HARD
Earn 100

Calculate limiting molar conductivity of CaSO4, limiting molar conductivities of calcium and sulphate ions are 119.0  and 106.0 S cm2 mol-1respectively?

Important Questions on Electrochemistry

EASY
Λm0 values for NaCl,HCl and CH3COONa are 126.4 S cm2 mol-1, 425.9 S cm2 mol-1 and 91.0 S cm2 mol-1 respectively. Λm0 value for CH3COOH is
MEDIUM
λ°m for NH4Cl, NaOH and NaCl are 130, 248 and 126. 5 ohm-1em2mo-1 respectively. The λ°m NH4OH will be,
MEDIUM
Given λ°Mg2+=106 S cm2 mole-1, λ°SO42-=160 S cm2 mole-1. The value of λ°MgSO4 (in S cm2 mole-1)  is
MEDIUM

The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol-1. What is the dissociation constant of acetic acid? Choose the correct option.

λH+=350 S cm2 mol-1λCH3COO-=50 S cm2 mol-1

HARD
State and explain kohlrausch's law of independent migration of ions.
EASY
State Kohlrausch’s law. Calculate the molar conductivity of an aqueous solution at infinite dilution for BaCl2, when ionic conductance of Ba2+ and Cl- ions are 127.30 S cm2 mol-1 and 76.34 S cm2 mol-1 respectively.
MEDIUM
Why can limiting molar conductivity of CH3COOH  not be determined experimentally ?
 
MEDIUM
The limiting molar conductivities of NaI,NaNO3 and AgNO3 are 12.7,12.0 and 13.3 mSm2 mol-1, respectively (all at 25 °C). The limiting molar conductivity of AgI at this temperature is____mSm2 mol-1.
EASY
The specific conductance K of 0.02 M aqueous acetic acid solution at 298 K is 1.65×10-4Scm-1. The degree of dissociation of acetic acid is

[Given: equivalent conductance at infinite dilution of H+=349.1Scm2mol-1 and CH3COO-=40.9Scm2  mol-1 ]
MEDIUM
The conductivity of a weak acid HA of concentration 0.001 mol L-1 is 2.0×10-5 S cm-1. If Λm0(HA)=190 S cm2 mol-1, the ionization constant Ka of HA is equal to ____ ×10-6
(Round off to the Nearest Integer)
EASY

The molar conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91.0 Scm2 mol-1 respectively. The molar conductance of CH3COOH at infinite dilution is. Choose the right option for your answer.

EASY
Define the following terms: Kohlrausch’s Law.
MEDIUM
The conductivity of 0.00241 M acetic acid is 7.896×10-5 S cm-1. Calculate its molar conductivity and its degree of dissociation. (Given Λm for acetic acid is 390.5 S cm2 mol-1)
MEDIUM
Calculate λm at 298 K of NH4OH at infinite dilution, given λOH-=174 S cm2 mol-1λCl-=66 S cm2 mol-1 and λNH4Cl=130 S cm2 mol-1.
MEDIUM
0.002M solution of a weak acid has an equivalent conductance (Λ)60ohm-1 cm2eq-1. What will be the pH ? (Given : Λ°=400ohm-1 cm2eq-1 ]
EASY
If the molar conductivities in S cm2 mol-1 of NaCl, KCl and NaOH  at infinite dilution are 126 and 150 and 250 respectively, the molar conductivity of KOH in S cm2 mol-1 is -
HARD
mo for NaCl,HCl and NaA are 126.4,425.9 and 100.5Scm2mol-1 respectively. If the conductivity of 0.001MHA is 5×10-5 S cm-1, degree of dissociation of HA is