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The EMF of the following cell is 0.86 volts `Ag|AgNO_(3)(0.0093M)||AgNO_(3)(xM)|Ag`. The value of x will beA. 82.8MB. 2.28MC. 0.228MD. 1.14M |
Answer» Correct Answer - C `E=(0.059)/(1)"log"(C_(2))/(C_(1))` `0.086=(0.059)/(1)"log"(x)/(0.0093)` `x=0.228M` |
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