Calculating molar mass of an oxidant by iodometry
The molar mass of a substance, as you said earlier, is:


  simply the mass of the substance divided by the number of moles in the same amount of substance


In this question, you can see that as per part b), we found that there were $7.553\times10^{-4} \text{ moles}$ of $\ce{M^{n+}}$ present.

Hence, the salt, i.e., $\ce{M_xA_y.ZH2O}$ contains $\ce{M_x}$ = $7.553\times 10^{-4} \text{ moles}$ $\ce{M^{n+}}$.

=> $\ce{M}$ = $ \frac{7.553\times 10^{-4}}{x} \text{ moles}$ of $\ce{M^{n+}}$

$\therefore$ taken moles of $\ce{M_xA_y.ZH2O}$ = $\frac{7.553\times 10^{-4}}{x} \text{ moles}$

But we know that the taken moles have a mass of $0.2642 g$.

Now, divide $0.2642 g$ by $\frac{7.553\times 10^{-4}}{x} \text{ moles}$ to get the molar mass, in terms of x, which is $349.8x$