What is the molarity of a solution made from dissolving 10.0 g of barium chlorid

Question

What is the molarity of a solution made from dissolving 10.0 g of barium chloride in 500. mL of water? What is the concentration of the barium ions in this solution? What is the concentration of the chloride ions in this solution? I think I have the first part: The moles of BaCl2 = (10.0 g)(1 mol BaCl2/208 g BaCl2) = 0.0481 mol BaCl2 Mass of water = (500 mL)(1 L/1000 mL) = 0.500 L Mass of BaCl2 = (10.0)(1 L/1000 mL) = 0.01 L This means the total mass of the solution is 0.510 L Molarity = moles of solute/L of solution = 0.481 mol BaCl2/0.510 L = 0.943 M First, is that correct? Second, how do I find the concentration of the ions?

Explanation / Answer

small error in calculation of concentration. volume of solution doesnot changes when solute is added to solvent. so vol of solvent = sol of solution. molarity = moles /vol (in liters) = 0.0481/0.5 = 0.0962 each BaCl2 has 1 Ba ion and 2 Cl- ions as per stochiometry. so conc of [Ba2+] = 0.0962 M conc of [Cl-]= 2 x 0.0962 = 0.1924 M

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