Suppose a student diluted and titrated a bleach unknown exactly as described in the experimental procedure, except only a single titration was performed which required 12.62 mL of 0.100 M Na_{2}S_{2}O_{3}

Calculate the number of moles of ClO in the sample titrated.

*The density of the original, undiluted bleach unknown was 1.042 g/mL.

This is a classic titration of hypochlorite in bleach. Usually done into a KI/ClO^- solution => I2 which is titrated with the S2O3^2- until all Red-Brown color is changed to clear solution. The ratio is 2 mole S2O3^2- to 1 mole ClO^-. So use Molarity x Volume in Liters => moles S2O3^2- used, then divide by 2 => moles of ClO^-.

To calculate the number of moles of ClO in the sample titrated, we need to use the balanced chemical equation for the reaction between ClO (bleach) and Na2S2O3 (sodium thiosulfate).

The balanced chemical equation for the reaction is:
2Na2S2O3 + ClO -> Na2S4O6 + NaCl

From the balanced chemical equation, we can deduce that the ratio of moles of Na2S2O3 to moles of ClO is 2:1.

Given that the volume of the Na2S2O3 solution used in the titration is 12.62 mL and its concentration is 0.100 M, we can calculate the number of moles of Na2S2O3 using the formula:

moles of Na2S2O3 = volume (in L) x concentration (in mol/L)

Converting the volume to liters:
12.62 mL = 12.62 mL x (1 L / 1000 mL) = 0.01262 L

Substituting the values into the formula:
moles of Na2S2O3 = 0.01262 L x 0.100 mol/L = 0.001262 mol

Since the ratio of moles of Na2S2O3 to moles of ClO is 2:1, the number of moles of ClO is half the number of moles of Na2S2O3:

moles of ClO = 0.001262 mol / 2 = 0.000631 mol

Therefore, the number of moles of ClO in the sample titrated is 0.000631 mol.