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In the laboratory you dissolve 14.2g of barium iodide in a volumetric flask and add water to a total volume of 500 mL. What is the-molarity of the solution? square m What is the concentration of the barium cation? square M. What is the concentration of the iodide anion? square M.

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In the laboratory you dissolve 14.2g of barium iodide in a volumetric flask and add water to a total volume of 500 mL.
What is the-molarity of the solution? square  m
What is the concentration of the barium cation? square  M.
What is the concentration of the iodide anion? square  M.

In the laboratory you dissolve 14.2g of barium iodide in a volumetric flask and add water to a total volume of 500 mL. What is the-molarity of the solution? square m What is the concentration of the barium cation? square M. What is the concentration of the iodide anion? square M.

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To find the molarity of the solution, we need to determine the number of moles of barium iodide (BaI2) dissolved in the solution.

Given information:
- Mass of barium iodide (BaI2) dissolved: 14.2 g
- Total volume of the solution: 500 mL (0.5 L)

Step 1: Calculate the molar mass of barium iodide (BaI2).
Molar mass of BaI2 = Molar mass of Ba + 2 × Molar mass of I
Molar mass of BaI2 = 137.33 g/mol + 2 × 126.90 g/mol = 390.13 g/mol

Step 2: Calculate the number of moles of barium iodide (BaI2) dissolved.
Number of moles of BaI2 = Mass of BaI2 / Molar mass of BaI2
Number of moles of BaI2 = 14.2 g / 390.13 g/mol = 0.0364 mol

Step: Calculate the molarity of the solution.
Molarity (M) = Number of moles of solute / Volume of solution in liters
Molarity (M) = 0.0364 mol / 0.5 L = 0.0728 M

Therefore, the molarity of the solution is 0.0728 M.

To find the concentration of the barium cation (Ba the solution, we the molarity of the solution directly, as each mole of barium iodide (BaI2) provides one mole of barium cation.

The concentration of the barium cation (Ba2+) is 0.0728 M.

To find the concentration of the iodide anion (I-) in the solution, we need to consider that each mole of barium iodide (BaI2) provides two moles of iodide anion.

The concentration of the iodide anion (I-) is 2 × 0.0728 M = 0.1456 M.

Therefore, the concentration of the iodide anion (I-) in the solution is 0.1456 M.
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