Answer:
47.68 mL
Explanation:
In this case, we have a dilution problem. So, we have to start with the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
We have to remember that in a dilution procedure we go from a higher concentration to a lower one. With this in mind, We have to identify the concentration values:
[tex]C_1~=~6.00~M[/tex]
[tex]C_2~=~1.53~M[/tex]
The higher concentration is C1 and the lower concentration is C2. Now, we can identify the volume values:
[tex]V_1~=~X[/tex]
[tex]V_2~=~187~mL[/tex]
The V2 value has "mL" units, so V1 would have "mL" units also. Now, we can include all the values into the equation and solve for "V1", so:
[tex]6.00~M*V_1=1.53~M*187~mL[/tex]
[tex]V_1=\frac{1.53~M*187~mL}{6.00~M}=47.68~mL[/tex]
So, we have to take 47.68 mL of the 6 M and add 139.31 mL of water (187-47.68) to obtain a solution with a final concentration of 1.53 M.
I hope it helps!
