A sterile solution of potassium chloride contains 2 mEq/mL. If a 20-mL vial of the solution is diluted to 1 liter, what is the percentage strength of the resulting solution?

Question

Rationale, and tips for what to look for in a problem like this.

Answer 1

Start with what is meant by % strength of the solution, since that is the question. Percent means parts per 100 parts. So, we are looking for grams of KCl per 100 ml of solution (for % weight/volume, or w/v).

We know the final volume is 1000 ml (1 L). So, we need to find the grams of KCl.

We start with what's given. 2 mEq / ml x 20 ml vial = 40 mEq KCl / 20 ml

40 mEq KCl x 1 mmol / mEq = 40 mmoles KCl in the 20 ml vial

40 mmole KCl x 74.6 mg/mmol = 2984 mg KCl in the 20 ml vial

Diluting this to 1 L (1000 ml) is a 50 fold dilution, so in the final 1 L we have 1/50 x 2984 mg = 59.7 mg

You could also do the math as ... 2984 mg/20 ml = x mg/1000 ml and x = 59.7 mg = 0.0597 g

So, now we have both parts to the question. We have the grams of KCl (0.0597 g) and the volume of 1000 ml

% strength = g / 100 ml = 0.0597 g / 1000 ml x 100 = 0.006% w/v (1 sig.fig.)