A student placed 15.0 gg of glucose (C6H12O6C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mLmL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 60.0 mLmL sample of this glucose solution was diluted to 0.500 LL. How many grams of glucose are in 100. mLmL of the final solution?

1.80 g

Explanation:

A student placed 15.0 g of glucose (C₆H₁₂O₆) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached.

The initial concentration of glucose is:

C₁ = 15.0 g/100 mL = 0.150 g/mL

A 60.0 mL sample of this glucose solution was diluted to 0.500 L (500 mL).

We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 0.150 g/mL × 60.0 mL / 500 mL

C₂ = 0.0180 g/mL

We can find the mass of glucose in 100 mL of the 0.0180 g/mL solution.

100 mL × 0.0180 g/mL = 1.80 g