When 0.025 moles of aluminum sulfate are
dissolved in enough water to make 115
milliliters of solution, what is the molarity
of sulfate ions?
Answer in units of M
To find the molarity of sulfate ions in the solution, we need to consider the dissociation of aluminum sulfate in water.
The chemical formula for aluminum sulfate is Al2(SO4)3.
From the formula, we can see that each mole of aluminum sulfate produces 3 moles of sulfate ions. Now we have 0.025 moles of aluminum sulfate.
Therefore, the moles of sulfate ions in the solution are 0.025 moles * 3 = 0.075 moles.
Next, we need to find the volume of the solution in liters. We are given that the volume is 115 milliliters, so we convert it to liters:
115 milliliters * (1 liter / 1000 milliliters) = 0.115 liters.
Finally, we can calculate the molarity of sulfate ions by dividing the moles of sulfate ions by the volume of the solution in liters:
Molarity = Moles / Volume
Molarity = 0.075 moles / 0.115 liters = 0.652 M.
Therefore, the molarity of sulfate ions in the solution is 0.652 M.
To find the molarity of sulfate ions, we need to use the given information about the moles of aluminum sulfate and the volume of the solution.
First, convert the given volume of the solution from milliliters (mL) to liters (L) since molarity is typically expressed in moles/liter.
115 milliliters = 115/1000 = 0.115 liters
Next, we calculate the molarity (M) using the formula:
Molarity (M) = moles of solute / liters of solution
Given that we have 0.025 moles of aluminum sulfate and a volume of 0.115 liters, we can substitute these values into the formula:
M = 0.025 moles / 0.115 liters
Calculating this division, we find:
M = 0.217 M
Therefore, the molarity of sulfate ions in the solution is 0.217 M.
M = mols/L soln = 0.025 mols/0.115 L soln = ? = M Al2(SO4)3
Then M SO4^2- is 3x that.