The active ingredient in many commercial liquid bleaches is sodium hypochlorite. The bottle lists the percentage of sodium hypochlorite as 6.0%. If the density of commercial bleach is 1.084 g/mL, how many mL of 0.150 M sodium thiosulfate is required to reach the end point in a titration similar to the one performed in this experiment, if a student analyzed a 2.0 mL sample of bleach.

23.3mL of 0.150M sodium thiosulfate

Explanation:

The  net reaction of sodium hypochlorite (NaClO) with sodium thiosulfate (Na₂S₂O₃) is:

NaClO + 2 Na₂S₂O₃ + 2H₃O⁺ → NaCl + 3 H₂O  +  Na₂S₄O₆ + 2 Na⁺

2.0mL of the sample of bleach are:

2.0mL ₓ (1.084g / mL) ₓ (6 / 100) ₓ (1 mol / 74.44g) = 1.75x10⁻³ moles of NaClO

As 1 mole of NaClO reacts with 2 moles of thiosulfate:

1.75x10⁻³ moles of NaClO ₓ (2 mol Na₂S₂O₃ / 1 mol NaClO) =

3.50x10⁻³ moles of Na₂S₂O₃

If you have a 0.150M solution of thiosulfate, mL of this solution you need to reach the end point of the titration are:

3.50x10⁻³ moles of Na₂S₂O₃ ₓ (1L / 0.150mol) = 0.0233L =

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

the answer is d. 74.02