# User:Allison K. Alix/Notebook/CHEM-581/2012/11/02

(Difference between revisions)
 Revision as of 14:57, 6 November 2012 (view source) (→Data)← Previous diff Revision as of 15:01, 6 November 2012 (view source) (→Data)Next diff → Line 24: Line 24: * $molality = \frac {mol solute}{kg solvent}$ * $molality = \frac {mol solute}{kg solvent}$ - $\frac {6 moles NaCl}{1 kg water} \times \frac {57.54 g NaCl}{moles NaCl}$ + + $\frac {6 moles NaCl}{1 kg water} \times \frac {57.54 g NaCl}{moles NaCl} = \frac {345.24g NaCl}{1 kg water}$ + + + $1 kg water \times \frac {1000 g water} {1 kg water} \times \frac {1 mL water} {1 g water} = 1000 mL water$ ==Notes== ==Notes==

## Revision as of 15:01, 6 November 2012

Experimental Chemistry Main project page
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## Objective

• Cross-link porphyrin/gluteraldehyde containing film
• Take IR spectra of all films synthesized thus far

## Description

• A 6 molal solution of NaCl was prepared (see calculations in "Data"). This solution was heated to 70°C at which point a film containing 1mL porphyrin and 1mL gluteraldehyde was added. The film was left to stir in solution until a visible color change occured, at which point we could assume the crosslinking was complete.
• IR spectral data was gathered for all films synthesized thus far. These spectra will later be analyzed to observe any apparent changes shifts in peaks due to the composition of the film

## Data

Preparation of 6m NaCl
• $molality = \frac {mol solute}{kg solvent}$

$\frac {6 moles NaCl}{1 kg water} \times \frac {57.54 g NaCl}{moles NaCl} = \frac {345.24g NaCl}{1 kg water}$

$1 kg water \times \frac {1000 g water} {1 kg water} \times \frac {1 mL water} {1 g water} = 1000 mL water$

## Notes

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

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