What is the concentration of methanol in a solution prepared by combining 8.65mL of a 3.669M methanol solution with 16.35mL of water?

Method 1 :

M₁V₁ = M₂V₂
i.e. M₂ = M₁ × (V₁/V₂)

Final concentration, M₂ = (3.669 M) × [8.65/(8.65+16.35)] = 1.269 M


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Method 2 :

Number of milli-moles of methanol = (3.669 mmol/mL) × (8.65 mL)
Volume of the final solution = (8.65 + 16.35) mL = 25.00 mL
Final concentration = moles/volume = (3.669 mmol/mL) × (8.65 mL) / (25.00 mL) = 1.269 M