To calculate the concentration in mol/L (molarity) of the barium acetate solution, we need to follow these steps:
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Determine the molar mass of barium acetate (Ba(C₂H₃O₂)₂):
- Barium (Ba): 137.33 g/mol
- Carbon (C): 12.01 g/mol × 4 = 48.04 g/mol (since there are 4 carbon atoms)
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol (since there are 6 hydrogen atoms)
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol (since there are 4 oxygen atoms)
Now, add these together to find the molar mass: \[ \text{Molar mass} = 137.33 , \text{g/mol} + 48.04 , \text{g/mol} + 6.048 , \text{g/mol} + 64.00 , \text{g/mol} = 255.418 , \text{g/mol} \]
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Calculate the number of moles of barium acetate in 28.3 g: \[ \text{Moles of Ba(C₂H₃O₂)₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{28.3 , \text{g}}{255.418 , \text{g/mol}} \approx 0.111 , \text{mol} \]
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Calculate the concentration in mol/L: The volume of the solution is 250 mL, which is equal to 0.250 L. Therefore, the concentration (C) can be calculated as follows: \[ C = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.111 , \text{mol}}{0.250 , \text{L}} \approx 0.444 , \text{mol/L} \]
Rounding this result to three significant digits, the concentration of the barium acetate solution is: \[ \boxed{0.444} , \text{mol/L} \]
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