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.

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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.