Logged out
Access All our Essays in an Instant
Swap your work for FREE access, or pay £4.99 for instant access
The determination of the Ar of Lithium.
Member rating: No Rating | Words: | Submitted: Thu Sep 04 2003
Page Preview
On the left is an image preview of every page of this document, and below are the first 150 words with formatting removed:
Chris Ellison The determination of the Ar of Lithium. The mass of Lithium that I used was 0.10g. Results In the first experiment I collected 186cm3 of hydrogen. Titration Attempt Result 1st 41.5cm3 2nd 35cm3 3rd 34.5cm3 Average 34.75cm3 The first result was anomalous so I did not include this in the average. Analysis The First Experiment 2Li(s) + 2H2O(g) ? 2LiOH(aq) + H2(g) In the first experiment there were 0.00775 moles of Hydrogen. Number of moles = Volume 24000 186cm3 =0.00775 moles 24000 There were 0.0155 moles of Lithium. Moles of Hydrogen x 2 = Moles of Lithium 0.00775 x 2 = 0.0155 moles of Lithium From this I can deduce that the Ar of Lithium in this experiment was 6.452 (Actual Ar of Lithium is 6.941) Mass = Ar Number of moles 0.10g = 6.452 0.0155 Titration LiOH(aq) + HCl(aq) ? LiCl(aq) +H2O(l) In the Titration there were 0.0139 moles of HCL (Hydrochloric Acid). Concentration x Litres = Number of moles 0.1 x 0.139 = 0.0139 moles There were 0.0139 moles of LiOH (Lithium Hydroxide). Concentration x Litres = Number of moles 0.1 x 0.139 = 0.0139...
Get instant access
- Instant, unlimited access to our documents in full
- Swap your work for free access, or pay £4.99
- To see the full version of this document and 149,410 others
