The specific conductance of a saturated solution of AgCl at `25^@C` is `1.821xx10^(-5)` mho `cm^(-1)`. What is the solubility of AgCl in water (in g `L^(-1)` ) , if limiting molar conductivity of AgCl is 130.26 mho `cm^(2) mol^(-1)` ?
A. `1.89xx10^(-3) g L^(-1)`
B. `2.78xx10^(-2) g L^(-1)`
C. `2.004xx10^(-2) g L^(-1)`
D. `1.43xx10^(-3) g L^(-1)`

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Correct Answer - C
Solubility =`(Kxx1000)/Lambda_m^@`
`=(1.821xx10^(-5)xx1000)/130.26=13.97xx10^(-5) mol L^(-1)`
`=13.97xx10^(-5) xx 143.5 ` (AgCl = 108 + 35.5 = 143.5 )
`=2.004xx10^(-2) g L^(-1)`

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