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The van't Hoff factor of BaCl2 at 0.01 M...

The van't Hoff factor of `BaCl_2` at 0.01 M concentration is 1.98 . The percentage of dissociations of `BaCl_2` at this concentration is:

A

49

B

69

C

89

D

98

Text Solution

Verified by Experts

The correct Answer is:
A
` BaCl_2 hArr Ba^(2+) + 2Cl^(-)`
initial ` " " 0.01 M`
At equilibrium `(0.01 -x) M x M2 x M`
` I = ((0.01-x)+x+2x)/(0.01)`
` = (0.01 +2x)/(0.01) = 1.98 `
` x= 0.0049`
` % alpha = (x)/(0.01) xx 100 = (0.0049 xx 100 )/(0.01) = 49% `
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