What is the mass (g) of solute in 175 mL of a 1.287 x 10^-3 M solution of Na3PO4? ?
回答 (1)
2020-09-30 4:34 pm
Molar mass of Na₃PO₄ = (23.0×3 + 31.0 + 16.0×4) g/mol = 164.0 g/mol
Moles of Na₃PO₄ in the solution = (1.287 × 10⁻³ mol/L) × (175/1000 L) = 2.252 × 10⁻⁴ mol
Mass of Na₃PO₄ in the solution = (2.252 × 10⁻⁴ mol) × (164.0 g/mol) = 0.0369 g
Each mole of Na₃PO₄ contains 3 moles of Na⁺ ions and 1 mole of PO₄³⁻ ions.
Molar concentration of Na⁺ ions = (1.287 × 10⁻³ M) × 3 = 3.861 × 10⁻³ M
Molar concentration of PO₄³⁻ ions = 1.287 × 10⁻³ M
Moles of Na₃PO₄ in the solution = (1.287 × 10⁻³ mol/L) × (175/1000 L) = 2.252 × 10⁻⁴ mol
Mass of Na₃PO₄ in the solution = (2.252 × 10⁻⁴ mol) × (164.0 g/mol) = 0.0369 g
Each mole of Na₃PO₄ contains 3 moles of Na⁺ ions and 1 mole of PO₄³⁻ ions.
Molar concentration of Na⁺ ions = (1.287 × 10⁻³ M) × 3 = 3.861 × 10⁻³ M
Molar concentration of PO₄³⁻ ions = 1.287 × 10⁻³ M
收錄日期: 2021-04-12 12:53:22
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