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
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