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3 years ago

Help me ASAP PLS

Chemistry

1 answer:

n200080 [17]3 years ago

6 0

Answer:

18 g

Explanation:

We'll begin by converting 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Next, we shall determine the number of mole of the glucose, C₆H₁₂O₆ in the solution. This can be obtained as follow:

Volume = 0.5 L

Molarity = 0.2 M

Mole of C₆H₁₂O₆ =?

Molarity = mole / Volume

0.2 = Mole of C₆H₁₂O₆ / 0.5

Cross multiply

Mole of C₆H₁₂O₆ = 0.2 × 0.5

Mole of C₆H₁₂O₆ = 0.1 mole

Finally, we shall determine the mass of 0.1 mole of C₆H₁₂O₆. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.1 mole

Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mass = mole × molar mass

Mass of C₆H₁₂O₆ = 0.1 × 180

Mass of C₆H₁₂O₆ = 18 g

Thus, 18 g of glucose, C₆H₁₂O₆ is needed to prepare the solution.

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