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Sagot :
Answer:
4.285 L of water must be added.
Explanation:
Hello there!
In this case, for this dilution-like problems, we need to figure out the final volume of the resulting solution so that we would be able to obtain the correct volume of diluent (water) to be added. In such a way, we can obtain the final volume, V2, as shown below:
[tex]M_1V_1=M_2V_2\\\\V_2=\frac{M_1V_1}{M_2}[/tex]
Thus, by plugging in the initial molarity, initial volume and final molarity (0.587 M) we obtain:
[tex]V_2=\frac{3.4M*0.895 L}{0.587M}\\\\V_2=5.18L[/tex]
It means we need to add:
[tex]V_{H_2O} ^{added}=5.18L-0.895L=4.285L[/tex]
Of diluent water.
Regards!
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