The solubility of aluminum sulfate in water at 20degrees Celcius is 26.7g/100L.

Calculate its molarity?

Note the correct spelling of celsius.

mols Al2(SO4)3 = grams/molar mass = ?
Then M = mols /L

Is that 26.7 g/100 L. That's hard to believe but check that unit before working the problem.

26.7g/100mL

Then M = mols from above/0.1L = ?

Not sure but this is what I came up with:

molarity= mol/mL

26.7/342.15=.078=.78M

To calculate the molarity of a solution, we need to know the moles of solute and the volume of the solution.

Step 1: Convert the given solubility from grams to moles:
The solubility of aluminum sulfate is given as 26.7g/100L. We can assume that this value represents the mass of aluminum sulfate in 100 liters of water. The molar mass of aluminum sulfate is calculated by adding up the molar masses of its constituent elements:
Aluminum (Al) = 26.98 g/mol
Sulfur (S) = 32.06 g/mol
Oxygen (O) = 16.00 g/mol (There are four oxygen atoms in one molecule of aluminum sulfate, so multiply the atomic mass by 4)
Therefore, the molar mass of aluminum sulfate (Al2(SO4)3) is:
2(26.98 g/mol) + 3(32.06 g/mol) + 12(16.00 g/mol) = 342.15 g/mol

To convert grams to moles, divide the given mass by the molar mass:
26.7 g / 342.15 g/mol = 0.078 moles

Step 2: Calculate the volume of the solution in liters:
The given solubility is for 100 liters of water. Therefore, the volume of the solution is also 100 liters.

Step 3: Calculate the molarity:
Molarity (M) is defined as moles of solute per liter of solution. Divide the moles of the solute by the volume of the solution in liters:
M = 0.078 moles / 100 L = 0.00078 M

Therefore, the molarity of aluminum sulfate in water is 0.00078 M.