Given that the molar mass of H3PO4 is 97.994 grams, determine the number of grams of H3PO4 needed to prepare 1.50 L of a 0.25 M H3PO4 solution. Show your work. (5 points)
(b) What volume, in liters, of a 0.25 M H3PO4 solution can be prepared by diluting 50 mL of a 1.75 M H3PO4 solution?
Given that the molar mass of H3PO4 is 97.994 grams, determine the number of grams of H3PO4 needed to prepare 1.50 L of 0.25 M H3PO4 solution?
2016-10-19 4:53 pm
回答 (1)
2016-10-19 5:04 pm
(a)
No. of moles of H₃PO₄ needed = (0.25 mol/L) × (1.50 L) = 0.375 mol
Mass of H₃PO₄ needed = (97.994 g/mol) × (0.375 mol) = 36.7 g (3 sig. fig.)
(b)
C₁V₁ = C₂V₂
(1.75 M) × (50 mL) = (0.25 M) × V₂
Final volume, V₂ = 1.75 × 50 / 0.25 mL = 350 mL = 0.35 L
No. of moles of H₃PO₄ needed = (0.25 mol/L) × (1.50 L) = 0.375 mol
Mass of H₃PO₄ needed = (97.994 g/mol) × (0.375 mol) = 36.7 g (3 sig. fig.)
(b)
C₁V₁ = C₂V₂
(1.75 M) × (50 mL) = (0.25 M) × V₂
Final volume, V₂ = 1.75 × 50 / 0.25 mL = 350 mL = 0.35 L
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