Out of the following pairs, predict with reason which pair will allow greater conduction of electricity:
(i) Silver wire at 30°C or silver wire at 60°C.
(ii) 0.1 M \(CH_3\)COOH solution or 1 M \(CH_3\)COOH solution.
(iii) KCl solution at 20°C or KCl solution at 50°C.
(i) Silver wire at 30°C or silver wire at 60°C.
(ii) 0.1 M \(CH_3\)COOH solution or 1 M \(CH_3\)COOH solution.
(iii) KCl solution at 20°C or KCl solution at 50°C.
(i) Silver wire at 30°C because as temperature decreases, resistance decreases so conduction increases.
(ii) 0.1 M \(CH_3\)COOH, because on dilution degree of ionization increases hence conduction increases.
(iii) KCl solution at 50°C, because at high temperature mobility of ions increases and hence conductance increases.
(ii) 0.1 M \(CH_3\)COOH, because on dilution degree of ionization increases hence conduction increases.
(iii) KCl solution at 50°C, because at high temperature mobility of ions increases and hence conductance increases.
The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 x 10⁻⁴ S cm⁻¹. Calculate molar conductivity () of the solution.
Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes:
(i) State the law which helps to determine the limiting molar conductivity of weak electrolyte.
(ii) Calculate limiting molar conductivity of CaSO₄ (limiting molar conductivity of calcium and sulphate ions are 119.0 and 160.0 Scm² mol⁻¹ respectively)
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution ?
The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 × 10³ ohm. Calculate its molar conductivity.
How is electrical conductance of a conductor related with length and area of cross-section of the conductor?
a. G = \(l. a.k^{-1}\)
b. G = \(k. l.a^{-1}\)
c. G = \(k.a. l^{-1}\)
d. G = \(k. l.a^{-2}\)
The conductivity of an aqueous solution of NaCl in a cell is 92 \(Ω^{−1}\) \(cm^{-1}\) the resistance offered by this cell is 247.8 Ω . Calculate the cell constant.
Calculate emf of the following cell
Cd/\(Cd^{2+}\) (.10 M)//\(H_+\) (.20 M)/\(H_2\) (0.5 atm)/Pt
[Given E° for \(Cd^{2+}\) /Cd = -0.403V]
Calculate the molar conductivity and degree of dissociation.
Conductivity of 2.5 × 10⁻⁴M methanoic acid is 5.25 × 10⁻⁵ Scm⁻¹.
Given : = 50.5Scm² mol⁻¹
(i) Complete the following equations :
(a) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ →
(b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state.
E°(Cr³⁺/Cr²⁺) = -0.4 V
E°(Mn³⁺/Mn²⁺) = +1.5 V
E°(Fe³⁺/Fe²⁺) = + 0.8 V
Out of the following pairs, predict with reason which pair will allow greater conduction of electricity:
(i) Silver wire at 30°C or silver wire at 60°C.
(ii) 0.1 M \(CH_3\)COOH solution or 1 M \(CH_3\)COOH solution.
(iii) KCl solution at 20°C or KCl solution at 50°C.
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 λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹.
The magnetic moment of few transition metal ions
are given below:
Complete the following reactions—
(i) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) 2CrO₄²⁻ + 2H⁺ →
(iii) 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ →