May 16, 2024, 10:50:11 PM
Forum Rules: Read This Before Posting


Topic: Molarity Question  (Read 11727 times)

0 Members and 1 Guest are viewing this topic.

webwheeler

  • Guest
Molarity Question
« on: November 19, 2005, 08:33:41 PM »
I am doing some research on crayfish diseases and came across this reference:

Quote
Rantamaki et al. (1992) found that, "Spore production of the fungus A. astaci was prevented in vitro by the addition of MgCl2 to a concentration of 20 mM or more into the lake water used as sporulation medium. When zoospores were used for infection at least 100 mM of MgCl2 was needed to protect the crayfish from infection. If crayfish were infected in lake water and then transferred to 25 mM or more of MgCl2, the fungus still grew within the cuticle but no sporangia were produced and the disease was not transmitted to uninfected A. astacus."

My question is how many grams of MgCl2 do I need to add to 1 liter of water to get a solution of 100 mM MgCl2?

I do not have a chemistry background, so I don't know whether to calculate using anhydrous MgCl2 or MgCl2.6H2O. I'm assuming anhydrous MgCl2 (does it even matter?), but please let me know if I'm wrong:

MgCl2 = 95.2104MW

1M MgCl2 = 95.2104g/l

100mM MgCl2 = 9.52104g/l

Not sure about these calculations as well???

Since I would like to discuss this on a crayfish forum, I would like to be sure the calculations are correct.

Thanks for your *delete me*

webwheeler

  • Guest
Re:Molarity Question
« Reply #1 on: November 20, 2005, 05:57:10 PM »
OK, I guess I better try to ask my questions a little better. ;D

I am an aquarist with an interest in keeping crayfish. It's been a long time since I've taken any chemistry courses; and, even then, I wasn't very good at it. :(

I don't know if there is any difference between a solution of anhydrous Magnesium Chloride and one of Magnesium Chloride hexahydrate, so

Question 1 - does it matter if I use MgCl2 or MgCl2.6H20 to achieve a 100mM concentration of MgCl2?

I make the assumption in my previous post, that mM means millimoles.

Question 2 - Does mM mean millimoles?

If so, the following quote seems to imply that it is a mistake to divide by moles by 1000 to get millimoles ???:

Quote
There is always the temptation to multiply or divide by 1000 in order to convert moles into millimoles or liters into milliliters. This might well be why algebra is seldom used in textbooks when discussing how to do titration calculations.

Here is the complete text for the above quote:

http://www.emporia.edu/chemist/davies/third.htm

And finally,

Question 3 - are my calculations in the previous post correct?

I would very much appreciate someone's *delete me*

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27682
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Molarity Question
« Reply #2 on: November 20, 2005, 05:58:18 PM »
100mM MgCl2 = 9.52104g/l

OK.

Note that there are several magnesium chlorides differing in number of water particles -  2, 4, 6, 8, 12 and anhydride. For every salt you will need differnt mass (for example 20.3 g of MgCl2.6H2O).

Try CASC for such calculations.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27682
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re:Molarity Question
« Reply #3 on: November 20, 2005, 06:04:19 PM »
Question 1 - does it matter if I use MgCl2 or MgCl2.6H20 to achieve a 100mM concentration of MgCl2?

No. You just have to take correct amount of the salt and add it not to 1000 mL, but to some smaller amount of water - say 700 mL, dissolve, and then fill up.

Quote
Question 2 - Does mM mean millimoles?

Yes.

Quote
Question 3 - are my calculations in the previous post correct?

Yes.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

webwheeler

  • Guest
Re:Molarity Question
« Reply #4 on: November 20, 2005, 06:08:11 PM »
Thanks a bunch, Borek! ;D ;D ;D

Sponsored Links