A 1.5L intraveneous (IV) solution contains 919 grams glucose (C6H12O6). What is the molarity of this solution?

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. in the first step, calcium carbide and water react to form acetylene and calcium hydroxide: cac2 (s) + 2h2o (g) → c2h2 (g) + caoh2 (s) =δh−414.kj in the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6c2h2 (g) + 3co2 (g) + 4h2o (g) → 5ch2chco2h (g) =δh132.kj calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. round your answer to the nearest kj .