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

Calculate the molarity of Na₂S₂O₃:

Given that 33.50 cm³ of Na₂S₂O₃ solution was used:

M = rac{n}{V} = rac{0.00333 ext{ mol}}{0.03350 ext{ L}} = 0.0999 ext{ mol/L}.

Use the reaction stoichiometry to find moles of NaClO:

From the equation, 2 moles of S₂O₃²⁻ react with 1 mole of ClO⁻:

n(ClO⁻) = rac{n(S₂O₃²⁻)}{2} = rac{0.00333}{2} = 0.001675 ext{ mol}.

Calculate the mass of NaClO:

The molar mass of NaClO is approximately 74.44 g/mol:

ext{mass of NaClO} = n imes ext{molar mass} = 0.001675 ext{ mol} imes 74.44 ext{ g/mol} = 0.124865 ext{ g}.

Find the mass of the original bleach sample:

Given that the diluted bleach solution was 25 cm³ of the original 100 cm³, we can calculate:

ext{mass}_{ ext{bleach}} = ext{density} imes ext{volume} = 1.20 ext{ g/cm}³ imes 100 ext{ cm}³ = 120 ext{ g}.

Calculate the percentage by mass of NaClO in the bleach solution:

rac{ ext{mass of NaClO}}{ ext{mass}_{ ext{bleach}}} imes 100 = rac{0.124865}{120} imes 100 ext{ which equals } 0.10405 ext{ g}.

ext{Final answer: } 0.10405 ext{ % rounded to 3 significant figures gives: } 0.104 %.