Chemistry Titration and Molarity Problem

can anyone explain to me how this problem is worked out? I have the answers listed, I just don't understand how to reach them...

A 4.36- sample of an unknown alkali metal hydroxide is dissolved in 100.0 of water.
An acid-base indicator is added and the resulting solution is titrated with 2.50 solution.
The indicator changes color signaling that the equivalence point has been reached after 17.0 of the hydrochloric acid solution has been added.


What is the Molar Mass of the metal Hydroxide?
(answer = 103 g/mol)

What is the identity of the alkali metal cation? (Li+, Na+, K+, Rb+, Cs+)
(answer = Rb+)

You are reacting the alkali metal hydroxide with HCl, so write a balanced eqaution.
Alkali metals can only have a charge of +1, so the formula must be MOH

MOH(aq) + HCl(aq) ----> MCl(aq) + H2O(l)

Now, the equation shows us that 1 mole of MOH is neutralised by 1 mole of HCl. The reaction is complete after addition of 17.0 ml of the HCl so there must be the same number of moles of MOH in 4.36 g as there are HCl in 17.0 ml of 2.50 M solution.

moles HCl = molarity x Litres
= 2.50 M x 0.0170 L
= 0.0425 mol

So moles MOH = 0.0425 mol
this many moles of MOH has a mass of 4.36 g

molar mass = mass / moles
= 4.36 g / 0.0425 mol
= 102.558 g/mol
= 103 g/mol (3 sig figs)

1 mole of MOH has
1 mole of M+ and 1 mole of OH-
So subtract the mass of 1 mole of OH- from the molar mass of MOH and you are left with the molar mass of M+

mass M+ = 103 g - 17 g = 86 g
atomic mass of M = 86 amu
which if we look on the periodic table is close to the atomic mass of Rb