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The Van't Hoff factor of Hg(2)Cl(2) in i...

The Van't Hoff factor of `Hg_(2)Cl_(2)` in its aqueous solution will be (`Hg_(2)Cl_(2)` is `80%` ionized in the solution)
a.`1.6` , b.`2.6` ,c.`3.6` ,d.`4.6`

A

`1.6`

B

`2.6`

C

`3.6`

D

`4.6`

Text Solution

Verified by Experts

The correct Answer is:
B

`HgCl_(2)hArr Hg_(2)^(2+)+2Cl^(-)`
n = 3
`alpha = (i-1)/(n-1)rArr 0.8 =(i-1)/(3-1)rArr i=2.6`
Alternate method to calculate (i)
i = (Number of ions `xx alpha`) `+(1-alpha)`
`=(3xx0.8)+(1-0.8)[alpha = 80% or 0.8]`
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