How many grams of oxalic acid dihydrate (COOH)2 x 2H2O, is needed to prepare 525 mL of a .570 M solution?

1 Answer
Jul 10, 2017

Answer:

Approx. #38*g......#

Explanation:

We require a MOLAR quantity of .............#525*mLxx10^-3*L*mL^-1xx0.570*mol*L^-1=0.299*mol# with respect to #"oxalic acid dihydrate"#.

#"Oxalic acid dihydrate"#, #"HO"_2"CCO"_2"H"*"2H"_2"O"#, has a formula mass of #126.07*g*mol^-1#.....

And thus we take the product....

#126.07*g*mol^-1xx0.299*mol-=??*g#.