HARD
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Given the following molar conductivities at 25 ο C :,   HCl ,   426   Ω - 1 cm 2 mol - 1 ;   NaCl,   126 Ω - 1 cm 2 mol - 1 ;   NaC   sodium crotonate ,   83   Ω - 1 cm 2 mol - 1 . What is the ionization constant of crotonic acid? If the conductivity of a 0.001 M crotonic acid solution is 3.83 × 10 - 5 Ω - 1 cm - 1 ?

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Important Questions on Electrochemistry

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
HARD
State and explain kohlrausch's law of independent migration of ions.
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

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

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.

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
Why can limiting molar conductivity of CH3COOH  not be determined experimentally ?
 
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
Define the following terms: Kohlrausch’s Law.
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 ]
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
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)
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)
EASY

The limiting molar conductivity and molar conductivity of acetic acid are 390.5 S cm2. mol-1 and 48.15 S cm2. mol-1 respectively. Calculate the degree acid are of dissociation of the weak acid?

MEDIUM
λ°m for NH4Cl, NaOH and NaCl are 130, 248 and 126. 5 ohm-1em2mo-1 respectively. The λ°m NH4OH will be,
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 -