hat is the molarity of a solution prepared by dissolving 12.0 g of ethylene glycol, C2H6O4, in water to make 250.0 mL of solution

0.512 M

Explanation:

We'll begin by calculating the number of mole in 12 g of C₂H₆O₄. This can be obtained as shown below:

Mass of C₂H₆O₄ = 12 g

Molar mass of C₂H₆O₄ = (2×12) + (6×1) + (4×16)

= 24 + 6 + 64

= 94 g/mol

Mole of C₂H₆O₄ =?

Mole = mass /molar mass

Mole of C₂H₆O₄ = 12 / 94

Mole of C₂H₆O₄ = 0.128 mole

Next, we shall convert 250 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

250 mL = 250 mL × 1 L / 1000 mL

250 mL = 0.25 L

Thus, 250 mL is equivalent to 0.25 L.

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Mole of C₂H₆O₄ = 0.128 mole

Volume = 0.25 L

Molarity =?

Molarity = mole /Volume

Molarity = 0.128 / 0.25

Molarity = 0.512 M

Thus, the molarity of the solution is 0.512 M

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