If conductivity of water used to make saturated of AgCl is found to be `3.1xx10^(-5)Omega^(-1)` `cm^(-1)` and conductance of the solution of `AgCl=4.5
If conductivity of water used to make saturated of AgCl is found to be `3.1xx10^(-5)Omega^(-1)`
`cm^(-1)` and conductance of the solution of `AgCl=4.5xx10^(-5)Omega^(-1)cm^(-1)`
if `lamda_(M)^(0)AgNO_(3)=200Omega^(-1)cm^(2)` `"mole"^(-1)`
`lamda_(M)^(0)NaNO_(3)=310Omega^(-1)cm^(2)` `"mole"^(-1)`
calculate `K_(SP)` of `AgCl`
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`lamda_(M)^(0)AgCl=140`
Total cunductance `=10^(-5)`
`S=(140xx4xx10^(-5)xx1000)/(140)`
`=(1.4xx10^(-4))/(14)`
`S=5.4xx10^(-4)`
`S^(2)=1xx10^(-8)`
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