How many moles and numbers of ions of each type are

Here we have to calculate how many moles and numbers of ions of each type are present in the following aqueous solutions.

Step 1 of 6

(a) \(88 \mathrm{~mL}\) of \(1.75 \mathrm{M}\) magnesium chloride

1st we have to write the chemical reaction for the dissociation of \(\mathrm{AlCl}_{3}\) (aluminum chloride)

\(\mathrm{MgCl}_{2}(\mathrm{~s}) \rightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})\)

In the above reaction it has been found that \(1 \mathrm{~mol}\) of \(\mathrm{Mg}^{2+}\) and \(2 \mathrm{~mole}\) of \(\mathrm{Cl}^{-}\) is dissolved per mole of \(\mathrm{MgCl}_{2}\)

In this question it has been given that molarity of \(\mathrm{MgCl}_{2}\) is \(1.75 \mathrm{M}\) and volume is \(88 \mathrm{~mL}\).

Mole of \(\mathrm{Gg}^{2+}\) formed from \(\mathrm{MgCl}_{2}:-\)

\(88 \mathrm{~mL}\) of \(\times \frac{1 L}{1000 \mathrm{~mL}} \times \frac{1.75 \mathrm{MMgCl}_{2}}{L} \times \frac{1 \text { molof } M g^{2+}}{1 \operatorname{mol} M g C l_{2}}\) \(=0.154 \mathrm{~mol}\) of \(\mathrm{Mg}^{2+}\)

Mole of \(\mathrm{Cl}^{-}\) formed fom \(\mathrm{MgCl}_{2}:\)

\(\begin{aligned}&88 \mathrm{~mL} \text { of } \times \frac{1 L}{1000 \mathrm{~mL}} \times \frac{0.45 \mathrm{MMgCl}_{2}}{L} \times \frac{2 \operatorname{molofCl}^{-}}{1 \mathrm{~mol} \mathrm{MgCl}_{2}} \\&=0.079 \mathrm{~mol} \text { of } \mathrm{Cl}^{-}\end{aligned}\)