Question #5641d

1 Answer
Aug 24, 2015

Answer:

The solution contains 8.17 g of #"H"_3"PO"_4#.

Explanation:

#"Equiv. mass" = "molar mass"/"no of ionizable H" = "98.00 g"/3 = "32.67 g"#

#"Mass of H"_3"PO"_4 = 0.500 color(red)(cancel(color(black)("L"))) × (0.500 color(red)(cancel(color(black)("eq"))))/(1 color(red)(cancel(color(black)("L")))) × "32.67 g"/(1 color(red)(cancel(color(black)("eq")))) = "8.17 g"#