2 years ago

Calculate the number of grams of ethanol nevessary to prepare 268g of a 9.76m solution of ethanol in water

1 answer:

7 0

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

<h3>Solution preparation</h3>

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

More on solution preparation can be found here: brainly.com/question/14667249

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