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Pure Acetic Acid,often Called Glacial Acetic Acid,is a Liquid with a Density

Question 20

Multiple Choice

Pure acetic acid,often called glacial acetic acid,is a liquid with a density of 1.049 g/mL.Which calculation correctly shows how to determine the volume of glacial acetic acid necessary to prepare 250 mL of 0.400 M CH₃CO₂H(aq) ?

A) $\left(\frac{0.400 \mathrm{~mol}}{\mathrm{~L}}\right) \left(\frac{60.05 \mathrm{~g}}{\mathrm{~mol}}\right) \left(\frac{1 \mathrm{~mL}}{1.049 \mathrm{~g}}\right) =$ $0.250 \mathrm{~L}$
B) $\left(\frac{1 \mathrm{~mol}}{0.400 \mathrm{~L}}\right) \left(\frac{60.05 \mathrm{~mol}}{\mathrm{~g}}\right) \left(\frac{1 \mathrm{~mL}}{1.049 \mathrm{~g}}\right) =$ $0.250 \mathrm{~L}$
C) $\left(\frac{0.400 \mathrm{~mol}}{\mathrm{~L}}\right) \left(\frac{1 \mathrm{~g}}{60.05 \mathrm{~mol}}\right) \left(\frac{1 \mathrm{~mL}}{1.049 \mathrm{~g}}\right) =$ $0.250 \mathrm{~L}$
D) $\left(\frac{0.400 \mathrm{~mol}}{\mathrm{~L}}\right) \left(\frac{1 \mathrm{~g}}{60.05 \mathrm{~mol}}\right) \left(\frac{1.049 \mathrm{~mL}}{1 \mathrm{~g}}\right) =$ $0.250 \mathrm{~L}$
E) $\left(\frac{1 \mathrm{~mol}}{0.400 \mathrm{~L}}\right) \left(\frac{60.05 \mathrm{~mol}}{\mathrm{~g}}\right) \left(\frac{1 \mathrm{~mL}}{1.049 \mathrm{~g}}\right) =$ $0.250 \mathrm{~L}$ .