1.00 mL of a 3.50 × 10^-4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. How many grams of oleic acid are in 5.00 mL of solution B? (molar mass for oleic acid = 282 g/mol) A) 9.87 × 10^-6 g B) 4.94 × 10^-5 g C) 4.94 × 10^-4 g D) 1.97 × 10^-6 g E) 7.00 × 10^-6 g

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Quizplus · Accepted Answer

The initial concentration of oleic acid is $3.50 \times 10^{-4} \, \text{M}$. After the first dilution, the total volume is $1.00 \, \text{mL} + 9.00 \, \text{mL} = 10.00 \, \text{mL}$, so the concentration becomes $3.50 \times 10^{-4} \, \text{M} \times \frac{1.00}{10.00} = 3.50 \times 10^{-5} \, \text{M}$. After the second dilution, the total volume is $2.00 \, \text{mL} + 8.00 \, \text{mL} = 10.00 \, \text{mL}$ again, so the concentration becomes $3.50 \times 10^{-5} \, \text{M} \times \frac{2.00}{10.00} = 7.00 \times 10^{-6} \, \text{M}$. In $5.00 \, \text{mL}$ of solution B, the amount of oleic acid is $7.00 \times 10^{-6} \, \text{M} \times 5.00 \times 10^{-3} \, \text{L} = 3.50 \times 10^{-8} \, \text{mol}$. The mass of oleic acid is $3.50 \times 10^{-8} \, \text{mol} \times 282 \, \text{g/mol} = 9.87 \times 10^{-6} \, \text{g}$.

100 ML of a 3