Calculate the percentage degree of dissociation of an electrolyte `XY_(2)` (Normal molar mass `=164`) in water if the observed molar mass by measuring elevation in boiling point is `65.6`:
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Correct Answer – `75%`
`{:(,XY_(2),hArr,X^(2+),+,2Y^(-)),(“Initially”,1,,0,,0),(“at equilibrium”,1-alpha,,alpha,,2alpha):}`
Total no. of moles `=1-alpha+alpha+2alpha=1+2alpha`
`i=(“Normal molar mass”)/(“Observed molar mass”), (1+2alpha)/(1)=(164)/(65.6)`
`therefore alpha=0.75,%alpha=75%`