You have a saturated solution of Mg ( C16H31O2)2 at 50 C
How many milligrams of Magnesiuum palmitate Mg ( C16H31O2)2 will precipitate from 965 ml of this solution when it is cooled at 25 C .
Ksp at 50 C = 4.8 x 10^ -12
Ksp at 25 C = 3.3 x 10^ -12
answer options (mg) 0.1 , 1000 , 100 , 10
what im doing is
Ksp = [Mg+] [2 P]^2 P stands for ( C16H31O2 - )
finding the concentration of ions in mols per litre at 50 c and at 25 C
and then subtracting . After that converting concentration in milligrams.
molar mass = 534
im getting
0.05674 gram / litre - 0.0500 gram / litr
0.00674 gram / litre
for 0.965 litre solution we have 0.00650 grams or 6.5 mg
but not getting the correct answer