Question

The oral pediatric maintenance solution Pedialyte liquid has the following electrolyte content per liter: 45 mEq of sodium, 20 mEq of potassium, 35 mEq of chloride, 30 mEq of citrate. Calculate the equivalents quantities of each in terms of milligrams. Na+ = 1035 milligrams (mg) K + = 780 milligrams (mg) Cl- = 1242.5 milligrams (mg) Citrate = 1890 milligrams (mg) i just need the work shown for the Citrate = 1890 milligrams i know how to do the rest. i belive the valance is (3) but i still dont know how they got that answer.

Given a 500-mL parenteral fluid containing 5% w/v sodium bicarbonate, a. Calculate milliequivalents per milliliter Answer: 0.595 mEq/mL b. Calculate total milliequivalents Answer: 297.62 mEq c. Calculate osmolarity Answer: 1,190.48 mOsmol/L i just need the explanation for Calculate osmolarity Answer: 1,190.48 mOsmol/L i was able to get the rest

thank you

Answer 1

Answer (1)

Atomic weight of sodium = 23
Equivalent weight of sodium = 23
mEq weight = 23 × 10-3 g of sodium
= 0.023 g of sodium
45 mEq of sodium = 45× 0.023
= 1.035 g of sodium
= 1.035 × 103 mg of Sodium
= 1035 mg of sodium

M.W of K+ = 39
Equivalent weight of K+ = 39
mEq weight = 39 × 10-3
= 0.039 g of K+
∴ mEq weight = 20 × 0.039 g of K+
= 0.78 g of K+
= 780 mg of K+

Atomic weight of chloride = 35.5
Equivalent weight chloride = 35.5
mEq weight of chloride = 35.5 × 10-3 g
= 35.5 g of chloride
35 mEq weight = 35 × 35.5 mg of Chloride
= 1242.5 mg of chloride

1035 mg/L sodium

1242.5 mg/L chloride

780 mg/L potassium

1890 mg/L citrate

Answer (2)

5 % w/v of NaHCO3 means
100 mL of parental fluid contain 5 g of NaHCO3
1 mL of parental fluid contain


of NaHCO3
= 0.05 g of NaHCO3
M.W of NaHCO3 = 84
Equivalent weight = 84
mEq = 84 ×10−3 g
= 0.084 g
0.084 g of NaHCO3 = 1 mEq
0.05 g of NaHCO3 =


× 0.05
= 0.595 mEq
500 mL of parental fluid contain


= 25 g
M.W of NaHCO3 = 84
mEq= 84 × 10−3
= 0.084 g of NaHCO3
0.084 g of NaHCO3 represent = 1 mEq
25 g of NaHCO3 represent =


Ans∶ 9 9 mEq
(c)
100 mL contain 5 mg
1000 mL contain


= 50 g
∴ g L
Mosmole/L =


= 1190.48 mosmole/L
Ans∶ 9 mosmole L

(b)
100 mL of parental fluid contain 5 g of NaHCO3


(a) 0.595 mEq/mL sodium bicarbonate

(b) 297.62 mEq sodium bicarbonate

(c) 1190.48 mOsmol/L