How many grams of AlCl3 (Molar mass = 133.5 g/mol) are needed to prepare 125 mL of a 0.150 M solution?

A) 20.0 g AlCl3

B) 16.7g AlCl3

C) 2.50 g AlCl3

2.50 g of AlCl3

Explanation:

Goodness, stoichiometry ...

So, what we need to find first is the amount of grams of AlCl3. To do this we look at the formula of molarity.

M = mols/L of solvent

So we know two parts of this formula. We have the Molarity (0.150) and the mL (125).

Now, we can't forget that we must convert 125 mL into liters so we have 0.125 L (I forgot and had to do the entire problem again ...)

So if we do the backwards equation we get:

0.150 = x/0.125

If we do math (fun ikr) we get 18.75 mols of the solution.

Now, we have to plug this wonderful number into stoichiometry

0.01875 mols | 133.5 g

| 1 mol AgCl3

If you are unfamiliar with what I'm doing, I'm basically going to multiply 0.01875*133.5 then divide that whole thing by 1.

So, I got 2.503125 g AlCl3

If your teacher is a stickler for significant figures there are 3 sig figs for this problem so your final answer should be

2.50 g of AlCl3

Hope you have a great day and fun with chemistry!