Question

If your lysed suspension contains 29.1 mL of liquid, how much in mL of 5 M NaClO4 do you need to add to obtain a final concentration of 1 M.

Answer 1

Ans. Given, Initial or stock [NaClO₄] = 5 M

            Final [NaClO₄] = 1 M

            Vol. of lysed suspension = 29.1 mL

Let the required vol. of stock NaClO₄ soln. be X mL

So, total volume of final/mixed soln. = 29.1 mL + X mL

Now, using    C1V1 (Stock soln.) = C2V2 (Final-mixed soln.)

            Or, 5M x V1 = 1M x (29.1 + X) mL

Note that V1 = X mL

            Or, 5M x X mL = 1M x (29.1 + X) mL

            Or, 5X = 29.1 + X

            Or, 5X – X = 4X = 29.1

            Or, X = 29.1 / 4 = 7.275

Hence, required volume of stock NaClO₄ soln. = X mL = 7.275 mL