1.

0.04 N solution of a weak acid has a specific conductance 4.23 xx10^(-4)" mho.cm"^(-1). The degree of dissociation of acid at this dilution is 0.0612. Calculate the equivalent conductance of weak acid at infinite solution.

Answer»

Solution :Specific CONDUCTANCE `kappa=4.23xx10^(-4)"mho.cm"^(-1)`
`lambda_(C )=(kappa1000)/(C )=(4.23xx10^(-4)xx1000)/(0.04)`
`=10.575"mho.cm"^(2).EQ^(-1)`.
`alpha=0.0612=(lambda_(C ))/(lambda_(oo))=(10.575)/(lambda_(oo))`
`therefore lambda_(oo)="172.79 mho. cm"^(2)."gm. equiv."^(-1)`.


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