{HOAC}}\] using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25°C.
Faradays constant is defined as ____________.
The equivalent conductance of `"M"/36` solution of a weak monobasic acid is 6 mho cm2 equivalent−1 and at infinite dilution is 400 mho cm2 equivalent−1. The dissociation constant of this acid is ____________.
A conductivity cell has been calibrated with a 0.01 M, 1 : 1 electrolytic solution (specific conductance (κ = 1.25 × 10−3 S cm−1) in the cell and the measured resistance was 800 Ω at 25°C. The cell constant is, ____________.
Conductivity of a saturated solution of a sparingly soluble salt AB (1 : 1 electrolyte) at 298 K is 1.85 × 10−5 S m−1. Solubility product of the salt AB at 298 K `(Λ_"m"^∘)_"AB"` = 14 × 10−3 S m2 mol−1.
Why does the conductivity of a solution decrease on dilution of the solution?
The conductivity of a 0.01 M solution of a 1 : 1 weak electrolyte at 298 K is 1.5 × 10−4 S cm−1.
i) molar conductivity of the solution
ii) degree of dissociation and the dissociation constant of the weak electrolyte
Given that
\[\ce{λ^∘_{cation}}\] = 248.2 S cm2 mol−1
\[\ce{λ^∘_{anion}}\] = 51.8 S cm2 mol−1
Which of 0.1 M HCl and 0.1 M KCl do you expect to have greater `Λ_"m"^∘` and why?
Arrange the following solutions in the decreasing order of specific conductance.
i) 0.01 M KCl
ii) 0.005 M KCl
iii) 0.1 M KCl
iv) 0.25 M KCl
v) 0.5 M KCl
Why is AC current used instead of DC in measuring the electrolytic conductance?
