An indigestion tablet contains a mass of 0.30 g of magnesium hydroxide [Mg(OH)₂] as its only basic ingredient - Leaving Cert Chemistry - Question 10 - 2005

The salt formed is magnesium chloride (MgCl₂). To find the mass:

1. Moles of Mg(OH)₂ = 0.0103 mol (from previous calculation).

2. Each mole of Mg(OH)₂ produces 1 mole of MgCl₂:

   • Thus, moles of MgCl₂ = 0.0103 mol.

3. Molar mass of MgCl₂ = 24.31 + 2(35.45) = 95.21 g/mol.

4. Mass of MgCl₂ = moles × molar mass:

   • Mass = $0.0103 \times 95.21 = 0.980 \approx 0.98 \, \text{g}$.

Since each formula unit of MgCl₂ contains one magnesium ion, the number of magnesium ions can be calculated:

1. Moles of MgCl₂ = 0.0103 mol (from previous part).

   • Number of ions = moles × Avogadro's number (approximately $6.022 \times 10^{23} \text{ ions/mol}$).

2. Total magnesium ions = $0.0103 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol} = 6.20 \times 10^{21} \, \text{Mg}^{2+} ext{ ions}$.

To find the volume of the second remedy that has a 6% (w/v) concentration, we need the amount of Mg(OH)₂ required:

1. We know from earlier that 0.60 g of Mg(OH)₂ is needed.

   • A 6% (w/v) solution means 6 g of solute in 100 cm³ of solution.

2. Therefore, in 1 cm³, the amount of Mg(OH)₂ is:

   • $\frac{6 \, \text{g}}{100 \, \text{cm}^3} = 0.06 \, \text{g/cm}^3$.

3. To find the required volume (V) that provides 0.60 g:

   • $V = \frac{0.60 \, \text{g}}{0.06 \, \text{g/cm}^3} = 10 \, \text{cm}^3$.