A) 1.520
B) 0.995
C) 19.00
D) 9.00
Correct Answer: D
Solution :
\[\frac{Wt.\text{ }of\text{ }metal\text{ }hydroxide}{Wt.\text{ }of\text{ }metal\text{ }oxide}\] \[=\frac{Eq.\text{ }wt.\text{ }of\text{ }metal+Eq.\text{ }wt.\text{ }of\text{ }O{{H}^{-}}}{Eq.\text{ }wt.\text{ }of\text{ }metal+Eq.\text{ }wt.\text{ }{{O}^{2-}}}\] \[=\frac{1.520}{0.995}=\frac{E+17}{E+8}\] \[1.520(E+8)=0.995(E+17)\] \[1.520\text{ }E+12.160=0.995E+16.915\] \[1.520E-0.995E=16.915-12.16\] or \[0.525E=16.915-12.16=4.755\] \[E=\frac{4.755}{0.525}=9.0\]
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