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The specific conductance of a saturated solution of silver bromide is `KScm^(-1)`. The limiting ionic conductivity of `Ag^(+)` and `Br^(-)` ions are x and y respectively. The solubility of silver bromide in g/L is (molar mass of `AgBr=188`)A. `(Kxx1000)/(x-y)`B. `(K)/(x+y)xx188`C. `(Kxx1000xx188)/(x+y)`D. `(x+y)/(k)xx(1000)/(188)` |
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Answer» Correct Answer - C `AgBr=x+y` `AgBr=(Kxx1000)/(M)` M-molarity of `AgBr` solution 8 (solubility in g/L)`=(Kxx1000)/((x+y))xx188` |
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