ElectrochemistryMedium
Question
Calculate $\Lambda_{m}^{\infty}$ (in Ω−1 cm2 mol−1) for SrCl2 at 25°C, from the following data:
Conc. 0.25 M 1.0 M
Λm (in Ω−1 cm2 mol−1) 260 250
Options
A.270
B.265
C.240
D.275
Solution
$\Lambda_{m} = \Lambda_{m}^{o} - A\sqrt{C}\left( \text{for strong electrolyte} \right)$
$\text{Or, 260 = }\Lambda_{m}^{o} - A \times \sqrt{0.25}(1) $$${250 = \Lambda_{m}^{o} - A \times \sqrt{1.00}(2) }{\therefore\Lambda_{m}^{o} = 270\text{ oh}\text{m}^{- 1}\text{ c}\text{m}^{2}\text{ mo}\text{l}^{- 1}}$$
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