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

(Difference between revisions)
 Revision as of 15:01, 6 November 2012 (view source) (→Data)← Previous diff Current revision (15:03, 6 November 2012) (view source) (→Data) (One intermediate revision not shown.) Line 29: Line 29: $1 kg water \times \frac {1000 g water} {1 kg water} \times \frac {1 mL water} {1 g water} = 1000 mL water$ $1 kg water \times \frac {1000 g water} {1 kg water} \times \frac {1 mL water} {1 g water} = 1000 mL water$ + + + $\frac {345.24g NaCl}{1 kg water} \times \frac {1 kg water} {1000 mL water} = \frac {34.524 g NaCl} {100 mL water}$ ==Notes== ==Notes==

## Current revision

Experimental Chemistry Main project page
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## Objective

• 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
• $molality = \frac {mol solute}{kg solvent}$

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

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

$\frac {345.24g NaCl}{1 kg water} \times \frac {1 kg water} {1000 mL water} = \frac {34.524 g NaCl} {100 mL water}$

## Notes

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

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