(type 5) mEq problems - randomized

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

Total tried:       Correct:       Wrong:

A solution contains 25 mg of sodium bicarbonate in each 5 mL (the MW of sodium bicarbonate is 84). How many milliequivalent of sodium bicarbonate will be present per liter of this solution?

Click on the button below to see the answer and explanations

Notice: Undefined variable: input_value1 in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 220
lb equals 59.52 mEq/L kg

    The molecular weight of sodium bicarbonate = 84.
    Valence or charge = 1.
    Therefore, we can write: 84 mg = 1 mEq.
    Since the question needs mEq/L as the final unit, we keep mEq in the numerator. So, 84 mg will be the denominator.
    Also, the solution contains 25 mg per 5 mL. To cancel out mg, we need to keep 25 mg in the numerator of second fraction. Therefore, the dimensional analysis will be like this:

    `(1 \quad mEq)/(84 \quad mg) � (25 \quad mg)/(5 \quad mL) � (1000 \quad mL)/(1 \quad L)=59.52\quad (mEq)/L` Ans.

Notice: Undefined index: TOTALTRY in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 675