If 56.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 554 mg sample of na2so4 (forming baso4), what is the molarity of the solution?

0.0687 m The balanced equation is BaCl2 + Na2SO4 = = > BaSO4 + 2 NaCl Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights. Atomic weight sodium = 22.989769 Atomic weight sulfur = 32.065 Atomic weight oxygen = 15.999 Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol Molarity is defined as moles per liter, so let's do the division. 0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m Rounding to 3 significant figures gives 0.0687 m