The solubility of LiCl is 83.5 g/100 g H2O at 20 oC , and is 89.8 g / 100 g H2O at 40 oC , then what is the required mass (in gram ) of LiCl to obtain a saturated solution in 300 g H2O at 25 oC ? please explain
thank a lot
2 answers
Please
Assuming you meant 25C and not 20C or 40C(for which data is available), you must draw a graph with these two points on it then read the solubility at 25 C.
Since the solubility varies only slightly between the two temperatures listed (at 20 and 40) you can come very close calculating the solubility at 25 this way.
83.5 g/100 g H2O at 20C
89.8 g/100 g H2O at 40C
Difference is 89.8-83.5 = 6.3g/100 for a 20C difference (that's 40C-20C). So we can estimate that the solubility for a 5 C difference is
[83.5 + (6.3 g x 5/20)] = about 83.5+1.6 = 85.1 g LiCl/100 g H2O.
Since you want a saturated solution in 300 g H2O that will obviously take 3*85.1 = ? g LiCl at 25C.
Since the solubility varies only slightly between the two temperatures listed (at 20 and 40) you can come very close calculating the solubility at 25 this way.
83.5 g/100 g H2O at 20C
89.8 g/100 g H2O at 40C
Difference is 89.8-83.5 = 6.3g/100 for a 20C difference (that's 40C-20C). So we can estimate that the solubility for a 5 C difference is
[83.5 + (6.3 g x 5/20)] = about 83.5+1.6 = 85.1 g LiCl/100 g H2O.
Since you want a saturated solution in 300 g H2O that will obviously take 3*85.1 = ? g LiCl at 25C.
3 answers