User:Matt Hartings/Notebook/AU Biomaterials Design Lab/2016/08/29: Difference between revisions

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## m = 22.5 mg
## m = 22.5 mg
# 10 mL of 2.5 mM AuCl<sub>3</sub>
# 10 mL of 2.5 mM AuCl<sub>3</sub>
 
## MW = 303.33 g/mol
## 0.0025 M = (m/303.33 g/mol)/(0.010 L)
## m = 7.58 mg


<u> sample solutions </u>
<u> sample solutions </u>
# 0.25 mM AuCl<sub>3</sub>, 0 mM fructose, pH 4 ([H<sup>+</sup>] = 1x10<sup>-4</sup> M), 5 mL total
## Volume of AuCl<sub>3</sub> stock
### M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub>
### (0.0025 M)V<sub>1</sub> = (0.00025 M)(5 mL)
### V<sub>AuCl<sub>3</sub> stock</sub> = 0.5 mL


==Data==
==Data==

Revision as of 09:39, 29 August 2016

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Objective

Calculations for stock solutions to be made tomorrow.

Description

stock solutions

  1. 1 L of 1 M HCl
    1. Concentrated HCl should be: 11.65 M
    2. M1V1 = M2V2
    3. (11.65 M)(V1) = (1M)(1L)
    4. V1 = 85.8 mL
  2. 1 L of 1 M HNO3
    1. Concentrated HNO3 should be: 15.8 M
    2. M1V1 = M2V2
    3. (15.8 M)(V1) = (1M)(1L)
    4. V1 = 63.2 mL
  3. 1 L of 1 M NaOH
    1. MW of NaOH = 39.997 g/mol
    2. M = (m/MW)/V
    3. 1 M = (m/39.997 g/mol)/1L
    4. m = 39.997 g
  4. 10 mL of 12.5 mM Fructose
    1. MW = 180.16 g/mol
    2. 0.0125 M = (m/180.16 g/mol)/(0.010 L)
    3. m = 22.5 mg
  5. 10 mL of 2.5 mM AuCl3
    1. MW = 303.33 g/mol
    2. 0.0025 M = (m/303.33 g/mol)/(0.010 L)
    3. m = 7.58 mg

sample solutions

  1. 0.25 mM AuCl3, 0 mM fructose, pH 4 ([H+] = 1x10-4 M), 5 mL total
    1. Volume of AuCl3 stock
      1. M1V1 = M2V2
      2. (0.0025 M)V1 = (0.00025 M)(5 mL)
      3. VAuCl3 stock = 0.5 mL

Data

  • Add data and results here...

Notes

This area is for any observations or conclusions that you would like to note.


Use categories like tags. Change the "Course" category to the one corresponding to your course. The "Miscellaneous" tag can be used for particular experiments, as instructed by your professor. Please be sure to change or delete this tag as required so that the categories remain well organized.




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