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The solubility product of `Pbl_(2)` is `7.2 xx 10^(-9)`. The maximum mass of `Nal` which may be added in `500ml` of `0.005M Pb (NO_(3))_(2)` solution without any precipitation of `Pbl_(2)` is `( = 127)`:A. `0.09g`B. `1.2 xx 10^(-3)g`C. `6 xx 10^(-4)g`D. `1.08 xx 10^(-5)g` |
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Answer» Correct Answer - A `[I^(-)] = sqrt(7.2 xx 10^(-9))/(0.005) = 1.2 xx 10^(-3) M` Maximum moles of Nal that can be added to `500 ml` of `Pb(NO_(3))_(2)` are `(1.2 xx 10^(-3))/(1000) xx 500 = 6 xx 10^(-4) "moles" rArr 0.09g` |
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