Can someone me with how to do this question? i have an exam tomorrow, and it’s greatly appreciated : )) in a solution containing 15.6 g of mgcl2 in 1.25l of solution, what is the concentration of cl- ions?

In a solution of MgCl₂, the ionic compound will dissociate in water into its ions:
                       MgCl₂ -> Mg²⁺ + 2 Cl⁻
Note that since in the formula Cl has a subscript of 2, this implies there are 2 Cl⁻ ions for every Mg²⁺ ion.
Then we find the moles of MgCl₂, given that we have 15.6 g and then we get to finding moles of Cl⁻
The molar mass of MgCl₂ is 24.3 + 2(35.5) = 95.3 g/mol
           moles MgCl₂ = grams MgCl₂ * 1 mol / 95.3 g
           moles MgCl₂ = 15.6 g * 1 mol / 95.3 g
           moles MgCl₂ = 0.16369 mol
Using stoichiometric ratios, we have 1 MgCl₂ for every 2 Cl⁻ so
                moles Cl⁻ = moles MgCl₂ * 2 Cl⁻ / 1 MgCl₂
                moles Cl⁻ = 0.16369 mol * 2  / 1
                moles Cl⁻ = 0.32738 mol
Concentration is number of moles over volume in liters so
                          [Cl⁻] = 0.32738 mol / 1.25 L
                                   = 0.262 M
The answer to the question is [Cl⁻] = 0.262 M

And a) oxygen i believe