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 carbonate solution exerts tonicity of 0.3799 mOsmol/ml. If the compound has a molecular weight of 138.2 g/mol, then express the concentration of the solution as percent w/v.
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lb equals 1.75 % kg
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For potassium carbonate , the 'golden formula' is:
1 mmol = `[138.2]` mg = `[2]` mEq = `[3]` mOsmol.
The solution contains 0.3799 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 0.3799 mOsmol is contained in 1 mL of solution, we keep the 0.3799 mOsmol in the numerator of the second fraction for getting the 'mOsmol' part cancelled out.
Using dimensional analysis:
`(138.2 \quad mg)/(3 \quad mOsmol) × (0.3799 \quad mOsmol)/(1 \quad mL) × (1 \quad g)/(1000 \quad mg)`
` × (100 \quad mL)/(100 \quad mL)= 1.75 \quad (g)/(100 \quad mL) => 1.75 % ` Ans.