Determine the volume of a 0.119 M AlPO4 solution that contains 21.3 grams of AlPO4
Number of moles of AlPO4 = mass of AlPO4 / molar mass of AlPO4
Number of moles of AlPO4 = 21.3 g / 121.9529 g/mol = 0.175 mole
molarity of AlPO4 = number of moles of AlPO4 / volume of solution in L
0.119 = 0.175 / volume of solution in L
volume of solution in L = 0.175 / 0.119 = 1.47 L
1 L = 1000 mL
1.47 L = 1470 mL
Therefore, the volume of AlPO4 = 1470 mL
Determine the volume of a 0.119 M AlPO4 solution that contains 21.3 grams of AlPO4
please help 8) Determine the volume ofa 0.0246 M Li3PO4 solution that contains 11.8 grams of LI3PO4. (3pts)
10.35 contains 0.119 mol of gas and has a volume of 2.83 L ½" 2.78
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