JEE MAIN - Chemistry (2006 - No. 31)
The molar conductivities $$ \wedge _{NaOAc}^o$$ and $$ \wedge _{HCl}^o$$ and at infinite dilution in water at 25oC are 91.0 and 426.2 Scm2/mol respectively. To calculate $$ \wedge _{HOAc}^o$$ , the additional value required is
$$ \wedge _{{H_2}O}^o$$
$$ \wedge _{KCl}^o$$
$$ \wedge _{NaOH}^o$$
$$ \wedge _{NaCl}^o$$
Uitleg
$$\Lambda _{C{H_3}COOH}^o$$ is given by the following equation
$$\Lambda _{C{H_3}COOH}^o = \left( {\Lambda _{C{H_3}COONa}^o + \Lambda _{HCl}^o} \right) - \left( {\Lambda _{NaCl}^o} \right)$$
Hence $$\Lambda _{NaCl}^ \circ $$ is required.
$$\Lambda _{C{H_3}COOH}^o = \left( {\Lambda _{C{H_3}COONa}^o + \Lambda _{HCl}^o} \right) - \left( {\Lambda _{NaCl}^o} \right)$$
Hence $$\Lambda _{NaCl}^ \circ $$ is required.