A 10.0 mL volvume of Ultra Bleach is diluted in a volumetric flask. A 25.0 ml sample of this solution is analyzed accordign to the procedure in this experiment. Given that 33.75 mL of 0.135 M Na2S2O3 are needed to reach the stochiometric point, answer the following.
a. How many grams of available Cl2 are in the titrated sample?
b. How many grams of Ultra Bleach are analyzed? Assume that the density of bleach is 1.084g/mL.
c. Calculate the percent available chlorine in the Ultra Bleach.
Don't we need to know the volume of the volumetric flask?
To answer these questions, we need to use stoichiometry and the given information. Let's go step by step.
a. How many grams of available Cl2 are in the titrated sample?
We know that 33.75 mL of 0.135 M Na2S2O3 is needed to react with the available Cl2. We can use the balanced chemical equation to find the molar ratio between Cl2 and Na2S2O3:
Cl2 + 2Na2S2O3 -> 2NaCl + Na2S4O6
From the equation, we can see that 1 mole of Cl2 reacts with 2 moles of Na2S2O3. Therefore, the moles of Cl2 can be calculated as follows:
moles of Cl2 = (0.135 mol/L) * (33.75 mL/1000 mL) * (1 L/1000 mL)
Next, we need to convert the moles of Cl2 to grams by multiplying with the molar mass of Cl2, which is 70.906 g/mol. Hence:
grams of Cl2 = moles of Cl2 * molar mass of Cl2
b. How many grams of Ultra Bleach are analyzed? Assume that the density of bleach is 1.084 g/mL.
We are given that 10.0 mL of Ultra Bleach was diluted to make the solution. To find the grams of Ultra Bleach in the titrated sample, we can use the density and volume of the solution.
grams of Ultra Bleach = (volume of solution in mL) * (density of bleach in g/mL)
c. Calculate the percent available chlorine in the Ultra Bleach.
The percent available chlorine can be calculated as:
percent available chlorine = (grams of Cl2 / grams of Ultra Bleach) * 100
By substituting the values we obtained from the previous steps, we can calculate the percent available chlorine in the Ultra Bleach.