A) 0.18 L
B) 17.98 L
C) 1.798 L
D) 0.36 L
Correct Answer: A
Solution :
\[Ksp=3.2\times {{10}^{-8}}\] \[PbC{{l}_{2}}P{{b}^{2+}}+2{{a}^{-}}\] \[O\le 52\] \[{{K}_{sp}}(5){{(25)}^{2}}\] \[{{K}_{sp}}={{45}^{3}}\] \[{{s}^{3}}=\frac{3.2\times {{10}^{-8}}}{4}\] \[{{s}^{3}}=-8\times {{10}^{-9}}\] \[s=2\times {{10}^{-3}}\] \[m.wtg=207+35.5\times 2\] \[PbC{{l}_{2}}=278\] \[\therefore 2\times {{10}^{-3}}=\frac{\frac{0.1}{278}}{x}\] \[\therefore x=\frac{0.1}{278\times 2\times {{10}^{-3}}}\] \[=\frac{100}{278\times 2}\] \[x=0.18L\]
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