What volume of 0.436 M barium nitrate solution is needed to prepare 240.0 mL of a solution that is 0.300 M in nitrate? A) 3.49 × 10² mL B) 8.26 × 10¹ mL C) 5.50 × 10¹ mL D) 1.65 × 10² mL E) 3.30 × 10² mL

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To solve this problem, we need to understand that barium nitrate, Ba(NO3)2, dissociates into one Ba2+ ion and two NO3- ions in solution. This means that the concentration of nitrate ions is twice the concentration of the barium nitrate solution. To prepare a 0.300 M nitrate solution, we need to adjust the volume of the original 0.436 M barium nitrate solution accordingly.First, calculate the moles of nitrate needed:$$0.300 \, \text{M} \times 0.240 \, \text{L} = 0.072 \, \text{moles NO}_3^-$$Since each mole of Ba(NO3)2 gives 2 moles of NO3-, the moles of Ba(NO3)2 needed are half of the moles of NO3-:$$0.072 \, \text{moles NO}_3^- / 2 = 0.036 \, \text{moles Ba(NO}_3)_2$$To find the volume of the 0.436 M solution needed to get these moles:$$0.036 \, \text{moles} / 0.436 \, \text{M} = 0.0826 \, \text{L} = 82.6 \, \text{mL}$$This corresponds to choice B when expressed in scientific notation as $8.26 \times 10^1 \, \text{mL}$.

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