Question

1. calculate the molar concentration of glucose complex present in each solution. enter the concentrations in the table.

$c_1v_1 = c_2v_2$
$v_1 = 5.00ml
ightarrow0.005l$
$\frac{0.0565m\times0.005l}{0.1l}=2.775\times10^{-3}m$
$\frac{0.0565m\times0.01}{0.1l}=5.65\times10^{-3}m$
$\frac{0.0565\times0.02}{0.1l}=1.13\times10^{-2}$
$\frac{0.0565\times0.025}{0.1l}=1.3875\times10^{-2}m$

1. prepare a beer’s law plot, using the data obtained in (2) and (3).
2. using your beer’s law plot

Explanation:

Step1: Identify the dilution - formula

The formula for dilution is $C_1V_1 = C_2V_2$, where $C_1$ is the initial concentration, $V_1$ is the initial volume, $C_2$ is the final concentration, and $V_2$ is the final volume. We need to solve for $C_2$. So, $C_2=\frac{C_1V_1}{V_2}$.

Step2: Substitute values for the first - case

Let's assume the first set of values: $C_1 = 0.0565M$, $V_1=0.005L$, and $V_2 = 0.1L$. Then $C_2=\frac{0.0565M\times0.005L}{0.1L}=2.825\times 10^{-3}M$ (there is a calculation error in the provided hand - written work where it says $2.775\times 10^{-3}M$).

Step3: Substitute values for the second - case

For another set of values (not clearly defined in the problem statement but following the pattern), if we assume different $V_1$ values and keep $V_2 = 0.1L$ and $C_1 = 0.0565M$. For example, if $V_1 = 0.01L$, then $C_2=\frac{0.0565M\times0.01L}{0.1L}=5.65\times 10^{-3}M$ (the hand - written work has $5.55\times 10^{-3}M$ which seems to be a calculation error).

Since the problem only asks to calculate molar concentrations and enter them in a table (not provided here), the general method for each calculation is as above.

Answer:

The molar concentrations are calculated using $C_2=\frac{C_1V_1}{V_2}$ for different values of $C_1$, $V_1$ and a fixed $V_2$. The correct values depend on the actual initial volumes and the fixed final volume and initial concentration values used in the experiment. For example, for $C_1 = 0.0565M$, $V_1 = 0.005L$, $V_2=0.1L$, $C_2 = 2.825\times 10^{-3}M$; for $C_1 = 0.0565M$, $V_1 = 0.01L$, $V_2 = 0.1L$, $C_2=5.65\times 10^{-3}M$.