How many milligrams of MgI2 must be added to 225.0 mL of .0858 M KI to produce a solution with I- = 0.1000M?!!
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How many milligrams of MgI2 must be added to 225.0 mL of .0858 M KI to produce a solution with I- = 0.1000M?!!

[From: ] [author: ] [Date: 12-03-07] [Hit: ]
PLEASE help if you can. Ill tell you what the answers ARENT. They are NOT - .0107 or .005369. PLEASE AND THANK YOU SO MUCH.......
Please, I've been stumped on this problem for quite a while, and I have an exam on Thursday...PLEASE help if you can. I'll tell you what the answers AREN'T. They are NOT - .0107 or .005369. PLEASE AND THANK YOU SO MUCH.

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This may be weird, but i think you might be in my chem class haha... cause i also have a chem test on thursday and had a problem like this assigned.... anyway....

0.0858 mol/L KI * 0.225 L = 0.019305 mol of KI, and since KI and I- and 1-to-1, that gives you 0.019305 mol I-.

225 mL of 0.1000 M KI contains 0.0225 mol of I- (by multiplying molarity by volume)

Subtract the two, and you get 0.003195 mol, which is the amount of I- you need to add. Divide this by two since you actually adding MgI2, which has double the concentration of I-.
so now you have 0.0015975 mol MgI2

Multiply this by the molar mass of MgI2 (278.113g), and then multiply again by 1000 to get your answer in mg.

Your final answer should be 444 mg after being rounded.
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