Advertisements
Advertisements
Question
Define the following terms: Molar conductivity (⋀m)
Advertisements
Solution
Molar conductivity (Λm) of a solution at a given concentration is the conductance of the volume V of the solution containing one mole of an electrolyte placed between two electrodes with the cross-sectional area (A) and the distance of unit length (l). Therefore
`^^_m=kA/l`
Since, l = 1 and A = V (volume containing one mole of electrolyte)
`^^_m=kV`
APPEARS IN
RELATED QUESTIONS
State Kohlrausch Law.
Define limiting molar conductivity.
Why conductivity of an electrolyte solution decreases with the decrease in concentration ?
The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol−1.
Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol−1.
(Given \[\ce{\lambda^{\circ}_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{\lambda^{\circ}_{(CH_3COO^-)}}\] = 40.95 S cm2 mol−1)
In the plot of molar conductivity (∧m) vs square root of concentration (c1/2), following curves are obtained for two electrolytes A and B:

Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of ∧m to concentration approaching zero for electrolytes A and B?
Kohlrausch law of independent migration of ions states ____________.
\[\ce{\Lambda^0_m(NH4OH)}\] is equal to ______.
\[\ce{Λ^0_m H2O}\] is equal to:
(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]
(ii) \[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaNO_3)} - \ce{Λ^0_m_{(NaOH)}}}}\]
(iii) \[\ce{Λ^0_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
The variation of molar conductivity with concentration of an electrolyte (X) m aqueous solution is shown in the given figure.

The electrolyte X is ______.
