Question

Calculate (a) the milliequivalents per milliliter, (b) the total milliequivalents, and (c) the osmolarity of a 500-mL parenteral fluid containing 5% w/v of sodium bicarbonate

Answer 1

(a) 5 % w/v of NaHCO3 means 5 grams of NaHCO3 in 100 mL

M.W of NaHCO3 = 84

1 mL of parental fluid contain calculation gives

1 mL ----------- ( 5 /100) = 0.05 g

0.05 g of NaHCO3 present in 1 mL

mEq = 84 ×10⁻³ g ( milli means 1/1000 th)
= 0.084 g
0.084 g of NaHCO3 = 1 mEq

now 0.084 g of NaHCO3 = 1 mEq
0.05 g of NaHCO3 = ?

(0.05 g x 1 mEq) / 0.084 g = 0.595 mEq

Answer for (a) the milliequivalents per milliliter = 0.595 mEq/mL

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(b) 500 mL parenteral fluid contains NaHCO3 = 500 mL x 0.05 g = 25 g NaHCO3

0.084 g of NaHCO3 represent = 1 mEq
25 g of NaHCO3 represent = (25 g x 1 mEq) / 0.084 = 297.62 mEq

Answer for (b) the total milliequivalents = 297.62 mEq

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(c) 100 mL contain 5 g NaHCO3
1000 mL contain = 50 g/ L    (1L = 1000 mL ) and for 100 mL
Mosmole/L = [(0.595 mEq/ mL) / 50 g ]x 100 mL x 1000 mL= 1190 mosmole/L

Answer for (c) the osmolarity of a 500-mL parenteral fluid containing 5% w/v of sodium bicarbonate = 1190 Mosmole/L

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Final answers are:

(a) 0.595 mEq/mL sodium bicarbonate

(b) 297.62 mEq sodium bicarbonate

(c) 1190 mOsmol/L

I would recommend to see module 8 for further understanding on how to calculate these fundamental units at https://static-documents.easygenerator.com/bfce45ef-52e9-40e8-9b69-f572f39f2750.pdf

Hope this helped you!

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