Question

The molar conductivities and at infinite dilution in water at 25⁰C are 91.0 and 426.2 Scm2/mol respectively. To calculate , the additional value required is :

• A. $\Lambda_{H_{2}O}^{0}$
• B. $\Lambda_{KCl}^{0}$
• C. $\Lambda_{NaOH}^{0}$
• D. $\Lambda_{NaCl}^{0}$

Answer 1

Answer: (D)

$\Lambda_{NaCl}^{0}$

Answer 2

$\text{C}\text{H}_{3}\text{COONa + HCI} \rightarrow \text{C}\text{H}_{3}\text{COOH + NaCI}$

From the reaction,

$\Lambda_{CH_{3}COONa}^{0} + \Lambda_{HCl}^{0} = \Lambda^{0}CH_{3}COOH + \Lambda_{NaCl}^{0}$or

$\Lambda_{CH_{3}COOH}^{0} = \Lambda_{CH_{3}COONa}^{0} + \Lambda_{HCl}^{0} - \Lambda_{NaCl}^{0}$

Thus to calculate the value of one should know the value of $\Lambda_{NaCl}^{0}$along with and $\Lambda_{HCl}^{0}$