(type 5) mEq problems - randomized

5. mEq and mOsmol 5.1) mEq 5.1.1) mEq - General mEq from weight/volume Normal 1

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Express the concentration of 0.35 % (w/v) calcium gluconate solution as milliequivalent per liter. The molecular weight of calcium gluconate is 430 g/mol

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lb equals 16.28 mEq/L kg

    The molecular weight of calcium gluconate = 430.
    Valence or charge = 2.
    Therefore, we can write: 430 mg = 2 mEq.
    Again, 0.35 % (w/v) means, 0.35 gram of solute is present per 100 mL of solution. 0.35 g equals 350 mg. Since the question needs mEq/L as the final unit, we keep mEq in the first numerator. So, 430 mg will be the denominator.
    The strength of the solution (350 mg per 100 mL) will be written such a way that mg is at the numerator of the second fraction.
    Therefore, the dimensional analysis will be like this:

    `(2 \quad mEq)/(430 \quad mg) � (350 \quad mg)/(100 \quad mL) � (1000 \quad mL)/(1 \quad L)=16.28\quad (mEq)/L` Ans.