Question #a24ac

1 Answer
Aug 8, 2015

Answer:

Here's how you can approach this problem.

Explanation:

You know that you're dealing with an infusion that is said to be #"0.5000% w/v"# #"KCl"#.

What you have to do here is use the mass of potassium that was found in the diluted sample to determine the mass of potassium chloride found in the diluted sample, then work backwards from dilution to dilution to see what the mass by volume percent concentration of potassium chloride was for the original sample.

So, start by determining the percent composition of potassium in potassium chloride - use the molar mass of potassium and the molar mass of #"KCl"#

#(39.0983color(red)(cancel(color(black)("g/mol"))))/(74.551339.0983color(red)(cancel(color(black)("g/mol")))) * 100 = "52.445%"#

Now, you know that you have 0.6670 mg of potassium per 100 mL of this solution; since you have a 200-mL sample, it means that you get

#200color(red)(cancel(color(black)("mL"))) * ("0.6670 mg K")/(100color(red)(cancel(color(black)("mL")))) = "1.334 mg K"#

This means that the mass of potassium you've found (which by the way is expressed in miligrams, not grams) will correspond to

#1.334 * 10^(-3)color(red)(cancel(color(black)("g K"))) * "100 g KCl"/(52.445color(red)(cancel(color(black)("g K")))) = 2.5436 * 10^(-3)"g KCl"#

Now that you know how much potassium chloride you have in the final solution, work your way backwards from dilution to dilution.

You know that this final solution was made by diluting a 10-mL sample until the final volume was 200 mL.

This means that this much potassium chloride was found in the 10-mL sample, since diluting a solution implies keeping the amount of solute constant.

So, you know that 10-mL of the solution that was first diluted contained #2.5436 * 10^(-3)"g"# potassium chloride.

SInce you made this first solution by diluting a 10-mL sample to 200 mL, you can say that it contained

#2.5436 * 10^(-3)color(red)(cancel(color(black)("g KCl"))) * (200color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = "0.050872 g KCl"#

Finally, this first solution was made by taking a 10-mL sample and diluting it to 200 mL, which means that the initial sample contained exactly the same amount of potassium chloride as this sample.

Therefore, the mass by volume percent concentration of potassium chloride in the stock solution is

#"0.050872 g K"/"10 mL" * 100 = color(green)("0.5087%w/v KCl")#

Now, to get the % stated content of potassium chloride, simply divide what you calculated by what was given to you and multiply by 100

#"% stated content" = (0.5087)/(0.5000) * 100 = color(green)("101.74%")#