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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

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

B

`1.045 xx10^(-9) S cm^(2) mol^(-1)`

C

`1.04 xx10^(-14) S cm^(2) mol^(-1)`

D

`1.04 xx10^(-14) S cm^(2) mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`wedge_(H_(2)O)^(@)=wedge_(H^(+))^(@)+wedge_(OH)^(@)=550 S cm^(2) "mol"^(-1) x =(wedge_(C ))/(wedge_(0))`
`therefore wedge_(C )=x xx wedge_(0)=1.9 xx 10^(-9) xx 550`
`=1.045 xx 10^(-6) S cm^(2) "mol"^(-1)`
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