The amount, in grams, of ethanol needed to prepare 268 grams of a 9.76 M solution of ethanol in water will be 120.50 grams
Solution preparation
In order to prepare 268 grams of solution, 268 mL of water would be needed because 1 mL of water weighs 1 gram.
Since we now know the volume of solution that we want to prepare, the amount of solute (ethanol) that would be required in order to make a solution of 9.76 M will be:
Mole required = 9.76 x 268/1000 = 2.62 moles
Mass of 2.62 moles ethanol = 2.62 x 46.07 = 120.50 grams
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