Question

You make up of a solution of anhydrous NaCl (FW = 58.44 g/mol) at a concentration of 14.61 mg mL-1, then dilute this solution by adding 2 mL of this to 48 mL of H₂O. What is the final concentration of the solution (mM) and what amount of solute is present in the 5 mL of dilute solution? Which is the most correct answer?​

Answer 1

To find the final concentration of the solution after dilution, we need to use the concept of molarity (M).

First, let's calculate the amount of NaCl in the initial solution. The concentration of the solution is given as 14.61 mg mL-1, which means that for every 1 mL of solution, there are 14.61 mg of NaCl.

Now, we have 2 mL of the initial solution, so the amount of NaCl in these 2 mL can be calculated as follows:

Amount of NaCl in 2 mL = 2 mL * 14.61 mg/mL = 29.22 mg

Next, we add these 2 m