Question 9.9: How many milliliters of a 0.75% (w/v) solution of the food p...
How many milliliters of a 0.75% (w/v) solution of the food preservative sodium benzoate are needed to obtain 45 mg?
ANALYSIS We are given a concentration and a mass, and we need to find the volume of solution by rearranging the equation for (w/v)% concentration. Remember that 45 mg = 0.045 g.
BALLPARK ESTIMATE A 0.75% (w/v) solution contains 0.75 g (750 mg) for every 100 mL of solution, so 10 mL contains 75 mg. To obtain 45 mg, we need a little more than half this volume, or a little more than 5 mL.
Since (w/v)\% = \frac{Mass of solute in g}{Volume of solution in mL} \times 100\%
then Volume of solution in mL = \frac{(Mass of solute in g) (100\%)}{(w/v)\%} \\ =\frac{(0.045 g)(100\%)}{0.75\%}= 6.0 mL
BALLPARK CHECK: The calculated answer is consistent with our estimate of a little more than 5 mL.