Question

Question #4056c

Answer 1

The value of n = 18.

The percentage of anhydrous aluminium sulfate must be 100 - 48.65 = 51.35 %

The Mr of aluminium sulfate is (2 x27) +[32 +(4 x16)] x 3 = 342

The Mr of water `(H_2O)` is (1+1) + 16 = 18

Ratio in moles is:

51.35/342 : 48.55/18

`rArr` 0.15 : 2.7

`rArr` 1 : 18

So the formula of hydrated aluminium sulfate , in this case, is

`Al_2(SO_4)_(3).18H_2O`