150.0 mL of a solution containing Ba2+ ions is analyzed by adding 0.18 M Na2SO4 solution until no more precipitate forms. The mass of BaSO4 precipitate obtained is 11.20 g. Select all the options that correctly reflect the steps required to calculate the concentration of Ba2+ in the original solution.- The concentration of Ba2+ is equal to 0.18 M.- The original solution contained 0.0270 moles of Ba2+ ions.- The concentration of Ba2+ ions is equal to 0.320 M.- The molar mass of BaSO4 is 233.4 g/mol.- The original solution contained 4.80 x 10-2 moles of Ba2+.- The concentration of Ba2+ ions is equal to 0.320 M.- The molar mass of BaSO4 is 233.4 g/mol.- The original solution contained 4.80 x 10-2 moles of Ba2+.



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