Sorry its not letting me post the entire thing in the other thread
.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols
= .000887972 mols S203^2-
Then M S2O3^-2 = .000887972 mols/0.02715 L.M = 0.032400
Therefore the Molar mass of Na2S2O3 is 0.032400
You don't mean the molar mass is 0.03240, you mean the molarity is 0.03240. I get the same figures you do. So the solution of Na2S2O3 is 0.03240 M. By the way, I think you are limited to just the one zero as I have it written and not 0.032400. Since the 27.15 mL, 25.00 mL aliquot, 0.1255 g KIO3 all have four significant figures, you can have no more than that in the answer.
Oh okay that you...and yes i meant molarity
thank*
To calculate the molar mass of Na2S2O3, you need to know the molar mass of each individual element (Na, S, and O) and add them together according to the molecular formula.
The atomic mass of sodium (Na) is 22.99 g/mol. Since there are two sodium atoms in Na2S2O3, the total contribution from sodium is 2 * 22.99 g/mol = 45.98 g/mol.
The atomic mass of sulfur (S) is 32.07 g/mol. There are two sulfur atoms in Na2S2O3, so the total contribution from sulfur is 2 * 32.07 g/mol = 64.14 g/mol.
The atomic mass of oxygen (O) is 16.00 g/mol. However, there are three oxygen atoms in Na2S2O3 since there is a subscript of 3 in the chemical formula. So the total contribution from oxygen is 3 * 16.00 g/mol = 48.00 g/mol.
Now, you can add up the contributions from all the elements:
45.98 g/mol (Na) + 64.14 g/mol (S) + 48.00 g/mol (O) = 158.12 g/mol
Therefore, the molar mass of Na2S2O3 is 158.12 g/mol.