Question

A 500mL iv bag contains 19mEq of Potassium Chloride (mol wt 75). How many mg of potassium chloride are present in the iv bag?

Answer 1

Amount of solution = 500 ml

Formula weight of KCL = 19mEq

Molecular weight of KCL = 75

1 mEq of KCL = 1/ 1000 * molecular weight of KCL

OR

mg = mEq* formula weight

Valence

Therefore , by applying first equation 1 mEq of KCL = 1 / 1000 * molecular weight of KCL

                                                                                   = 1 / 1000 * 75g

                                                                                   = 0.075 gm

                                                                                    = 75 mg

                                               Then, 19 mEq of KCL = 19 * 75

                                                                                   = 1425 mg/ml

Therefore, the amount of KCL in 500 ml = 1425 mg *500 ml

                                                                 = 712500 mg