1 / 7

Comprehensive Analysis of Sodium Chloride and Glucose Concentrations in Various Aqueous Solutions

This study explores the concentrations and interactions of sodium chloride (NaCl) and glucose (Cglu) in various aqueous solutions. Multiple titration methods are employed to determine the molarity of NaCl and glucose across different volumes and conditions. Emphasis is placed on the effects of additional components, including water and other solutes. The findings aim to provide insights into the relationships between concentration, volume, and molarity, relevant for applications in biochemistry and medicinal chemistry.

reegan
Télécharger la présentation

Comprehensive Analysis of Sodium Chloride and Glucose Concentrations in Various Aqueous Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Vestiges titre TD 00

  2. H2O D V= 20 mL V= 20 mL V= 100 mL + m NaCl = 0,2 g A D C D = 1 / 15 B E CNa = ? CNa = ? titre Vi = ? + m glucose = 3 g V= ? mL V= ? mL + H2O D VD = 10 mL Cglu= 0,5 mol. L-1 Cglu = ? mol.L-1 Cglu = 0,4 mol. L-1

  3. V= 100 mL E B titre Cglu = 0,4 mol. L-1

  4. V= 100 mL + mNaCl = ? V= 50 mL F E B titre H2O j CNa = 0,154 mol.L-1 + Vj = 50 mL CNa = ? + Ve = 10 mL Cglu = ? Cglu = 0,4 mol. L-1

  5. V= 100 mL + mNaCl = ? V= 50 mL G F E B titre H2O j CNa = 0,154 mol.L-1 + Vj = 50 mL V = ? CNa = ? + Ve = 10 mL + Veau j = 10 mL + mglu = 1,62 g Cglu = ? CNa = 0,128 mol.L-1 Cglu = 0,4 mol. L-1 Cglu = 0,067 mol.L-1

  6. V= 50 mL F G + V H2OD = ? titre mNa = ? H2O j CNa = 0,154 mol.L-1 mglu = ? + Vj = 50 mL V = ? CNa = ? + Veau j = 10 mL + mglu = 1,62 g Cglu = ? CNa = 0,128 mol.L-1 Cglu = 0,067 mol.L-1

  7. CNa = ? G I Cglu = ? H VH = 20 mL + VC = 10 mL titre VG = 60 + 10 = 70 mL + Vi= ? V= 150 mL CNa = 0,130 mol.L-1 CNa = ? D = 1 / 10 avec H2O j Cglu = 0,186 mol.L-1 Cglu = ?

More Related