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6.025 × 10 20 molecules of acetic acid are present in 500 ml of its solution. The concentration of solution is

Question

6.025 × 10 20 molecules of acetic acid are present in 500 ml of its solution. The concentration of solution is

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Solution

To find the concentration of the solution, we first need to find the number of moles of acetic acid.

  1. The molecular weight of acetic acid (CH3COOH) is approximately 60 g/mol.

  2. Avogadro's number tells us that there are approximately 6.022 x 10^23 molecules in one mole.

  3. So, 6.025 x 10^20 molecules of acetic acid is (6.025 x 10^20) / (6.022 x 10^23) = 0.001 moles.

  4. The concentration of a solution is defined as the amount of solute (in moles) divided by the volume of the solution (in liters).

  5. 500 ml is equal to 0.5 liters.

  6. Therefore, the concentration of the solution is (0.001 moles) / (0.5 liters) = 0.002 M (Molar).

So, the concentration of the acetic acid solution is 0.002 M.

This problem has been solved

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