Question

If hydrochloric acid is obtained commercially at a concentration of 12.1M, how many milliliters of 12.1M HCl(aq) must be used to prepare 2.00x103mL of 0.500M HCL(aq)?

Answer 1

Answer:

$V_1=82.6mL$

Explanation:

Hello there!

In this case, according to this question, we will need to deal with this dilution problem, because it is asking for the volume of a 12.1-M stock solution of HCl. In such a way, we can use the following equation, under the assumption of no change in the number of moles in the solution:

$M_2V_2=M_1V_1$

Thus, we solve for the initial volume, V1, as shown below:

$V_1=\frac{M_2V_2}{M_1}$

And plug in the initial concentration and final concentration and volume to obtain:

$V_1=\frac{2000mL*0.500M}{12.1M}\\\\V_1=82.6mL$

Regards!