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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`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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