This problem is providing information about the volume and concentration of HCl as 505 mL and 0.160 M respectively, and it is asking for the water that have to be added to prepare a 0.100-M solution.
In such a way, we work over the assumption of constant moles in dilution processes, so that we are able to write:
[tex]M_1V_1=M_2V_2[/tex]
Which relates de volume and concentration at the beginning and end of the experiment. This means we can solve for the resulting volume of the solution as a first calculation:
[tex]V_2=\frac{M_1V_1}{M_2}\\\\V_2=\frac{0.160M*505mL}{0.100M}=808mL[/tex]
Hence, the following amount of water must be added:
[tex]V_w=808mL-505mL=303mL[/tex]
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