Question

The equivalent conductivity of KCl at infinite dilution is 130 S cm2eq–1. The transport number of Cl–1 ion in KCl at the same temperature is 0.505. The limiting ionic mobility of K+ ion, is

• A. $\text{6.67 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ c}\text{m}^{2}\text{ se}\text{c}^{\text{-1}}\text{ vol}\text{t}^{\text{-1}}$
• B. $\text{5.01 }\text{×}\text{ 1}\text{0}^{\text{-3}}\text{ c}\text{m}^{2}\text{ se}\text{c}^{\text{-1}}\text{ vol}\text{t}^{\text{-1}}$
• C. $\text{3.22 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ c}\text{m}^{2}\text{ se}\text{c}^{\text{-1}}\text{ vol}\text{t}^{\text{-1}}$
• D. $\text{2.00 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ c}\text{m}^{2}\text{ se}\text{c}^{\text{-1}}\text{ vol}\text{t}^{\text{-1}}$

Answer 1

Answer: (A)

$\text{6.67 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ c}\text{m}^{2}\text{ se}\text{c}^{\text{-1}}\text{ vol}\text{t}^{\text{-1}}$

Answer 2

$\lambda_{k +}^{0} = F \times U_{K +}$ or $U_{K +} = \frac{(130 - 65.65)}{96500}cm^{2}volt^{- 1}\sec^{- 1}{}$

$\text{= 6.67 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ c}\text{m}^{\mathbf{2}}\text{ vol}\text{t}^{\text{-1}}\text{ se}\text{c}^{\text{-1}}\mathbf{.}$