A student placed 11.5 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 25.0 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?

Given: Mass of glucose: w₁ = 11.5 g, Volume of solution 1: V₁ =100 mL, Volume of solution 2: V₂ = 0.5 L = 0.5 × 100 = 500 mL, Mass of glucose: w₂ = ? g

Molar mass of glucose; m = 180.16 g/mol

As, Molarity = (given mass × 1000) ÷ (molar mass × volume of solution in mL)

Molarity of glucose solution 1: M₁ = (w₁ × 1000) ÷ (m × V₁) = (11.5 g × 1000) ÷ (180.16 g/mol × 100 mL) = 0.638 M

Dilution of 25.0 mL 0.638 M solution to 500 mL:

According to the Dilution equation: M₁ × V₁ = M₂ × V₂

0.638 M × 25 mL = M₂ × 500 mL

M₂ = 0.638 M × 25 mL ÷ 500 mL

M₂ = 0.0319 M

Molarity of glucose solution 2: M₂ = 0.0319 M = (w₂ × 1000) ÷ (m × V₂)

⇒ mass of glucose: w₂ = (M₂ × m × V₂) ÷ (1000)

⇒ w₂ = (0.0319 M × 180.16 g/mol × 100) ÷ (1000) = 0.5747 g

Therefore, the mass of glucose in 100 mL of final solution: w₂ =0.5747 g