Question

a chemist must prepare 225. ml of 220. mm aqueous barium chlorate (ba(clo₃)₂) working solution. shell do this by pouring out some 0.282 mol/l aqueous barium chlorate stock solution into a graduated cylinder and diluting it with distilled water.
calculate the volume in ml of the barium chlorate stock solution that the chemist should pour out. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert the concentration of the working solution to mol/L

Since \(1\space mM = 10^{- 3}\space mol/L\), for a \(220\space mM\) solution, \(C_2=220\times10^{-3}\space mol/L = 0.220\space mol/L\)

Step2: Use the dilution formula \(C_1V_1 = C_2V_2\)

We know that \(C_1 = 0.282\space mol/L\), \(C_2=0.220\space mol/L\), and \(V_2 = 225\space mL\). Rearranging the formula for \(V_1\) gives \(V_1=\frac{C_2V_2}{C_1}\)
Substitute the values: \(V_1=\frac{0.220\space mol/L\times225\space mL}{0.282\space mol/L}\)

$$V_1=\frac{49.5}{0.282}\space mL$$

$$V_1 = 176\space mL$$

Answer:

\(176\space mL\)