dilution
- If I have 340 mL of a 0.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?
\((340 mL)(0.5 M) = (900 mL)(x M)\)
\(x = \frac{340 * 0.5}{900}\)
\(x \approx 0.189 M\)
- If I dilute 250 mL of 0.10 M lithium acetate solution to a volume of 750 mL, what will the concentration of this solution be?
\((250 mL)(0.10 M) = (750 mL)(x M)\)
\(x = \frac{250*0.10}{750}\)
\(x \approx 0.03 M\)
- If I leave 750 mL of 0.50 M sodium chloride solution uncovered on a windowsill and 150 mL of the solvent evaporates, what will the new concentration of the sodium chloride solution be?
\((750 mL)(0.50 M) = (600 mL)(x M)\)
\(x = \frac{750*0.50}{600}\)
\(x = 0.625 M\)
- To what volume would I need to add water to the evaporated solution in problem 3 to get a solution with a concentration of 0.25 M?
\((600 mL)(0.625 M) = (x mL)(0.25 M)\)
\(x = \frac{600*0.625}{0.25}\)
\(x = 1500 mL\)
add 900mL
- Calculate the volume of a 17.5M stock solution of NaOH needed to prepare 200. mL of a 2.5M solution.
\((200 mL)(2.5M) = (xmL)(17.5)\)
\(x = \frac{200*2.5}{17.5}\)
\(x \approx 28.57 mL\)
add 171.43 ml water to soln to make 200 mL 2.5 M soln
Last update:
June 5, 2023
Created: June 5, 2023
Created: June 5, 2023