Type 2: Find the weight of the solute from mOsmol and volume

5. mEq and mOsmol 5.2) mOsmol 5.2.1) mOsmol - General mOsmol - General Normal 1

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How much NaCl will you need to dissolve in 1 mL in order to obtain the osmolarity of 615.38? The molecular weight of NaCl is 58.5 g/mol.

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lb equals 18 mg/1mL kg

    For NaCl , the 'golden formula' is:
    1 mmol= `[58.5]` mg = `[1]` mEq = `[2]` mOsmol.

    Osmolarity is the number of mOsmol per 1 L (1000 mL).
    Since we need mg quantity in the numerator of the desired unit, we begin with keeping it in the numerator of the first fraction.
    Again, since 615.38 mOsmol is contained in 1000 mL of solution, we keep the 615.38 mOsmol in the numerator of the second fraction for getting the 'mOsmol' part cancelled out.
    Using dimensional analysis:
    `(58.5 \quad mg)/(2 \quad mOsmol) × (615.38 \quad mOsmol)/(1000 \quad mL) × (1 \quad mL)/(1 \quad mL)= 18 \quad (mg)/(1 \quad mL)` Ans.

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