1.

van't Hoff factor of an electrolyte A_(2)B_(2) assuming that it ionizes 75% in the solution is

Answer»

SOLUTION :`{:(,A_(2)B_(3),hArr,2A^(3+),+,3B^(2-)),("Initial","1 mole",,0,,0),("After IONIZATION",1-0.75,,1.5,,"2.25 MOLES"),(,= 0.25,,,,):}`
`"Total moles after dissociation "=0.25+1.5+2.25`
= 4
`therefore I = 4`


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