Answer
121.62 g/mol.
Work Step by Step
1. Write the equation of the titration to know the proportions.
*Normally, the Group 2 Metals (X) have +2 as charge of their Ions.
$X^{+2}_{}(O_{}H_{})^{-}_{2}$ + $H^{+}Cl^{-}$
1 $X^{+2}_{}(O_{}H_{})^{-}_{2}$ + 2 $H^{+}Cl^{-}$
2. Find the amount of mols (HCl) needed for the titration:
molarity = nº of moles (solute) ÷ volume (solution)(L)
2.5 = nº of moles ÷ 0.0569
nº of moles = 2.5 * 0.0569
nº of moles = 0.14225
3. Knowing that the proportion is 1 to 2. Find the nº of moles of the Hydroxide that reacted.
1 Mol (Hydroxide) -- 2 Moles (HCl)
x Mol (Hydroxide) -- 0.14225 Moles (HCl)
2 * x = 1 * 0.14225
x = 0.14225 / 2
x = 0.071125
4. With the number of moles and the mass, let's calculate the molar mass of the hydroxide.
molar mass = mass(g) ÷ nº of moles
molar mass = 8.65 ÷ 0.071125
molar mass $\approx$ 121.62
