Question

what is the concentration of the CO3 anion in grams per liter of an unknown solution of a 10 mL aliquot yields 0.8352g of calcium carbonate

Answer 1

CaCO3 mass = 0.8352 g

CaCO3 moles = mass of CaCO3 / ( molar mass of CaCO3) = ( 0.8352g) / ( 100 g/mol) = 0.008352 mol

we got 0.008352 mol of CaCO3 when we had taken 10 ml solution

per 1000 ml ( i.e 1L) solution we will get CaCO3 moles = ( 1000 ml ) x ( 0.008352 mol / 10 ml) = 0.8352 mol

CaCO3 gives 1Ca2+ and 1CO3^2- ions per dissociation.

Hence CO3^2- moles formed = CaCO3 moles dissociated = 0.8352

CO3^2- mass = CO3^2- moles x molar mass of CO3^2- = 0.8352 mol x 60 g/mol = 50.1 g

Thus CO3^2- anion in g/L is given by 50.1 g/L