What volume of .225 M aluminum chloride solution contains 5.0 g solute?

mols in 5.0 g AlCl3 = grams/molar mass = ?

Then M = mols/L. You know M and mols, solve for L and convert to whatever unit you want.

Well, let's see. If we have a .225 M solution of aluminum chloride, that means we have 0.225 moles of aluminum chloride per liter of solution. To find the volume of solution containing 5.0 grams of solute, we need to convert the grams of solute to moles.

Now, the molar mass of aluminum chloride is 133.34 grams/mol. So, if we divide 5.0 grams by 133.34 grams/mol, we find that we have approximately 0.0375 moles of aluminum chloride.

Since we know that the concentration of the solution is 0.225 M, we can use the formula: concentration (M) = moles of solute / volume (L) to solve for volume.

By rearranging the formula, we find that volume (L) = moles of solute / concentration (M). Plugging in our values, we get volume (L) = 0.0375 moles / 0.225 M, which is approximately 0.167 L or 167 mL.

So, to answer your question, approximately 167 mL of the 0.225 M aluminum chloride solution would contain 5.0 grams of solute. But remember, all this math doesn't take into account the fact that I'm just a clown bot, not a lab technician. Proceed with caution!

To determine the volume of a .225 M aluminum chloride solution containing 5.0 g of solute, you need to use the formula:

Moles = Mass / Molar Mass

First, you need to determine the molar mass of aluminum chloride (AlCl₃). The atomic mass of aluminum (Al) is 26.98 g/mol, and the atomic mass of chlorine (Cl) is 35.45 g/mol. Since there are three chlorine atoms in one aluminum chloride molecule, the molar mass of aluminum chloride can be calculated as:

(1 * 26.98 g/mol) + (3 * 35.45 g/mol) = 133.33 g/mol

Next, plug in the given mass and molar mass into the formula to calculate the moles of solute:

Moles = 5.0 g / 133.33 g/mol

Moles ≈ 0.0375 mol

Finally, to find the volume of the solution, you can use the formula:

Volume = Moles / Molarity

Volume = 0.0375 mol / 0.225 M

Volume ≈ 0.167 L

Therefore, approximately 0.167 liters (or 167 milliliters) of the 0.225 M aluminum chloride solution contains 5.0 g of solute.

To find the volume of a solution, we need to use the formula:

Volume (V) = Mass of solute (m) / Concentration (C)

Here, we are given the mass of the solute as 5.0 g and the concentration of the solution as 0.225 M.

First, convert the mass of solute from grams to moles using the molar mass of aluminum chloride (AlCl3).

The molar mass of aluminum chloride is calculated by adding up the atomic masses of its components. Aluminum (Al) has a molar mass of 26.98 g/mol, and Chlorine (Cl) has a molar mass of 35.45 g/mol. Since there are three chloride ions in aluminum chloride, we multiply the molar mass of chlorine by 3:

Molar mass of AlCl3 = (1 x molar mass of Al) + (3 x molar mass of Cl)
= (1 x 26.98 g/mol) + (3 x 35.45 g/mol)
= 26.98 g/mol + 106.35 g/mol
= 133.33 g/mol

Next, convert the mass of solute to moles:

Number of moles (n) = Mass of solute / Molar mass
= 5.0 g / 133.33 g/mol
= 0.0375 mol

Finally, use the formula for volume:

V = m / C
V = 0.0375 mol / 0.225 mol/L
V = 0.1667 L

The volume of the 0.225 M aluminum chloride solution that contains 5.0 g of solute is 0.1667 liters or 166.7 milliliters.