The specific conductance of 0.1 M NaCl solution is 1.06 × 10–2 ohm–1 cm–1. Its molar conductance in ohm–1 cm2 mol–1 is
1.06 × 102
1.06 × 103
1.06 × 104
53 × 102
Additional Questions
|
Among the following cells Primary cells are |
Answer |
|
The hydrogen electrode is dipped in a solution of pH 3 at 25°C. The potential would be (the value of 2.303 RT/F is 0.059 V) |
Answer |
|
When same quantity of electricity is passed for half an hour, the amount of Cu and Cr deposited are respectively 0.375 g and 0.30 g. Ratio of electrochemical equivalent of Cu and Cr is |
Answer |
|
Given the data at 25°C, Ag + I– → AgI + e– ; E° = 0.152 V Ag → Ag+ + e–; E° = –0.800 V What is the value of log Ksp for AgI? $\left(2.303dfrac{ ext{RT}}{ ext{F}} = 0.059 ext{ V} ight)$ |
Answer |
|
The quantity of electricity needed to separately electrolyze 1 M solution of ZnSO4, AlCl3 and AgNO3 completely is in the ratio of |
Answer |
|
An infinite dilution of aqueous solution of BaCl2, molar conductivity of Ba2+ and Cl– ions are = 127.32 S cm2/moland 76.34 S cm2/mol respectively. What is Λ∞ |
Answer |
|
Aqueous copper sulphate solution is electrolyzed using platinum electrodes. The electrode reaction occurring at cathode is |
Answer |
|
Zn | Zn2+ (a = 0.1M) || Fe2+ (a = 0.01M) | Fe. The emf of the above cell is 0.2905 V. Equilibrium constant for the cell reaction is |
Answer |
|
Calculate the reduction potential of a half–cell containing platinum electrode immersed in 2.0 M Fe2+ and 0.02 M Fe3+. Given Eo Fe3+ + e– → Fe2+ |
Answer |
|
Consider the following four electrodes, P = Cu2+ (0.0001 M) | Cu(s) Q = Cu2+ (0.1 M) | Cu(s) R = Cu2+ (0.01 M) | Cu(s) S = Cu2+ (0.001 M) | Cu(s) If the standard reduction potential of Cu2+/Cu is +0.34 V, the reduction potentials in volts of the above electrodes follow the order |
Answer |