Question 7.C-L.5: a. A solution is made by dissolving 0.900 g of salt in 100.0...
a. A solution is made by dissolving 0.900 g of salt in 100.0 mL of water. Assume that each milliliter of water weighs 1.00 g and that the final solution volume is 100.0 mL. Calculate the %(w/w) and %(w/v) for the solution using the assumptions as necessary.
b. An alcoholic beverage is labeled 90 proof, which means the alcohol concentration is 45% (v/v). How many milliliters of pure alcohol would be present in 1 oz (30 mL) of the beverage?
a. To calculate %(w/w), the mass of solute contained in a specific mass of solution is needed. The mass of solute is 0.900 g, and the mass of solution is equal to the solvent mass (100 g) plus the solute mass (0.900 g):
\%(w/w) = \frac{solute mass }{solution mass} × 100 = \frac{0.900 \cancel{g}}{100.9 \cancel{g}} × 100 = 0.892 \%(w/w)To calculate %(w/v), the number of grams of solute must be known along with the number of milliliters of solution. The mass of solute is 0.900 g, and the solution volume is 100 mL:
\%(w/v) = \frac{g of solute}{mL of solution} × 100 = \frac{ 0.900 g}{100 mL} × 100 = 0.900 \%(w/v)b. The given quantity is 30 mL of beverage, and the desired quantity is milliliters of pure alcohol. The %(v/v) provides the necessary factor:
30 \cancel{mL} \cancel{beverage} × \frac{45 mL alcohol}{100 mL beverage} = 14 mL alcohol