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If `0.04 M Na_(2)SO_(4)` solutions at `300K` is found to be isotonic with `0.05M NaCl` (`100%` dissociation) solutions. Calculate degree of dissociation of sodium sulphide. |
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Answer» `i_(l)C_(1)RT=i_(2)C_(2)RT` `i_(l)C_(1)=i_(2)C_(2)` `0.04(1+2alpha)=0.05xx2` `alpha=0.75=75%` |
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