A stock solution of Is prepared and found to contain 13.5 mol L^-1 of (aq).If 25.0 mL of the stock solution is diluted with water to a final volume of 0.500 L,what is the concentration of the diluted solution? A)0.270 mol L^-1 B)1.48 mol L^-1 C)0.675 mol L^-1 D)675 mol L^-1 E)270 mol L^-1

Question

Quizplus · Accepted Answer

First, we need to find the number of moles of the stock solution present in the 25.0 mL sample: moles = concentration x volume moles = 13.5 mol/L x 0.0250 L moles = 0.3375 mol Next, we need to find the concentration of the diluted solution, knowing that the final volume is 0.500 L and the volume of the stock solution used is 25.0 mL (0.0250 L): concentration = moles / volume concentration = 0.3375 mol / 0.500 L concentration = 0.675 mol/L Therefore, the concentration of the diluted solution is 0.675 mol L^-1, which corresponds to choice C.

A Stock Solution of Is Prepared and Found to Contain