0.5118 g sample of CaCO3 is dissolved in 6M HCl, and the resultingsolution is di

Question

0.5118 g sample of CaCO3 is dissolved in 6M HCl, and the resultingsolution is diluted to 250.0 ml in a volumetric flask. How many moles of CaCO3 are in the sample? (formula mass =100.1)
What is the molarity of the Ca2+ ion in the 250 ml ofsolution?
How many moles of Ca2+ ion are in a 25.00 ml aliquot of thesolution? How many moles of CaCO3 are in the sample? (formula mass =100.1)
What is the molarity of the Ca2+ ion in the 250 ml ofsolution?
How many moles of Ca2+ ion are in a 25.00 ml aliquot of thesolution?

Explanation / Answer

MW of CaCO3 = 100.087 g/mol (0.5118 g) / 100.087 g/mol = 0.0051 mols molarity = mols / liter = 0.0051 mols / .25 L = 0.0205 M 0.0205 M * 0.025 L = 0.0005125 mols of CaCO3 = 0.0005125 mols ofCa2+ ions

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