**Answer:**

The mass of glucose in 100 mL of final solution: <u>w₂ =0.5747 g</u>

**Explanation:**

**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)

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

<u>Dilution of 25.0 mL 0.638 M solution to 500 mL:</u>

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)

⇒ <u>mass of glucose: w₂</u> = (M₂ × m × V₂) ÷ (1000)

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

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