(type 5) mEq problems - randomized
5. mEq and mOsmol 5.1) mEq 5.1.1) mEq - General
5.1.1.1) mEq from weight/volume Normal 1
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A solution contains 25 mg of sodium bicarbonate in each 5 mL (the MW of sodium bicarbonate is 84). How many milliequivalent of sodium bicarbonate will be present per liter of this solution?
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lb equals 59.52 mEq/L kg
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The molecular weight of sodium bicarbonate = 84.
Valence or charge = 1.
Therefore, we can write: 84 mg = 1 mEq.
Since the question needs mEq/L as the final unit, we keep mEq in the numerator. So, 84 mg will be the denominator.
Also, the solution contains 25 mg per 5 mL. To cancel out mg, we need to keep 25 mg in the numerator of second fraction. Therefore, the dimensional analysis will be like this:
`(1 \quad mEq)/(84 \quad mg) � (25 \quad mg)/(5 \quad mL) � (1000 \quad mL)/(1 \quad L)=59.52\quad (mEq)/L` Ans.