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The van't Hoff factor for BaCl(2) at 0.0...

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 :

A

49

B

69

C

89

D

98

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(,BaCl_(2),hArr,Ba^(2+),+,2Cl^(-)),("Initially",1,,0,,0),("After dissociation",1-alpha,,alpha,,2alpha):}`
Total `= 1 - alpha + alpha + 2alpha = 1 + 2 alpha`
`alpha = (i-1)/(m-1) = (1.98 - 1)/(3 - 1) = (0.98)/(2) = 0.49`
m = number of particles in solution
For a mole `alpha = 0.49`
For 0.01 mole `alpha = (0.49)/(0.01) = 49`.
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