Question

Based on this data, I find it hard to the calculations and analysis and discussion
Beer's Law Calculations: A. Dilution Calculations 1. Use the actual mass of cobalt(II) nitrate hexahydrate you used to calculate the concentration of your stock solution. 2. Calculate the concentration of your diluted solutions. The easiest way to do this is with the equation MiV2-M2V2. Show the work for the calculation of the concentration of at least one of your dilute solutions. 3. Make a table showing the concentration and absorbance of all five of your known solutions. B. Standard Calibration Curve Using Excel, enter your concentrations in Column A and your absorbances in Column B Enter a label for each data column in the first box. Include units with your label. Generate an Excel graph of the absorbance (y-axis) as a function of concentration (x-axis) for the standard solutions using the Scatter plot option Add a linear trendline fit to the data and have Excel display both the equation and R2 value on the graph. Label each of the axes with units (where appropriate) and give the graph a title. Record the equation in the calculations section of your lab notebook. The can either print your graph or copy it into your lab notebook. Beer's Law An example Beer's Law plot is shown to the right: Standard C. Analysis of Unknown Concentrations Contatto Use the absorbance of the unknown solution and the equation for the linear trendline you found in part B to determine the concentrations of your unknown solution. Show your work in your lab notebook! Clearly report the concentration of your unknown solution Concentration molti

Answer 1

1) Molecular weight of cobalt nitrate hexahydrate is 291.03 g/mol

For preparation of stock solution 2.183 gm of Co(NO3)2. 6H2O is used.

To calculate concnetration of stock solution,

1 Molar Co(NO3)2. 6H2O = 291.03 /1000 mL

x Molar Co(NO3)2. 6H2O = 2.183 / 50mL

x * 291.03/1000 = 1 * 2.183/50

x = 0.15 Molar of stock solution

2) Concentration of diluted solutions

i) 2 mL of stock solution + 8 mL of H2O

M1V1 = M2V2

0.15*2 = M2*10

M2 = 0.03 M

ii) M2 = 0.15*4/10

M2 = 0.06

iii) M2 = 0.09 M

iv) M2 = 0.12 M

v) M2 = 0.15 M

Concentration (M)	Absorbance
0.03	0.169
0.06	0.359
0.09	0.567
0.12	0.759
0.15	0.765

y = 5.3067x + 0.0462 R2 = 0.9443 Absorbance 0 0.02 0.04 0.12 0.14 0.16 0.06 0.08 0.1 Concentration (Molar)

From the above graph we got equation y = 5.3067x+0.0462

For absorbance 0.356 we can write

x = (0.356-0.0462)/ 5.3067

x = 0.058 M

1) To calculate ϵ,

A = ϵbC

ϵ = 0.356/(1*0.058)

ϵ = 6.13 Lmol-1cm-1

2) To calculate volume

M1V1 = M2V2

0.50* V1 = 0.15*10

V1 = 3mL