1.

The limiting molar conductivities of HCl, CH_(3)COONa and NaCl are respective 425, 190 and 150 mho cm^(2)mol^(-1) at 25^(@)C. The molar conductivity of 0.1 M acetic acid is 9.2 mho cm^(2)mol^(-1). The degree of dissociation of 0.1 M acetic acid is

Answer»

`0.10`
0.02
0.19
0.03

Solution :`wedge_(CH_(3)COOH)^(@)=wedge_(CH_(3)COONA)^(@)+wedge_(HCl)^(@)-wedge_(NaCl)^(@)`
`""=190+425-150=465`
`alpha=wedge_(m)^(C)/wedge_(m)^(@)=9.2/465=0.19`


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