$1.52 \mathrm{g}$ of the hydroxide of a metal gave on ignition, $0.995 \mathrm{g}$ of the oxide. The equivalent weight of the metal is :

$\frac{Wt. of metal hydroxide}{Wt. of metal oxide}$

                   $\begin{array}{l}=\frac{\text{ Eq. wt. of metal + Eq. wt. of }{\mathrm{OH}}^{-}}{\text{Eq. wt. of metal + Eq. wt. of }{\mathrm{O}}^{-2}}\\ \frac{1.520\mathrm{g}}{0.995\mathrm{g}}=\frac{\mathrm{E}+17}{\mathrm{E}+8};\frac{1520}{995}=\frac{\mathrm{E}+17}{\mathrm{E}+8}\\ 1520\mathrm{E}+1520\times 8=995\mathrm{E}+995\times 17;1520\mathrm{E}+12160\\ =995\mathrm{E}+16915\\ 1520\mathrm{E}-995\mathrm{E}=16915-12160;525\mathrm{E}=4755\\ \therefore \mathrm{E}=\frac{4755}{525}=9.06\end{array}$

So, the correct answer is (b).