Question

A chemist must prepare 400 mL of 1.00M aqueous aluminum sulfate working solution. He'll do this by pouring out 1.82 mol/L aqueous aluminum sulfate stock solution into a graduated cylinder and diluting it with distilled water. How many mL of the aluminum sulfate stock solution should the chemist pour out?

Answer 1

Explanation:

The given data is as follows.

       $V_{1}$ = 400 ml,    $M_{1}$ = 1.00 M

       $V_{2}$ = ?,             $M_{2}$ = 1.82 M

Hence, the relation between molarity and volume is as follows.

             $M_{1}V_{1} = M_{2}V_{2}$

             $1.00 M \times 400 ml = 1.82 \times V_{2}$

                $V_{2}$ = 219.8 ml

Thus, we can conclude that 219.8 ml of the aluminum sulfate stock solution should the chemist pour out.

Answer 2

Answer:

219.78 mL of the stock solution are needed

Explanation:

First, we take a look at the desired Al2(SO4)3 working solution. We are told that the we need 400 mL of an aqueous aluminum sulfate solution 1.0 M. Let's see how many moles of the compound we have in the desired volume:

1000 mL Al2(SO4)3 solution ----- 1 mole of Al2(SO4)3

400 mL Al2(SO4)3 solution ----- x = 0.4 moles of Al2(SO4)3

To reach the desired concentration in the working solution we need 0.4 moles of Al2(SO4)3 in 400 mL, so we calculate the volume of the stock solution needed to prepare the working solution:

1.82 moles Al2(SO4)3/L = 1.82 M → This is the molar concentration of the stock solution.

1.82 moles of Al2(SO4)3 ----- 1000 mL

0.4 moles of Al2(SO4)3 ----- x = 219.78 mL

So, if we take 219.78 mL of the 1.82 M stock solution, we put it in a graduated cylinder and we dilute it to 400 mL, we would obtain a 1.0 M Al2(SO4)3 solution.