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.
The solubility of aluminum sulfate in water at 20degrees Celcius is 26.7g/100L.
Calculate its molarity?
4 answers
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
molarity= mol/mL
26.7/342.15=.078=.78M