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Suppose 25.0 g of solid NaOH is added to 1.50 L of an aqueous solution that is already 2.40 M in NaOH. Then water is added until the nal volume is 4.00 L. Determine the concentration of the NaOH in the resulting solution.

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Answer:

1.06 M

Explanation:

Lets answer this question using the definition for molarity of a solution, M.

M = mol of solute/ Volume of solution in L

We are adding more moles of NaOH to a solution which is already 2.40, so the final number of moles are

25 g x 1 mol /  40 g/mol = 0.625 mol

mol NaOH in 1.5 L of 2.40 M NaOH:

1.5 L x 2.40 mol/L = 3.6 mol

Total # mol = 0.625 mol + 3.6 mol = 4.23 mol

The final volume is 4.00 L so the new molarity is:

M = 4.23 mol / 4 L = 1.06 M

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