Asked by Erica
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
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
Answers
Answered by
DrBob222
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.
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.
Answered by
Erica
thank you so much for the explanation
Answered by
ALAA
THAAAAAAAAAAANK YOOOOOOU !!!!
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