Question

1. the determination of glucose concentration in blood serum is often based on the formation of a blue - green complex of glucose and o - toluidine in glacial acetic acid. this reaction is shown at the bottom of this page.

the analytical wavelength for the complex is 635 nm. a set of standard solutions was prepared by taking a known volume of a stock glucose solution, containing 1.000 g glucose dissolved in 100 ml of distilled water, and diluting to 100 ml with distilled water. these solutions and the blood samples were treated with the o - toluidine reagent, and the percent transmittance of each sample was read against a reference solution. the data obtained are

stock glucose, ml	concentration of glucose complex, m	% t	a
10.00		59.16	0.228
20.00		35.08	0.455
25.00		27.10	0.567
30.00		20.65	0.685
unk. solution		46.17	0.350

Explanation:

Step1: Calculate molar mass of glucose

The molar mass of glucose ($C_6H_{12}O_6$) is $M = 6\times12.01+12\times1.01 + 6\times16.00= 180.18\ g/mol$.

Step2: Calculate concentration of stock solution

The stock solution has $1.000\ g$ glucose in $100\ mL$ water. The molarity of stock solution $C_{stock}=\frac{n}{V}$, where $n=\frac{m}{M}=\frac{1.000\ g}{180.18\ g/mol}\approx0.00555\ mol$ and $V = 0.1\ L$. So $C_{stock}=\frac{0.00555\ mol}{0.1\ L}=0.0555\ M$.

Step3: Use dilution formula $C_1V_1 = C_2V_2$

For each standard - solution, $C_1$ is the concentration of stock solution, $V_1$ is the volume of stock solution taken, and $C_2$ is the concentration of the diluted solution with $V_2 = 100\ mL=0.1\ L$.
For $V_1 = 5.00\ mL = 0.005\ L$, $C_2=\frac{C_1V_1}{V_2}=\frac{0.0555\ M\times0.005\ L}{0.1\ L}=2.775\times 10^{-3}\ M$.
For $V_1 = 10.00\ mL = 0.01\ L$, $C_2=\frac{0.0555\ M\times0.01\ L}{0.1\ L}=5.55\times 10^{-3}\ M$.
For $V_1 = 20.00\ mL = 0.02\ L$, $C_2=\frac{0.0555\ M\times0.02\ L}{0.1\ L}=1.11\times 10^{-2}\ M$.
For $V_1 = 25.00\ mL = 0.025\ L$, $C_2=\frac{0.0555\ M\times0.025\ L}{0.1\ L}=1.3875\times 10^{-2}\ M$.
For $V_1 = 30.00\ mL = 0.03\ L$, $C_2=\frac{0.0555\ M\times0.03\ L}{0.1\ L}=1.665\times 10^{-2}\ M$.
For the unknown solution, we can use Beer - Lambert law $A=\epsilon bc$. First, we need to establish a calibration curve from the known standard solutions (concentration vs absorbance $A$). Since $A = 2-\log(\%T)$, we have absorbance values for standards as given. Then, we can find the concentration of the unknown solution from the calibration curve. But if we assume a linear relationship $A = kC$ (where $k$ is the slope of the calibration curve), we first find the slope from the standard - solution data.
Let's assume we have a linear regression of the data points $(C_i,A_i)$ for $i = 1,\cdots,5$ (standard solutions). After calculating the slope $k$ (using least - squares method or by hand - calculation of slope between two points if linearity is assumed), and knowing $A$ of the unknown solution ($A = 0.350$), we can find $C=\frac{A}{k}$.
However, if we assume a simple linear relationship between $A$ and $C$ and use two - point method. Let's take two points $(C_1 = 2.775\times 10^{-3}\ M,A_1 = 0.114)$ and $(C_2=1.11\times 10^{-2}\ M,A_2 = 0.455)$. The slope $k=\frac{A_2 - A_1}{C_2 - C_1}=\frac{0.455 - 0.114}{1.11\times 10^{-2}-2.775\times 10^{-3}}=\frac{0.341}{8.325\times 10^{-3}}\approx41$.
Then for $A = 0.350$ of the unknown solution, $C=\frac{0.350}{41}\approx8.54\times 10^{-3}\ M$.

The concentrations for the blanks in the table are:
For $5.00\ mL$ stock glucose: $2.775\times 10^{-3}\ M$
For $10.00\ mL$ stock glucose: $5.55\times 10^{-3}\ M$
For $20.00\ mL$ stock glucose: $1.11\times 10^{-2}\ M$
For $25.00\ mL$ stock glucose: $1.3875\times 10^{-2}\ M$
For $30.00\ mL$ stock glucose: $1.665\times 10^{-2}\ M$
For unknown solution: $8.54\times 10^{-3}\ M$

Answer:

stock glucose, mL	concentration of glucose complex, M	% T	A
10.00	$5.55\times 10^{-3}$	59.16	0.228
20.00	$1.11\times 10^{-2}$	35.08	0.455
25.00	$1.3875\times 10^{-2}$	27.10	0.507
30.00	$1.665\times 10^{-2}$	20.65	0.685
unk. solution	$8.54\times 10^{-3}$	46.17	0.350