Answer:
[tex]M_2=2.15M[/tex]
Explanation:
Hello!
In this case, since the dilution processes are characterized by the addition of extra solvent, in this case water, to an initial solution in order to decrease the concentration of the solute, in this case sugar; as water is added, the moles of sugar remain unchanged, so the following equation can be written:
[tex]n_1=n_2\\\\M_1V_1=M_2V_2[/tex]
Whereas 1 accounts for the initial 250.0 mL of 4.00 M sugar and 2 for the concentration at the new volume which is:
[tex]V_2=250.0mL+216.1mL=466.1mL[/tex]
Thus, the final concentration is:
[tex]M_2=\frac{M_1V_1}{V_2} =\frac{250.0mL*4.00M}{466.1mL} \\\\M_2=2.15M[/tex]
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