How many grams of glucose (C₆H₁₂O₆) are required to prepare 50.0 mL of 6.30% (m/v) glucose-water solution?

Question

Quizplus · Accepted Answer

First, we need to calculate the mass of glucose required to prepare 50.0 mL of a 6.30% (m/v) solution.

6.30% (m/v) means 6.30 g of glucose is present in 100 mL of the solution.

So, the mass of glucose required to prepare 50.0 mL of the solution would be:

Mass of glucose = 6.30 g/100 mL x 50.0 mL = 3.15 g

Therefore, the correct choice is C.

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