Question

If you dilute 675 mL of a 3.00 mol/L solution of MgCl2 to 2.00 L, what is the concentration of the new solution? When converting to liters, keep three decimal places.

Answer 1

To find the concentration of the diluted solution, you can use the formula for dilution, which is:

\[
C_1 V_1 = C_2 V_2
\]

where:
- \(C_1\) = initial concentration (3.00 mol/L),
- \(V_1\) = initial volume (675 mL),
- \(C_2\) = final concentration (what we want to find),
- \(V_2\) = final volume (2.00 L).

First, convert \(V_1\) from milliliters to liters:

\[
V_1 = 675 \text{ mL} = \frac{675}{1000} \text{ L} = 0.675 \text{ L}
\]

Now we can plug the values into the dilution equation:

\[
(3.00 \text{ mol/L})(0.675 \text{ L}) = C_2 (2.00 \text{ L})
\]

Calculating the left side:

\[
(3.00)(0.675) = 2.025 \text{ mol}
\]

So we have:

\[
2.025 \text{ mol} = C_2 (2.00 \text{ L})
\]

Now, solve for \(C_2\):

\[
C_2 = \frac{2.025 \text{ mol}}{2.00 \text{ L}} = 1.0125 \text{ mol/L}
\]

Now rounding to three decimal places, the concentration of the new solution is:

\[
\boxed{1.013} \text{ mol/L}
\]