Question

the concentration of pb²⁺ in a commercially available standard solution is 1.00 mg/ml. what volume of this solution should be diluted to 6.0×10² ml to produce a solution in which the concentration of pb²⁺ is 0.050 mg/l?

Explanation:

Step1: Convert units

First, convert the final - volume to liters and the initial concentration to mg/L. The final volume $V_2 = 6.0\times10^{2}\text{ mL}=0.60\text{ L}$. The initial concentration $C_1 = 1.00\text{ mg/mL}=1000\text{ mg/L}$, and the final concentration $C_2 = 0.050\text{ mg/L}$.

Step2: Use the dilution formula

The dilution formula is $C_1V_1 = C_2V_2$. We want to find $V_1$, so we can re - arrange the formula to $V_1=\frac{C_2V_2}{C_1}$.
Substitute the values: $C_2 = 0.050\text{ mg/L}$, $V_2 = 0.60\text{ L}$, and $C_1 = 1000\text{ mg/L}$ into the formula.
$V_1=\frac{0.050\text{ mg/L}\times0.60\text{ L}}{1000\text{ mg/L}}$
$V_1 = 3.0\times10^{-5}\text{ L}$

Step3: Convert the volume back to mL

Since $1\text{ L}=1000\text{ mL}$, then $V_1=3.0\times10^{-5}\text{ L}\times1000\text{ mL/L}=0.030\text{ mL}$

Answer:

$0.030$