User:Ashish Bhattarai/Notebook/EBC 571/2017/08/29: Difference between revisions

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The test tubes with the 10 ml solutions is put in oven at 80º C for 4 hours.  
The test tubes with the 10 ml solutions is put in oven at 80º C for 4 hours.  


2) Measurement practice  
Measurement practice  


a) Table of water densities vs Temperature  
1) Table of water densities vs Temperature  
[[Image:Measurement practice.jpg]]
[[Image:Measurement practice.jpg]]
2a) Preparation of 1.71mM NaCl from solid:
MM of NaCl: 58.44 g/mol
Volume: 0.1L
Amount of NaCl needed:
                1.71*10^-3 M *(0.1l)*(58.44g/mol) = 0.00999 g NaCl in 100 ml water
2b) Preparation of 10 mL of 13.5 µM dye from stock and wet flasks
Molarity of die* Volume of die = Volume of solution * Molarity of solution
161* 10^-3 * (x) = 0.01L * 13.5*10^-6M
(x) = 8.38*10^-4L
3)





Latest revision as of 02:22, 27 September 2017

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August 29, 2017

Preparation of HAuCl4 and BSA stock solutions

The target concentration for HAuCl4 is 2.5mM and for BSA solution is 250µM. The molar mass of AuCl4 is 337.79g/mol and BSA is 65000 g/mol. We prepared 50 ml for each stock solution. To reach the target concentration for 50 ml solution, we needed 0.424g Au. The amount of BSA we needed to make the 50 ml stock solution is 0.00132 grams.

From the stock solution we made Gold Nanoparticles solution that would yield both gold nanoparticles and gold protein fibers. In order to form these gold nanoparticles we worked with 2 different Au:BSA ratios (80:1 and 160:1) for 10 ml solutions. The constraint for this solution was the 3.6 pH criteria. pH of 3.6 is what we want our final solution to be. To fulfill the pH criteria, we need 1ml of HAuCl4. Through calculations the amount of stock BSA that was needed for the solutions was determined.

Volume of stock BSA required: 80:1 ---> 0.125 ml ----> 8.875 ml water added 160 :1 ---> 0.0625 ---> 8.9375 ml water added

The test tubes with the 10 ml solutions is put in oven at 80º C for 4 hours.

Measurement practice

1) Table of water densities vs Temperature

2a) Preparation of 1.71mM NaCl from solid: MM of NaCl: 58.44 g/mol Volume: 0.1L Amount of NaCl needed:

                1.71*10^-3 M *(0.1l)*(58.44g/mol) = 0.00999 g NaCl in 100 ml water 

2b) Preparation of 10 mL of 13.5 µM dye from stock and wet flasks

Molarity of die* Volume of die = Volume of solution * Molarity of solution 161* 10^-3 * (x) = 0.01L * 13.5*10^-6M (x) = 8.38*10^-4L

3)