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igomit [66]
2 years ago
6

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

Chemistry
1 answer:
stiv31 [10]2 years ago
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

#SPJ3

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The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
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(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

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brainly.com/question/14356286

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