Question

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?

Answer 1

`C_2=1.25 M`

Step-by-step 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:

`n_1=n_2\\\\C_1V_1=C_2V_2`

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

`C_2=(C_1V_1)/(V_2)=(3M*500\mu L)/(1200\mu L)\\ \\C_2=1.25 M`

Best regards.