0.535m is 0.535 molal = 0.535 mols KCl/kg solvent.
0.535 mols KCl = 74.55 x 0.535 = 39.88 g KCl.
Therefore a 0.535 m soln consists of 39.88 g KCl + 1000 g H2O for a total of 1039.88 g solution.
You want 107 g of soln; therefore, you want to take
39.88 x (107/1038.88) = about 4.1g KCl for 107 g of 0.535m soln. (The problem didn't ask for the solvent but you would take 107-4.1 = about 103 g H2O.)
Check my thinking. Redo the problem and clean up any estimates I made. Also look at the correct number of significant figures at the end.
I need help please I am so confused.
How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?
I know the MM of KCl is 74.55
107g of 0.535m KCl
3 answers
thank you so much for the explanation
THAAAAAAAAAAANK YOOOOOOU !!!!