Question

a chemist makes 0.960 l of magnesium fluoride (mgf₂) working solution by adding distilled water to 0.290 l of a 0.000413 m stock solution of magnesium fluoride in water. calculate the concentration of the chemists working solution. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Identify the dilution - formula

The dilution formula is $M_1V_1 = M_2V_2$, where $M_1$ is the initial concentration, $V_1$ is the initial volume, $M_2$ is the final concentration, and $V_2$ is the final volume.

Step2: Assign the given values

We are given that $M_1=0.000413\ M$, $V_1 = 0.290\ L$, and $V_2=0.960\ L$.

Step3: Solve for $M_2$

Rearranging the dilution formula for $M_2$ gives $M_2=\frac{M_1V_1}{V_2}$. Substitute the values: $M_2=\frac{0.000413\ M\times0.290\ L}{0.960\ L}$.
First, calculate the numerator: $0.000413\times0.290 = 0.000413\times(0.3 - 0.01)=0.000413\times0.3-0.000413\times0.01=0.0001239 - 0.00000413 = 0.00011977$.
Then, divide by the denominator: $M_2=\frac{0.00011977}{0.960}\ M\approx0.000125\ M$.

Answer:

$0.000125$