Question

a chemist must prepare 0.250 l of 1.00 m aqueous nickel(ii) chloride ($nicl_2$) working solution. hell do this by pouring out some 1.82 $\frac{mol}{l}$ aqueous nickel(ii) chloride stock solution into a graduated cylinder and diluting it with distilled water. calculate the volume in l of the nickel(ii) chloride stock solution that the chemist should pour out. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Use dilution formula

The dilution formula is $M_1V_1 = M_2V_2$, where $M_1$ is the molarity of the stock - solution, $V_1$ is the volume of the stock - solution, $M_2$ is the molarity of the working solution, and $V_2$ is the volume of the working solution.
We are given that $M_1=1.82\ mol/L$, $M_2 = 1.00\ mol/L$, and $V_2=0.250\ L$.

Step2: Solve for $V_1$

Rearranging the dilution formula for $V_1$ gives $V_1=\frac{M_2V_2}{M_1}$.
Substitute the given values: $V_1=\frac{1.00\ mol/L\times0.250\ L}{1.82\ mol/L}$.
$V_1=\frac{0.250}{1.82}\ L\approx0.137\ L$.

Answer:

$0.137$