(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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The normal blood level of K+ is 3.5 - 5.5 mEq/L. A patient is suffering from hypokalemia , and the blood level is found to be 2.86 mEq/L. Express this in mmol/dL unit.
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lb equals 0.286 mmol/dL kg
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The term deci liter (dL) means tenth of a liter, or, 100 mL. For K+ , the atomic weight is 39.1 and the valence is 1. Therefore, we can write:
39.1 mg = 1 mEq.
39.1 mg = 1 mmol.
`therefore` 1 mmol = 1 mEq.
This fraction `(1 \quad mmol)/(1 \quad mEq)` will be used in this problem. Using dimensional analysis, we can write: `(2.86 \quad mEq)/(1 \quad L) � (1 \quad mmol)/(1 \quad mEq) � (1 \quad L)/(1000 \quad mL) � (100 \quad mL)/(1 \quad dL) `
`= (0.286 \quad mmol)/(dL)` Ans.