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:

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