Chemistry Dilution?

1.
Stock solution: C₁ = 2.200 M, V₁ = ? mL
Diluted solution: C₂ = 0.5104 M, V₂ = 3.50 L = 3500 mL

For dilution: C₁V₁ = C₂V₂
Then, V₁ = V₂ × (C₂/C₁)
Volume of stock solution, V₁ = (3500 mL) × (0.5104/2.200) = 812 mL

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2.
Original solution: C₁ = 5.639 M, V₁ = 20.00 mL
Diluted solution: C₂ = ? M, V₂ = 250.00 mL

For dilution: C₁V₁ = C₂V₂
Then, C₂ = C₁ × (V₂/V₁)
Concentration of diluted solution, C₂ = (5.639 M) × (20.00/250.00) = 0.4511 M

Molar mass of NaCl = (23.0 + 35.5) g/mol = 58.5 g/mol
Mass of NaCl = (0.4511 mol/L) × (4.00/1000 mL) × (58.5 g/mol) = 0.106 g

OR:
Moles of NaCl in original solution = (5.639 mol/L) × (20.00/1000 L) = 0.1128 mol
Moles of NaCl in 250.00 mL of diluted solution = 0.1128 mol
Moles of NaCl in 4.00 mL of diluted solution = (0.1128 mol) × (4/250) = 0.001805 mol
Mass of NaCl in 4.00 mL of diluted solution = (0.001805 mol) × (58.5 g/mol) = 0.106 g