How much of MgCl2 required to make the solution of 1000ppm Mg from MgCl2 in 5 ml water? Is it molecular weight should be used in the calculation?
For 1000 ppm (1gm/l) in one litre you require 95 ( mol wt of MgCl2) / 24 (mol wt of Magnesium). It comes to 3.958 gm for one litre stock of 1000 ppm.
If you need 5 litre stock multiply it by 5, it comes to 19.7916 gm.
Considering MgCl2 is salt content, you can used the formula
FW of salt = 95.211.
At. wt. of Mg = 24.30
1g Mg in relation to FW of salt = 95.211/ 24.30 = 3.918g.
Hence, weigh out 3.918g MgCl2 and dissolve in 1 liter volume to make a 1000 ppm Mg standard.
In case 5ml, reduce amount to 200 times less, which require 0.019 g MgCl2 in 5 ml
Thank you everyone. I am not sure about the correct calculation. But, I have chose the calculation that you write up to me. Luckily, it was same and I can proceed with the next stock solution.
This example can also help you. we need to prepare 500 ml stock of water with 115 ppm.
(115 mg/lit) * (1 gram Mg /1000 mg Mg) * (95.211/24.30) * 0.5 lit
it means we have to add 0.225 g of MgCl2 in 500 ml of water
Hope you understand
Regards
Muqeet
- Badges
- Science topic