If you want this done with equivalents, then
mL x M x (molar mass KIO3/6000) = grams KIO3
The mass of potassium iodate (KIO3) contained in an impure sample was determined by titration with sodium thiosulfate (0.1005 M). The impure KIO3 was dissolved in 50 cm 3 of water, an excess of potassium iodide (KI) and 5 cm^3 of dilute sulphuric acid was then added. At the starch endpoint, 30.65 cm^3 of thiosulfate solution had been added. Calculate the mass of KIO3 in the sample.
I - → ½I2 + e -
IO3- + 6H+ + 5e - → ½I2 + 3H2O
The liberated iodine reacts as follows:
I2 + 2e - → 2I-
2S2O3 2- → S4O62- + 2e-
Hint: 6 equivalents of S2O32- are equivalent to IO3-.
5 answers
How do I find mL?
The problem tells you.
30.65 cm^3 = 30.65 cc = 30.65 mL.
1 cc = 1 mL = 1 cubic centimeter = 1 cm^3.
30.65 cm^3 = 30.65 cc = 30.65 mL.
1 cc = 1 mL = 1 cubic centimeter = 1 cm^3.
I got 0.1099g
Is that correry?