How much #KI# is present in .836 L of a 1.82 M solution, assuming the solvent is #H_2O#?

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Answer 1 · 2016-03-11T22:15:28

The solution contains 252 g of #"KI"#.

Step 1. Calculate the moles of #"KI"#.

Your solution contains 1.82 mol of #"KI"# in 1 L of solution.

So, in 0.836 L of solution you have

#0.836 color(red)(cancel(color(black)("L solution"))) × "1.82 mol KI"/(1 color(red)(cancel(color(black)("L solution")))) = "1.522 mol KI"#

Step 2. Convert moles of #"KI"# to grams of #"KI"#

#"1 mol KI"# has a mass of #"(39.10 + 126.90) g = 166.00 g"#.

Hence,

#"Mass of KI" = 1.522 color(red)(cancel(color(black)("mol KI"))) × "166.00 g KI"/(1 color(red)(cancel(color(black)("mol KI")))) = "252 g KI"#