### Type 3: Determine percent strength from osmolarity of the solution

5. mEq and mOsmol 5.2) mOsmol 5.2.1) mOsmol - General

5.2.1.1) mOsmol - General Normal 1

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A potassium bicarbonate solution exerts tonicity of 0.32 mOsmol/ml. If the compound has a molecular weight of 100 g/mol, then express the concentration of the solution as percent w/v. Click on the button below to see the answer and explanations

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lb equals 1.6 % kg

For potassium bicarbonate , the 'golden formula' is:
1 mmol =  mg =  mEq =  mOsmol.

The solution contains 0.32 mOsmol per 1 mL.
Since we need g quantity in the numerator of the desired unit, we begin with keeping the mg unit in the numerator of the first fraction.
Again, since 0.32 mOsmol is contained in 1 mL of solution, we keep the 0.32 mOsmol in the numerator of the second fraction for getting the 'mOsmol' part cancelled out.
Using dimensional analysis:

(100 \quad mg)/(2 \quad mOsmol) × (0.32 \quad mOsmol)/(1 \quad mL) × (1 \quad g)/(1000 \quad mg)

 × (100 \quad mL)/(100 \quad mL)= 1.6 \quad (g)/(100 \quad mL) => 1.6 %  Ans.

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