If the solubility of alum is 1g @ 1 degree celcius and 1.7 g at 6 degree celcius how much alum (KAl(SO4)2.12H20) would you expect to dissolve at 5.5% C?
Without any other information this maybe a simple linear relationship, especially over the small temperature range.
1 g at 1C
1.7 g at 6C
so solubility is given by
1 g + 0.7g/5 (T-1)
so if T=5.5C
solubility is
1 g + 0.7g/5 (5.5-1)
=1 g + 0.63 g = 1.63 g