Question

table 1: standard solution preparation and absorbance data
calculate the concentration of each standard solution. the absorbance values have been provided for you.

	standard 1	standard 2	standard 3	standard 4	standard 5	unknown sample data
distilled h₂o (ml)	8.0	6.0	4.0	2.0	0
concentration (m)
absorbance	0.205	0.399	0.592	0.785	0.979	0.667

the absorbance of your unknown cuso₄ solution was measured to be 0.667.

Explanation:

Step1: Calculate total volume for each standard

For each standard, the total volume $V_{total}$ is the sum of the volume of $0.40\ M\ CuSO_4$ and the volume of distilled $H_2O$. For example, for Standard 1: $V_{total1}=2.0 + 8.0=10.0\ mL$.

Step2: Use dilution formula $C_1V_1 = C_2V_2$

The dilution formula is $C_1V_1 = C_2V_2$, where $C_1$ is the initial concentration ($0.40\ M$), $V_1$ is the volume of the initial solution, $V_2$ is the total volume of the diluted solution, and $C_2$ is the concentration of the diluted solution.
For Standard 1:
$C_1 = 0.40\ M$, $V_1=2.0\ mL$, $V_2 = 10.0\ mL$.
$C_2=\frac{C_1V_1}{V_2}=\frac{0.40\ M\times2.0\ mL}{10.0\ mL}=0.08\ M$.
For Standard 2:
$C_1 = 0.40\ M$, $V_1 = 4.0\ mL$, $V_2=4.0 + 6.0=10.0\ mL$.
$C_2=\frac{0.40\ M\times4.0\ mL}{10.0\ mL}=0.16\ M$.
For Standard 3:
$C_1 = 0.40\ M$, $V_1=6.0\ mL$, $V_2=6.0 + 4.0 = 10.0\ mL$.
$C_2=\frac{0.40\ M\times6.0\ mL}{10.0\ mL}=0.24\ M$.
For Standard 4:
$C_1 = 0.40\ M$, $V_1=8.0\ mL$, $V_2=8.0+2.0 = 10.0\ mL$.
$C_2=\frac{0.40\ M\times8.0\ mL}{10.0\ mL}=0.32\ M$.
For Standard 5:
$C_1 = 0.40\ M$, $V_1 = 10.0\ mL$, $V_2=10.0+0=10.0\ mL$.
$C_2=\frac{0.40\ M\times10.0\ mL}{10.0\ mL}=0.40\ M$.

Standard	Concentration (M)
Standard 2	0.16
Standard 3	0.24
Standard 4	0.32
Standard 5	0.40

Answer:

Standard	Concentration (M)
Standard 2	0.16
Standard 3	0.24
Standard 4	0.32
Standard 5	0.40