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For pure water degree of dissociation ...

For pure water degree of dissociation of water is `1.9xx10^(-9)`
`wedge_(m)^(infty)(H^(+))=350 Scm^(2) mol^(-1)`
`wedge_(m)^(infty)(OH^(-))=200 S cm^(2) mol^(-1)`
Hence molar conductance of water is

A

`3.8xx10^(-7)" S "cm^(2)mol^(-1)`

B

`5.7xx10^(-7)" S "cm^(2)mol^(-1)`

C

`9.5xx10^(-7)" S "cm^(2)mol^(-1)`

D

`1.045xx10^(-6)" S "cm^(2)mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Degree of dissociation `(alpha)=(wedge_(m)^(c))/(wedge_(m)^(0))`
`wedge_(m)^(c)=alphaxxwedge_(m)^(0)=(1.9xx10^(-9))xx(200+350)" S "cm^(2)mol^(-1)`
`=1.9xx550xx10^(-9)=1.045xx10^(-6)" S "cm^(2)mol^(-1)`
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