A) 3
B) 4
C) 5
D) 6
Correct Answer: A
Solution :
Centre of circle = (2, 1) and radius \[{{V}_{2}}>{{V}_{4}}\] \[{{V}_{3}}\] \[MgC{{l}_{2}}\to 2NaOH\to Mg{{(OH)}_{2}}\downarrow +2NaCl\] \[Mg{{(OH)}_{2}}\] \[M{{g}^{2+}}\] = 10 \[O{{H}^{-}}\] At point (10, 7), \[{{K}_{sp}}\] \[Mg{{(OH)}_{2}}\] The point P is outside the given circle. Now, required largest distance = PA \[Mg{{(OH)}_{2}}\,\underset{s}{\mathop{M{{g}^{2+}}}}\,+\underset{2s}{\mathop{2O{{H}^{-}}}}\,\] PA = AB + BP = 2 radius + (OP - radius) \[{{K}_{sp}}=4{{s}^{3}}\] \[s=\sqrt[3]{\frac{12\times {{10}^{-12}}}{4}}=1.44\times {{10}^{-4}}\] \[\therefore \] \[[O{{H}^{-}}]=2s=2\times 1.44\times {{10}^{-4}}=2.88\times {{10}^{-4}}\] (given) \[1\,d{{m}^{3}}=1\,L=1000\,mL\] \[\therefore \]You need to login to perform this action.
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