Question

The aluminium sulphate hydrate contains 8.10 percent Al by mass. Calculate x, that is, the number of water molecules associated with each unit

Answer 1

Al2(SO4)3 = 342 grams / mol
H2O = 18 grams / mol
so
342 +18 x grams / mol total
Al = 27 so Al2 = 54 grams /mol
so
.081 = 54 / (342+18x)
answer is x times Avagadro's number

Answer 2

Damon has given you the correct answer strictly from a math viewpoint. I must say it is much simpler as a math problem than a chemistry problem; however, let me give you the chemical view. Remember either method should give the same/correct answer. Solving the math equation Damon gave gives x = 18. Here is the chemical approach.
Take a 100 grams sample of Al(SO4)3.xH2O. It will contain 8.1 g Al. How many mols is that? mols Al = g/atomic mass = 8.1/27 = 0.3.
Convert that to mols Al2(SO4)3.xH2O. That will be
0.3 mols Al x (1 mol Al2(SO4)3.xH2O/2 mols Al) = 0.15
Convert that to grams Al2(SO4)3. That is 0.15 x molar mass Al2(SO4)3 = 0.15 x 342 = 51.3 g. The sample weighed 100 g and this weighs 51.3 so 100-51.3 = 48.7 g H2O.
mols Al2(SO4)3 = 51.3/342 = 0.15 (but we already knew that from above)
mols H2O = 48.7/18 = 2.70. Now find the ratio. The easy way to do that is to divide both numbers by the smaller of the two. So,
mols Al(SO4)3 = 0.15/0.15 = 1.0
mols H2O = 2.70/18 = 18.0 and that makes the formula Al2(SO4)3.18H2O.
I must say the math way puts the chemistry way, at least in terms of simplicity and ease of working the problem, to shame.