1.

The van't Hoff factor for BaCl_(2) at 0.01 M concentration is 1.98. The percentage dissociation of BACl_(2) at this concentration is

Answer»

49
69
89
98

Solution :`{:(,BaCl_(2),HARR,BA^(2+),+,2CL^(-)),("Initial","1 MOLE",,0,,0),("After disso. ",1-ALPHA,,alpha,,2alpha","):}`
"Total no. of particles "=1+2alpha
`i=1+2alpha`
`or alpha=(i-1)/(2)=(1.98-1)/(2)=(0.98)/(2)=0.4949%`.


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