Question

A student placed 12.5 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 flask was then shaken until the solution was uniform. A 55.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

There are 1.38 grams of glucose in 100. mL of the final solution

Explanation:

1) Concentration of glucose in the initial solution:

• Mass of glucose: 12.5 g (given)
• Volume os solution: 100. mL (given)
• Formula: % = (mass of solute / volume of solution)×100
• % glucose = (12.5g / 100. mL)×100 = 12.5%
• I will name this concentration C₁, i.e. C₁ = 12.5%

2) Dilution

• Formula: C₁ V₁ = C₂V₂

• Data:

        C₁ = 12.5%

        V₁ = 55.0 mL

        C₂ = ?

        V₂ = 0.500 L = 500. mL

• Concentration of the final solution, C₂:

        C₁ V₁ = C₂V₂ ⇒ C₂ = C₁ V₁ / V₂ = 12.5% × 55.0 mL / 500. mL = 1.375%

3) Grams of glucose in 100 mL of the final solution

• Formula: % = (mass of solute / volume of solution)×100

• Solve for mass of solute: mass of solute = (% / 100) × volume of solution

• mass of glucose = (1.375% / 100) × 100 mL = 1.375 g

• Round to three significant figures: 1.38 g