A sample of 6.65 grams of NaOH is dissolved

into 621 mL of aqueous 0.250 M NaOH (as-
sume no volume change). This solution is
then poured into 2.16 gallons of water. (You
may assume that the two volumes can be
added.) What is the concentration of NaOH
in the final solution?
Answer in units of M

2.16 gallons = ? L. Make that conversion; 1 gallon about 3.7854 L.

mols solid NaOH = grams/molar mass = ?
mols liquid NaOH = M x L = ?
M final soln = total mols/total L.

To find the concentration of NaOH in the final solution, we need to calculate the number of moles of NaOH in the initial solution and the final volume of the solution after mixing.

First, let's calculate the number of moles of NaOH in the initial solution.

We have the mass of NaOH and its molar mass. The molar mass of NaOH is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H), which is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol

Now, we can calculate the moles of NaOH using the formula:
moles = mass / molar mass

moles of NaOH = 6.65 g / 40.00 g/mol ≈ 0.16625 mol

Next, let's calculate the final volume of the solution in liters.

We know that 1 gallon is equal to 3.78541 liters.

Volume of water = 2.16 gallons * 3.78541 L/gal = 8.18 L

Since the initial solution is poured into the water, the final volume of the solution will be the sum of the initial volume (621 mL) and the volume of water (8.18 L).

Final volume of the solution = 621 mL + 8.18 L = 8.18 L + 0.621 L = 8.801 L

Now we can calculate the concentration of NaOH in the final solution using the formula:
concentration = moles / volume

concentration = 0.16625 mol / 8.801 L ≈ 0.0189 M

Therefore, the concentration of NaOH in the final solution is approximately 0.0189 M.