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Given the ionic equivalent conductivitie...

Given the ionic equivalent conductivities for the following ions `:`
`lambda^(@)._(eq)K^(o+)=73.5cm^(2)ohm^(-1)eq^(-1)`
`lambda^(@)._(eq)Al^(3+)=149cm^(2)ohm^(-1)eq^(-1)`
`lambda^(@)._(eq)SO_(4)^(2-)=85.8cm^(2)ohm^(-1)eq^(-1)`
The `wedge_(eq)^(@)` for potash alum `(K_(2)SO_(4).Al_(2)(SO_(4))_(3).24H_(2)O)` is `:`

A

`215.92`

B

`348.3`

C

`368.2`

D

`108.52`

Text Solution

Verified by Experts

The correct Answer is:
a
`(Eq=` Charge on the iion `//` Total charge `)`
`[K^(o+)]=(1)/(8)mol xx 2=(1)/(4) mol =(1)/(4)Eq`
`[Al^(3+)]=(6)/(8)=(3)/(4)Eq`
`[SO_(4)^(2-)]=(8)/(8)=1Eq`
`wedge_(eq)^(@)K_(2)SO_(4).Al_(2)(SO_(4))_(3).24H_(2)O`
`=lambda_(eq)^(@)(K^(o+))+lambda_(eq)^(@)(Al^(3+))+lambda_(eq)^(@)(SO_(4)^(2-))`
`-=(1)/(4)xx73.5+149xx(3)/(4)+85.8xx1`
`-=18.375+141.75+85.8`
`-=215.92`
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