A student does an experiment to determine the percentage by mass of sodium chlorate(I), NaClO, in a sample of bleach solution - AQA - A-Level Chemistry - Question 8 - 2020 - Paper 1

To find the percentage by mass of NaClO in the original bleach solution, we first need to determine how many moles of NaClO are in the original bleach solution sample.

1. Calculate the concentration of the Na₂S₂O₃ solution used:

   Given that 33.50 cm³ (or 0.03350 L) of Na₂S₂O₃ was used, and from the stoichiometry of the reaction, we can derive that:

   $ext{Moles of Na₂S₂O₃} = 2 imes ext{Moles of ClO⁻}$

   2 moles of Na₂S₂O₃ react with 1 mole of I₃⁻, therefore:

   $n( ext{I}₃⁻) = 0.03350 imes 2 = 0.06700 ext{ moles}$

2. The number of moles of NaClO is half that of Na₂S₂O₃:

   $n( ext{NaClO}) = 0.06700 imes 1 = 0.03350 ext{ moles}$

   The mass of NaClO can then be found using the molar mass of NaClO, which is approximately 74.44 g/mol:

   $ext{mass of NaClO} = n( ext{NaClO}) imes ext{Molar mass} = 0.03350 imes 74.44 ext{ g} = 2.49 ext{ g}$

3. To find the mass of the bleach solution:

   Since the density of the original bleach solution is 1.20 g/cm³ and we started with 10.0 cm³:

   $ext{mass of bleach} = ext{density} imes ext{volume} = 1.20 imes 10.0 = 12.0 ext{ g}$

4. Now we can calculate the percentage by mass of NaClO in the bleach solution:

   ext{Percentage by mass} = rac{ ext{mass of NaClO}}{ ext{mass of bleach}} imes 100 = rac{2.49}{12.0} imes 100 = 20.74 ext{%}

   Rounding to three significant figures gives us:

   ext{Percentage by mass} = 20.7 ext{%}