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The molar solubility of AgCl in `1.8 M AgNO_(3)` solution is (`K_(sp)` of `AgCl = 1.8 xx 10^(-10)`)A. `10^(-5)`B. `10^(-10)`C. `1.8 xx 10^(-5)`D. `1.8 xx 10^(-10)` |
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Answer» Correct Answer - B `1.8 M AgNO_(3) "……" 1.8 xx 10^(-10)` `1M AgNO_(3) "…….." ? ` Molar sububility `= (1.8 xx 10^(-10))/(1.8) = 10^(-10)` |
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