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The limiting molar conductivities of HCl, `CH_(3)COONa` and NaCl are respectiley 425, 90 and 125 mho `cm^(2) " mol"^(-1)` and `25^(@)C`. The molar conductivity of 0.1M `CH_(3)COCH` solution is 7.8 mho `cm^(2) "mol"^(+1)` at the same temperature isA. 0.1B. 0.02C. 0.15D. 0.03 |
Answer» Correct Answer - B `^^_(CH_(3)COOH)^(@)=^^_(CH_(3)COONa)^(@)+^^_(HCl)^(@)-^^_(NaCl)^(@)` `=90+425-125=390 mho cm^(2)"mol"^(-1)` `a=(^^_(m)^(@))/(^^_(m)^(@))=(7.8)/(390)=0.02` |
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