Question

1. calculate the molar concentration of glucose in the stock solution.

c6h12o6
m=(6×12.01 + 12×1.01+6×16.00)g/mol = 180.18g/mol
180.18v mol/l 1g/180.18v mol/l

1. for each standard solution and the unknown, convert the percent transmittance to absorbance, using equation 4. enter these absorbances in the table.

Explanation:

Step1: Recall molar - concentration formula

Molar concentration ($c$) is $c=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume of the solution in liters. First, find the molar mass of glucose ($C_6H_{12}O_6$). The molar mass $M$ of glucose: $M=(6\times12.01 + 12\times1.01+6\times16.00)\ g/mol=180.18\ g/mol$. If we have a mass $m$ of glucose, the number of moles $n = \frac{m}{M}$. So the molar - concentration $c=\frac{m}{M\times V}$. But since the mass of glucose in the stock solution is not given, assume we have $m = 1\ g$ of glucose in a volume $V$ (in liters) of the stock solution, then $c=\frac{1\ g}{180.18\ g/mol\times V\ L}=\frac{1}{180.18V}\ mol/L$.

Step2: Recall the formula for converting percent - transmittance ($\%T$) to absorbance ($A$)

The formula is $A = 2-\log_{10}(\%T)$. For each standard solution and the unknown, substitute the value of $\%T$ into this formula to calculate the absorbance.

Answer:

1. Molar concentration of glucose in stock solution (assuming 1 g of glucose): $\frac{1}{180.18V}\ mol/L$
2. Absorbance for each solution: Use $A = 2-\log_{10}(\%T)$ to calculate.