2. You deposit the 500 ul from #1 into a solution with a final volume of 1200 uL. What is the final concentration of NaCl in molar? In molar?

Respuesta :

Answer:

[tex]C_2=1.25 M[/tex]

Explanation:

Hello,

In this case, since the concentration in #1 is 3M, during a dilution process, the moles of the solute (NaCl) remains the same, just the concentration and volume change as shown below:

[tex]n_1=n_2\\\\C_1V_1=C_2V_2[/tex]

In such a way, as the final volume is 1200 microliters, the resulting concentration turns out:

[tex]C_2=\frac{C_1V_1}{V_2}=\frac{3M*500\mu L}{1200\mu L}\\ \\C_2=1.25 M[/tex]

Best regards.