Identification of a metal from group 2
By definition molecular weight is mass $m$ divided by amount $n$, e.g. for the unknown metal $\ce{M}$:

$$M(\ce{M}) = \frac{m(\ce{M})}{n(\ce{M})} \tag{1} \label{eq:1}$$

From the chemical reaction of hydroxide formation one can deduce that $\pu{1 mol}$ of $\ce{M}$ results in $\pu{1 mol}$ of $\ce{M(OH)2}$. However, the amount of hydroxide-anions $\ce{OH-}$ in the solution (assuming complete dissociation) is twice as high as the amount of $\ce{M(OH)2}$. Hence

$$n(\ce{M}) = n(\ce{M(OH)2}) = \frac{n(\ce{OH-})}{2} = \frac{C(\ce{OH-}) \cdot V}{2} \tag{2} \label{eq:2}$$

where $C$ – concentration and $V$ – volume.

Plugging $n(\ce{M})$ from \eqref{eq:2} in \eqref{eq:1}, a final expression for the molecular weight can be established. Remember, before plugging in the numbers in make sure the units are unified.


   $$M(\ce{M}) = \frac{2 \cdot m(\ce{M})}{C(\ce{OH-}) \cdot V} = \frac{2 \cdot \pu{0.115 g}}{\pu{6.56e-3 mol dm-3} \cdot \pu{0.4 dm3}} = \pu{87.6 g mol-1} \tag{3} \label{eq:3}$$
 The unknown metal appears to be strontium $\ce{Sr}$.