11. Answer two of the parts (a), (b) and (c) - Leaving Cert Chemistry - Question 11 - 2002

To express a 5% (w/v) solution in grams per liter, we first note that 5% (w/v) means there are 5 grams of sodium hypochlorite in 100 cm³ of solution. Therefore, in 1 liter (1000 cm³), the concentration can be calculated as follows:

• 5 grams per 100 cm³ implies that for 1000 cm³:

  = 5 g × (1000 cm³ / 100 cm³)

  = 50 g

So, the concentration of the sodium hypochlorite solution is 50 g/L.

From the balanced equation:

$$ext{OCl}^- + ext{Cl}^- + 2 ext{H}^+ ightarrow ext{Cl}_2 + ext{H}_2 ext{O}$$

For every mole of OCl⁻, one mole of Cl₂ is produced. The amount of OCl⁻ present in the 100 cm³ of 5% solution was calculated earlier. First, we find the moles of OCl⁻:

• From the 50 g of NaOCl, the number of moles is given by:

  = rac{50 ext{ g}}{74.5 ext{ g/mol}} ext{ (molar mass of NaOCl)}

  = 0.674 ext{ mol}

Given the stoichiometry, the moles of Cl₂ produced will be 0.067 moles (following rounding from calculations). Therefore, the molecules liberated can be calculated as:

• Number of molecules = 0.067 moles × Avogadro's number (approximately $6.022 × 10^{23}$)

Thus, the total number of molecules of chlorine gas liberated is:

$ext{Number of molecules} = 0.067 imes 6.022 imes 10^{23} \approx 4 imes 10^{22} ext{ molecules}$

At standard temperature and pressure (s.t.p.), 1 mole of an ideal gas occupies a volume of 22.4 liters. To find the volume of chlorine gas produced under these conditions, we use the previously calculated moles of Cl₂:

• Volume at s.t.p. = Number of moles × 22.4 L/mol

  = 0.067 moles × 22.4 L/mol

  = 1.5 L

Therefore, the volume occupied by this quantity of chlorine gas at s.t.p. is 1.5 liters.