# How many milliliters of an aqueous solution of 0.165 M aluminum chloride is needed to obtain 13.2 grams of the salt?

## Convert 13.2 g AlCl3 to mols. mols = grams/molar mass = ?

Then M = mols/L solution. You know mols and M, solve for L and convert to mL.

## 599,9 mL

## To find the volume of the aqueous solution needed, we can use the equation:

moles = mass / molar mass

First, let's calculate the moles of aluminum chloride:

moles of AlCl3 = mass of AlCl3 / molar mass of AlCl3

The molar mass of aluminum chloride (AlCl3) can be calculated by adding the atomic masses of aluminum (Al) and three times the atomic mass of chlorine (Cl). The atomic masses are:

Aluminum (Al) = 26.98 g/mol

Chlorine (Cl) = 35.45 g/mol

So, the molar mass of aluminum chloride (AlCl3) would be:

Molar mass of AlCl3 = (26.98 g/mol) + (3 * 35.45 g/mol)

Now that we know the molar mass of aluminum chloride, we can calculate the moles of AlCl3:

moles of AlCl3 = 13.2 g / (molar mass of AlCl3)

Next, we can use the molarity (0.165 M) of the aluminum chloride solution to relate moles to volume. The definition of molarity is:

Molarity (M) = moles / volume (in liters)

Rearranging the equation, we have:

Volume (in liters) = moles / molarity

Since we want the volume in milliliters, we can convert liters to milliliters by multiplying the volume by 1000:

Volume (in milliliters) = (moles / molarity) * 1000

Finally, we substitute the values we calculated into the equation to find the volume of the solution needed:

Volume (in milliliters) = (moles of AlCl3 / molarity) * 1000

Now, we can calculate the volume of the aqueous solution needed.