Question

if you take a 12.0 - ml portion of the stock solution and dilute it to a total volume of 0.700 l, what will be the concentration of the final solution? m = m

Explanation:

Step1: Convert volume units

First, convert the volume of the stock - solution portion from mL to L. Since 1 L = 1000 mL, then $V_1=12.0\ mL = 12.0\times10^{- 3}\ L$. Let the initial concentration be $M_1$ (not given in the problem, assume it is $M_1$) and the final volume $V_2 = 0.700\ L$, and the final concentration be $M_2$.

Step2: Use the dilution formula

The dilution formula is $M_1V_1 = M_2V_2$. We want to find $M_2$, so we can re - arrange the formula to $M_2=\frac{M_1V_1}{V_2}$. Substituting the values of $V_1 = 12.0\times10^{-3}\ L$ and $V_2 = 0.700\ L$ into the formula, we get $M_2=\frac{M_1\times12.0\times10^{-3}}{0.700}=\frac{12.0\times10^{-3}}{0.700}M_1$. If we assume the initial concentration $M_1 = 1\ M$ (since it is not given, and the ratio is what matters for the dilution calculation), then $M_2=\frac{12.0\times10^{-3}}{0.700}\ M\approx0.0171\ M$.

Answer:

If the initial concentration is assumed to be 1 M, the final concentration is approximately $0.0171\ M$