A stock solution is prepared by dissolving 12.5 g of NaCI in enough water to prepare 150.0 mL of solution. What volume of this stock solution will be used to prepare a diluted solution that is 250.0 mL of a 0.500 M solution of NaCI?

8.75 mL

Explanation:

First, we calculate the molar mass of NaCl = molar mass of Na + molar mass of Cl. Molar mass of Na = 23 g/mol, molar mass of Cl = 35.5 g/mol.

So molar mass NaCl = (23 + 35.5) g/mol = 58.5 g/mol. The number of moles, n of NaCl in 12.5g is n = mass of NaCl / molar mass NaCl = 12.5 g / 58.5 g/mol = 0.214 mol.

The molarity, M of 150 mL M = number of moles / volume = 0.214 mol / 150 mL = 1.427 M.

We now calculate the number of moles of NaCl in 250 mL of 0.500 M.

Number of moles, n = molarity * volume. molarity = 0.500 M, volume = 250 mL. So n = 0.500 * 250 = 0.125 moles. Since we have 0.125 moles in the dilute 250 mL solution, the volume of the 150 mL 1.43 M solution required is number of moles in 250 mL solution/molarity of 150 mL solution = 0.125 mol / 1.427 M = 0.0875 L = 8.75 mL