Question

1. how many milliliters of the approximately 0.012 $m$ copper(ii) sulfate solution are needed to prepare 100.00 milliliters of a 0.00024 $m$ copper(ii) sulfate solution?

_______ ml

Explanation:

Step1: State dilution formula

$C_1V_1 = C_2V_2$
Where $C_1=0.012\ M$, $C_2=0.00024\ M$, $V_2=100.00\ mL$, solve for $V_1$.

Step2: Rearrange for $V_1$

$V_1 = \frac{C_2V_2}{C_1}$

Step3: Substitute values

$V_1 = \frac{0.00024\ M \times 100.00\ mL}{0.012\ M}$

Step4: Calculate result

$V_1 = \frac{0.024}{0.012}\ mL = 2.00\ mL$

Answer:

2.00 mL