chemistry
- Labbie
- Posts: 3243
- Joined: 28 Nov 2006, 10:00
- Job Title: Retired
- Suburb: At Home
- State/Location: NSW
Re: chemistry
Taken from the Lab book page 4.
Normal solution (N)
A Normal solution contains one gram-equivalent weight of solute per litre. An equivalent weight is the amount of substance that is able to combine with or displace 1 atom of hydrogen. A gram-equivalent is the amount of substance in grams that is able to combine with or displace 1 gram-atom of hydrogen (1.008g.) Solutions of the same normality will react completely with each other if equal volumes are used. The gram-equivalent of a substance depends on the reaction in which it is involved. For this reason it is now common practice to express concentrations in molarity. Older books still use normal solutions.
a 2N-solution contains 2 gram-equivalents per litre of solution.
a 5N 5
0.1N 0.1g
Hope this helps
Normal solution (N)
A Normal solution contains one gram-equivalent weight of solute per litre. An equivalent weight is the amount of substance that is able to combine with or displace 1 atom of hydrogen. A gram-equivalent is the amount of substance in grams that is able to combine with or displace 1 gram-atom of hydrogen (1.008g.) Solutions of the same normality will react completely with each other if equal volumes are used. The gram-equivalent of a substance depends on the reaction in which it is involved. For this reason it is now common practice to express concentrations in molarity. Older books still use normal solutions.
a 2N-solution contains 2 gram-equivalents per litre of solution.
a 5N 5
0.1N 0.1g
Hope this helps
Regards Labbie
Lab Manager/Lab Tech, mind reading etc etc
Now retired
Lab Manager/Lab Tech, mind reading etc etc
Now retired
Re: chemistry
Thanks for your reply Sue
Re: chemistry
I used to have use Normal [N] solutions when I worked in research.. what a PAIN! Molar solutions so easy to calculate
Re: chemistry
i noticed this is some of the labels in chemwatch - but i made the solutions to M, so didn't worry - but thanks for the explanation