there are 1000 cubic centimeters which is 1000 grams of water (close enough) in a liter
which is 10 cm * 10 cm * 10 cm
now I do not know what chemical you are using that is 62 grams/mol
You better give me the whole problem carefully because unless you know M, the molarity, you can not go from grams/mol to grams/liter
molarity is mols/liter
and
62 grams/mol means 6*10^23 molecules has mass of 62 grams
so I have this chemistry problem which involves math and basically I need to convert 62 g/mol into mol/L but I don't know how.
like do you DIVIDE the g/mol by 1000 to convert g to L? and then how to you get the mol to become the numerator?
8 answers
oh, sorry! yes, M= 62g/mol in my case and I need to go from there to mol/L.
and the question is: The solubility of magnesium fluoride, MgF2(s), is 1.72x10^-3 g/100mL at 25 degrees. What is the Ksp value for Magnesium fluoride at 25 degrees?
and the question is: The solubility of magnesium fluoride, MgF2(s), is 1.72x10^-3 g/100mL at 25 degrees. What is the Ksp value for Magnesium fluoride at 25 degrees?
so
you have
1.72 * 10^-3 g /100 mL
which is
17.2 * 10^-3 g/liter ten times as much because 1000mL is 1 liter
how many mols is 17.2 * 10^-3 g
17.2*10^-3 g/L / 62 g/mol
= .277 * 10^-3 mol/Liter
2.77 *10^-3 mols/Liter or M
Now to get from there to Ksp
I suggest looking at
https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/copy-of-solubility-equilibria-mcat/v/solubility-product-constant-from-the-solubility
you have Mg ++
and 2 F-
you have
1.72 * 10^-3 g /100 mL
which is
17.2 * 10^-3 g/liter ten times as much because 1000mL is 1 liter
how many mols is 17.2 * 10^-3 g
17.2*10^-3 g/L / 62 g/mol
= .277 * 10^-3 mol/Liter
2.77 *10^-3 mols/Liter or M
Now to get from there to Ksp
I suggest looking at
https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/copy-of-solubility-equilibria-mcat/v/solubility-product-constant-from-the-solubility
you have Mg ++
and 2 F-
I am just plodding through this
http://www.khanacademy.org/science/chemistry/acid-base-equilibrium/copy-of-solubility-equilibria-mcat/v/solubility-product-constant-from-the-solubility
Solubility MgF2 = 1.72E-3 g/100 mL. The molar mass is closer to 62.3 than to 62.
mols = g/molar mass = 1.72E-3/62.3 = 2.76E-5 mols amd that is in 100 mL; therefore, M = 2.76E-5 mols/0.1L = 2.76E-4.
.......MgF2 ==> Mg^2+ + 2F^-
I......solid....0........0
C......solid....x.......2x
E......solid....x.......2x
Ksp = (Mg^2+)(F^-)^2
(Mg) = x molar from above.
(F^-) = 2x molar from above.
Ksp = (x)(2x)^2 = 4x^3
Substitute the solubility (from above) of 2.76E-4 M and calculate Ksp at 25 C.
mols = g/molar mass = 1.72E-3/62.3 = 2.76E-5 mols amd that is in 100 mL; therefore, M = 2.76E-5 mols/0.1L = 2.76E-4.
.......MgF2 ==> Mg^2+ + 2F^-
I......solid....0........0
C......solid....x.......2x
E......solid....x.......2x
Ksp = (Mg^2+)(F^-)^2
(Mg) = x molar from above.
(F^-) = 2x molar from above.
Ksp = (x)(2x)^2 = 4x^3
Substitute the solubility (from above) of 2.76E-4 M and calculate Ksp at 25 C.
Already on it. :-)
thank you dr.bob and damon! :)