(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

    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.


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