(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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If potassium acetate solution is expressed as 15.31 mEq/L, then how many milligram of potassium acetate will be present in each 1 mL of the solution? Round your answer to the nearest tenth. The molecular weight of potassium acetate is 98 g/mol.
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lb equals 1.5 mg/1mL kg
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The molecular weight of potassium acetate = 98.
Valence or charge = 1.
Therefore, we can write: 98 mg = 1 mEq.
Since the question needs mg/1 mL as the final unit, we keep 98 mg in the starting numerator. So, 1 mEq will be the denominator.
Also, the solution contains 15.31 mEq per L; therefore, the mEq is used as the numerator for the second fraction, in order to cancel out. The dimensional analysis will be like this:
`(98 \quad mg)/(1 \quad mEq) � (15.31 \quad mEq)/(1 \quad L)� (1 \quad L)/(1000 \quad mL) � (1 \quad mL)/(1 \quad mL)`
` = 1.5\quad (mg)/(1 \quad mL)` Ans.