1.

The van't Hoff factor for 0.1 M Ba(NO_(3))_(2) solution is 2.74 . The degree of dissociation is :

Answer»

`91.3%`
`87%`
`100%`
`74%`

Solution :`BA(NO_(3))_(2)TOBA^(2+)+NO_(3)^(-)`
`alpha=(i-1)/(n-1)=(2.74-1)/(3-1)`
`=(1.74)/2=0.87`
DEGREE of dissociation =0.87xx100=87%


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