How many mL of 0.750 M solution of lithium phosphate solution must be used to deliver 10.3 g of lithium?
回答 (1)
2017-05-27 2:26 pm
Molar mass of Li = 6.94 g/mol
Number of moles of Li = (10.3 g) / (6.94 g/mol) = 1.484 mol
1 mole of Li₃PO₄ contains 3 moles of Li.
Number of moles of Li₃PO₄ needed = (1.484 mol) × (1/3) = 0.4947 mol
Molarity of Li₃PO₄ solution = 0.750 mol/L
Volume of Li₃PO₄ = (0.4947mo) / (0.750 mol/L) = 0.660 L = 660 mL
Number of moles of Li = (10.3 g) / (6.94 g/mol) = 1.484 mol
1 mole of Li₃PO₄ contains 3 moles of Li.
Number of moles of Li₃PO₄ needed = (1.484 mol) × (1/3) = 0.4947 mol
Molarity of Li₃PO₄ solution = 0.750 mol/L
Volume of Li₃PO₄ = (0.4947mo) / (0.750 mol/L) = 0.660 L = 660 mL
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